Front Cover of the Book “One Grain of Rice”

Synopsis

In the story, a girl named Rani is given an amount of rice each day for thirty days: one grain of rice on day one, two grains of rice on day two, four grains of rice on day three, eight grains of rice on day four, etc. In other words, on each day after the first, she receives double the amount of rice she received on the previous day. Since the doubling starts from one grain of rice, the amounts received correspond to the first 30 nonnegative powers of two.

The story progresses through select days, describing the number of grains received, the capacity and number of containers used to hold them, and the total received to each point. The amounts of each measure of rice are not spelled out for each day, so I filled in the details.

Containers

To manage the description of the number of grains of rice, and to put the growth in perspective, the author (Demi) uses these containers: bowls, bags, baskets, and storehouses:

• 1 bowl = 4,096 grains
• 1 bag = 8 bowls = 32,768 grains
• 1 basket = 32 bags = 256 bowls = 1,048,576 grains
• 1 storehouse = 128 baskets = 4,096 bags = 32,768 bowls = 134,217,728 grains

(I inferred some of these relationships from the story.)

The units of the containers themselves progress through powers of two — not only through nonnegative powers of two, but through negative powers of two as well. Nonnegative powers of two represent whole number units, and negative powers of two represent fractional units. For example, 1,024 grains of rice is 1/4 of a bowl and 2,048 grains is 1/2 a bowl.

A sequence of powers of two is a geometric sequence, also known as a geometric progression; in this case, each number in the sequence is double the previous.

Rice Received, By Day

I made a table showing the progression of powers of two by day, by measure; negative powers of two amounts are shown in gray (I generated this table, and the ones that follow, with PARI/GP scripts I wrote):

(The thirteen days mentioned explicitly in the story are shown in bold).

Here’s the same table, but using exponents to make the powers of two progressions more obvious:

Finally, here’s the same information, condensed into simple formulas:

• Number of grains on day d = 2^d-1
• Number of bowls on day d = 2^d-13
• Number of bags on day d = 2^d-16
• Number of baskets on day d = 2^d-21
• Number of storehouses on day d = 2^d-28

These relationships are simple: each measure doubles each day, so the exponent of each power of two increases by one each day. The offset in the exponent of the container formulas shows where each transitions from negative powers of two to nonnegative powers of two.

Cumulative Rice Received, By Day

I made a table showing the cumulative number of grains of rice by day, by measure. The amounts are not powers of two, although they follow a pattern, which I’ll explain below. The table is broken into two parts (to fit on this page), with fractional amounts less than one shown in gray:

The pattern is easier to see if you rewrite the values using exponents, and write mixed numbers as improper fractions:

Each value is simply the number of grains of rice accumulated, divided by the size of the corresponding measure in grains:

• Number of grains through day d: 2^d – 1
• Number of bowls through day d: (2^d – 1)/2¹²
• Number of bags through day d: (2^d – 1)/2¹⁵
• Number of baskets through day d: (2^d – 1)/2²⁰
• Number of storehouses through day d: (2^d – 1)/2²⁷

The sum of grains is a geometric series, being the sum of numbers from a geometric sequence — in this case, the first d nonnegative powers of two. Conveniently, we don’t need to add them to know their total; this well-known formula gives their sum directly:

(Numbers of the form 2ⁿ – 1 are called Mersenne numbers.)

For example, on day 15, 2¹⁵ – 1 = 32767 grains of rice have accumulated, an amount that would fill (2¹⁵ – 1)/2¹² = 32767/4096 = 7 4095/4096 bowls.

The Formulas for Representation as Mixed Numbers

The simple formulas above, expressed as improper fractions, hide the pattern that lies in the mixed number representation of the values of the containers.

Each value is a sum of consecutive powers of two. Some values are sums of negative powers of two, and some are sums of both negative and nonnegative powers of two. Values for measures that start out accumulating in fractional units (each measure except grains) are sums of negative powers of two. At some point, when whole number units start to add in, sums of nonnegative powers of two become part of the total. This is when the value becomes an improper fraction, or mixed number.

These formulas take into account the point at which the values become improper fractions, but expressing them as mixed numbers:

• Number of bowls through day d:

  If day ≤ 12: (2^d – 1)/2¹²

  If day > 12: (2^d-12 – 1) + (2¹² – 1)/2¹²

• Number of bags through day d:

  If day ≤ 15: (2^d – 1)/2¹⁵

  If day > 15: (2^d-15 – 1) + (2¹⁵ – 1)/2¹⁵

• Number of baskets through day d:

  If day ≤ 20: (2^d – 1)/2²⁰

  If day > 20: (2^d-20 – 1) + (2²⁰ – 1)/2²⁰

• Number of storehouses through day d:

  If day ≤ 27: (2^d – 1)/2²⁷

  If day > 27: (2^d-27 – 1) + (2²⁷ – 1)/2²⁷

Fractional values accumulate as a Mersenne number over a power of two; once whole number values begin to add in, the accumulated fractional part remains unchanged while the whole number part accumulates as a Mersenne number.

It’s easy to show that the mixed number formulas are equivalent to the improper fraction formulas. For example, take the mixed number formula for bowls after day 12:

You can rewrite this, using simple algebra and the laws of exponents, as follows:

This is the the same formula as for before day 12, and thus the same formula for all days.

In the Classroom

This is a good story upon which to base a math lesson, and you can vary the mathematical sophistication according to grade level. I read it to first and third graders, so I had to keep it simple. My goal was to show how numbers grow through doubling, and to introduce place value of large numbers.

First Graders

When I read the story to a first grade class, I got lots of “oohs” and “aahs” whenever I read a big number from the text and whenever they saw an illustration of the many animals needed to transport the rice. At the end of the story, I wrote 1,073,741,823 on the board to show them how it looks in numerals. Although I knew they’d only been taught up to the hundreds place, I went through each place and named it. I finished by saying “when you get to third grade I can explain the math a little more.”

Third Graders

For the third grade class, I read the story and spent about fifteen minutes afterwards discussing the numbers. First I wrote 1,073,741,823 on the board and went through the places. Even though they had been taught only up to the thousands place, most of them knew to the billions place (and beyond). Next, I asked them to say the number in words, and many had difficulty — particularly the “seven-hundred forty one thousand” part (I don’t know why it was harder than “seventy-three million” — I would have thought the zero in the hundred millions place would be confusing).

Next, I wrote the number 1 and had them tell me the next numbers in the sequence. A couple of kids answered correctly up to 8,192 (2¹⁴). When I asked one boy if he had them memorized, he said “No, I can add them in my head.”

I continued the sequence using a pocket calculator. I had the students take turns computing the next number with it (they found it very hard to recite the numbers back to me, because there were no commas). By design, one unlucky kid had 2²⁷ (134,217,728) for his turn, which caused the calculator to overflow. I told the class that the numbers get so big that you need a computer to figure them out.

At the end, I referred to these numbers as “powers of two,” and told them they would be useful someday if they wanted to work with computers (can you blame me for plugging my profession?).

One boy asked a good question: “what if you got 20 grains every minute for 30 days — would that be more?” I said “Great question: I think you’d still get more by doubling though.” The calculator showed him this to be true — his method accumulated only 864,000 grains.

Lesson Plan Resources

If you search the Internet for “One Grain of Rice” you’ll find other ideas on how to incorporate this story into a math lesson. For example, this worksheet uses graphs and logarithms.

Errors in the Story

I found two errors and one inconsistency in the story:

Before the Rice Is Delivered

Rani describes her plan for receiving the rice as follows:

“Today, you will give me a single grain of rice. Then, each day for thirty days you will give me double the rice you gave me the day before. Thus, tomorrow you will give me two grains of rice, the next day four grains of rice, and so on for thirty days.”

The doubling occurs for twenty-nine days, not thirty days.

On Day 16

On day 16 it says

“On the sixteenth day, Rani was presented with a bag containing thirty-two thousand, seven hundred and sixty-eight grains of rice. All together she had enough rice for two full bags”

Through day 16 she had 65,535 grains of rice, one grain less than two full bags; she actually had 1 32767/32768 = 1.999969482421875 bags.

On Day 9

On day 9 it says

“On the ninth day, Rani was presented with two hundred and fifty-six grains of rice. She had received in all five hundred and eleven grains of rice, only enough for a small handful.”

Technically, there’s nothing wrong with this, but the measure “small handful” — 511 grains — is not a power of two. It doesn’t fit neatly into the scheme of other units (and hence is why I didn’t include it in the tables).

By Rick Regan (Copyright © 2008-2012 Exploring Binary)

1,073,741,823 Grains of Rice

Binary Tree Showing Nested Ending Digit Patterns of the Positive Powers of Five

The tree layout shows that certain pairs of ending digits are related, and that these pairs differ by five in their starting digits. I will show why this is true.

Ending Digit Cycles

The ending digit cycles of the positive powers of five can be determined by incremental calculation modulo powers of ten. The resulting calculations can be tabulated as follows:

• 1-digit endings
  • 5¹: 5
• 2-digit endings
  • 5²: 25
• 3-digit endings
  • 5³: 125
  • 5⁴: 625
• 4-digit endings
  • 5⁴: 0625
  • 5⁵: 3125
  • 5⁶: 5625
  • 5⁷: 8125
• 5-digit endings
  • 5⁵: 03125
  • 5⁶: 15625
  • 5⁷: 78125
  • 5⁸: 90625
  • 5⁹: 53125
  • 5¹⁰: 65625
  • 5¹¹: 28125
  • 5¹²: 40625
• …

Each group of m-digit endings repeats in a cycle. Each cycle is a power of two in length, and for m ≥ 2, equals 2^m-2.

Ending digits are labeled with specific powers of five: those that make up the first occurrence of each cycle. Each cycle — from the 3-digit endings on — contains the ending digits of 5^m through 5^(m+2m-2-1).

Ending Digit Cycles In Tree Form

The pairing of ending digits in the binary tree shows a relationship between endings in successive cycles; in particular, for level 3 and higher:

• Each pair of m digit endings has a common m-1 digit suffix.
• The starting digits of each pair of m digit endings differ by five.

Why Pairs of Siblings Have the Same Suffix

Why is there always a pair of siblings, and why do siblings match up as they do? The answer lies in the nesting of power of two sized cycles of ending digits. For example, the last four digits repeat in a cycle of four, and the last five digits repeat in a cycle of eight. To form the eight unique five digit endings, the set of four digit endings must be repeated; duplicate four digit endings will have different starting digits prefixed.

Endings with the same suffix are a half cycle length apart, and are paired as siblings. For example, in the length eight cycle of five ending digits, the following endings are paired: 1st and 5th; 2nd and 6th; 3rd and 7th; 4th and 8th.

In terms of powers of five, sibling pairs differ in their exponents by half a cycle length. You can see this by looking at the underlying powers of five in the tree:

Binary Tree Showing Powers of Five Corresponding to Ending Digits

Why the Starting Digits of Siblings Differ by Five

To show why the starting digits of siblings differ by five, let’s look at an example. In level 5 of the tree — the 5-digit endings — let’s compute the difference between each pair of powers of five (I’ll go in cycle order, not in tree order):

• 5⁹ – 5⁵ = 1950000
• 5¹⁰ – 5⁶ = 9750000
• 5¹¹ – 5⁷ = 48750000
• 5¹² – 5⁸ = 243750000

You can see that each differs by five in their fifth digit — the algebra behind the subtraction shows why:

• 5⁹ – 5⁵ = 5⁵(5⁴ – 1) = 5⁰·5⁵(5⁴ – 1)
• 5¹⁰ – 5⁶ = 5⁶(5⁴ – 1) = 5¹·5⁵(5⁴ – 1)
• 5¹¹ – 5⁷ = 5⁷(5⁴ – 1) = 5²·5⁵(5⁴ – 1)
• 5¹² – 5⁸ = 5⁸(5⁴ – 1) = 5³·5⁵(5⁴ – 1)

Let’s analyze these expressions by evaluating them in three stages:

1. 5⁴ – 1

   This expression has a factor of 2⁴. I’ve shown that 5²ⁿ – 1 has exactly n+2 factors of 2; that is, a factor of 2ⁿ⁺².

2. 5⁵(5⁴ – 1)

   This expression has a factor of 5⁵·2⁴ = 5·10⁴ = 50000.

3. 5ⁱ·5⁵(5⁴ – 1), 0 ≤ i ≤ 3

   This expression ends in 50000. For i > 0, 5ⁱ ends in the digit 5; for i = 0, 5ⁱ = 1. In either case, multiplying by 5ⁱ leaves the 5 in the fifth digit unchanged.

Generalizing

The example generalizes to show that every pair of endings in every level m of the tree, m ≥ 3, differ by 5 in their starting digits:

• Cycle length is 2^m-2
• Level m includes the powers of five 5^m through 5^(m+2m-2-1)
• Siblings are the powers of five p and p·5^2m-3
• The difference between each sibling pair is 5ⁱ·5^m·(5^2m-3 – 1), for each i between 0 and 2^m-3 – 1
• The factor 5^2m-3 – 1 itself has a factor of 2^m-1
• The factor 5^m·(5^2m-3 – 1) itself has a factor of 5^m·2^m-1 = 5·10^m-1
• 5·10^m-1 is an m-digit number consisting of the digit 5 followed by m-1 zeros
• The factor 5ⁱ leaves the last m digits unchanged: the digit 5 followed by m-1 zeros

By Rick Regan (Copyright © 2008-2012 Exploring Binary)

Ending Digits of Powers of Five Form a Binary Tree

• Part 1 shows that the powers of five mod 2^m cycle with period 2^m-2, m ≥ 2, starting at 5⁰.
• Part 2 shows that the powers of five mod 10^m cycle with the same period as the powers of five mod 2^m, starting at 5^m.

The highlight of my proof is in part 1, where I derive a formula to show that the period, or order, of 5 mod 2^m is 2^m-2. While it is in general not possible to derive a formula for the order of a number, I’ll show it is possible for the powers of five mod 2^m — due to a hidden, binary structure I’ve uncovered.

To understand the proof, you’ll need to know some algebra and the basics of modular arithmetic.

Part 1: Period mod 2^m is 2^m-2

Part 1 shows that the order of 5 mod 2^m is 2^m-2. I break it into three steps, showing that

• The order of 5 mod 2^m is a power of two. This constrains the values for the order so that only expressions of the form 5²ⁿ – 1 need be considered.
• 5²ⁿ – 1 can be factored to isolate its factors of two. This is my key insight, made possible by recursive factoring of differences of squares.
• 5²ⁿ – 1 is divisible by 2ⁿ⁺². This shows that the order of 5 mod 2ⁿ⁺² is 2ⁿ, or equivalently, that the order of 5 mod 2^m is 2^m-2.

In addition, I show that the powers of five mod 2^m cycle starting at 5⁰; this is important because it will show that the cycle mod 10^m always starts after the cycle mod 2^m.

The Order of 5 mod 2^m is a Power of Two

The order of a mod b is related to , so let’s start there. is Euler’s phi function, the number of positive integers less than or equal to b that are relatively prime to b (for b a power of two, this is just the count of the odd numbers less than b). We want to compute , which we can do with this simple formula: . For p=2, this gives = 2^m-1.

The order of a mod b must divide , so the order of the powers of five mod 2^m must divide , or 2^m-1. In other words, the order must be a power of two — a power of two less than or equal to 2^m-1 to be specific — because the divisors of a positive power of two are all the positive powers of two less than or equal to it.

Let 2^e represent any of the candidates for the order of 5 mod 2^m: 1, 2, 4, 8, … , 2^m-2, 2^m-1. By definition, if the order is 2^e, , or equivalently, . That is, 5^2e – 1 is divisible by 2^m. Now all we have to do is figure out which e makes this true.

5²ⁿ – 1 Can Be Factored to Isolate its Factors of Two

You can factor 5²ⁿ – 1 into (5^2n/2 + 1)(5^2n/2 – 1), since it is a difference of two squares. You can factor 5^2n/2 – 1 similarly, into (5^2n/4 + 1)(5^2n/4 – 1). You can do this n times, since log₂(2ⁿ) = n. Here’s an example:

1. 5¹⁶ – 1 = (5⁸ + 1)(5⁸ – 1)
2. 5⁸ – 1 = (5⁴ + 1)(5⁴ – 1)
3. 5⁴ – 1 = (5² + 1)(5² – 1)
4. 5² – 1 = (5¹ + 1)(5¹ – 1)
5. 5¹ – 1 = 4

Recursive Factoring of Differences of Squares

Combining these recursively you get 5¹⁶ – 1 = 4(5¹ + 1)(5² + 1)(5⁴ + 1)(5⁸ + 1).

This generalizes to 5²ⁿ – 1 = 4(5¹ + 1)(5² + 1)(5⁴ + 1)···(5^2n/2 + 1).

There is a factor of two in each 5^2f + 1, as I’ll show in the next section.

5²ⁿ – 1 is divisible by 2ⁿ⁺²

The factor of four in the factored expression shows there are at least two factors of two in 5²ⁿ – 1. There is also exactly one factor of two in each of the factors 5^2f + 1. I’ll show it in two steps:

• 5^2f + 1 is divisible by 2.

  Every positive power of five ends in 5, so it is odd. This means 5^2f + 1 is even, meaning it’s divisible by 2.

• 5^2f + 1 is not divisible by any power of two greater than 2.

  To show this, it suffices to show that 5^2f + 1 is not divisible by 4. If it’s not divisible by 4, it’s not divisible by any higher power of two (because the divisors of a positive power of two are all positive powers of two less than or equal to it). So if 2² is not a factor, 2^s, s > 1, is not a factor.

  Let’s show 5^2f + 1 is not divisible by 4 by breaking it in two cases:

  • 5²⁰ + 1. 5¹ + 1 = 6, which obviously is not divisible by 4.
  • 5^2f + 1, f ≥ 1. 5^2f ends in 25, so 5^2f + 1 ends in 26. An integer is divisible by 4 if and only if its last two digits are divisible by 4. Since 26 is not divisible by 4, neither is 5^2f + 1.

So the example expression 4(5¹ + 1)(5² + 1)(5⁴ + 1)(5⁸ + 1) has a factor of 2⁶: 2² times four factors of 2.

This generalizes to show that 5²ⁿ – 1 is divisible by 2ⁿ⁺². Recasting this from the perspective of the power of five to the modulus, let’s substitute m-2 for n. This gives the equivalent statement 5^2m-2 – 1 is divisible by 2^m. Just to be explicit, this implies that no 5^2e, e < 2^m-2, is divisible by 2^m. So, the candidate power of two we were seeking is 2^m-2, the order of 5 mod 2^m.

The Powers of Five Mod 2^m Cycle Starting at 5⁰

Let the period p = 2^m-2. As we proved above, . We also know, trivially, that . The difference between the exponents p and 0 is p, showing a full cycle occurs starting at 5⁰.

Part 2: Period mod 10^m is 2^m-2

Part 2 shows, using the definition of modular arithmetic, the laws of exponents, and simple algebra, that the powers of five mod 10^m have the same period as the powers of five mod 2^m. It’s broken into two halves, showing that

• The period mod 2^m is a cycle mod 10^m.
• The cycle mod 10^m is the period mod 2^m.

The first half only shows that the period mod 2^m is a multiple of the period mod 10^m. The second half shows that the period mod 10^m is in fact the same as the period mod 2^m.

I’ll use the variable p for the period mod 2^m, as above, to simplify the notation in the proof that follows.

The period mod 2^m is a cycle mod 10^m

For i ≥ 0, what we’ve shown is that . That is, powers of five with exponents a period apart are congruent mod 2^m.

Congruence mod 2^m means that 5^i+p – 5ⁱ is a multiple of 2^m; that is, 5^i+p – 5ⁱ = n·2^m, for some nonnegative integer n.

Factoring out 5ⁱ we get 5ⁱ(5^p – 1) = n·2^m. When i≥m, n is a multiple of 5^m, so n = n’·5^m.

Substituting n’·5^m for n we get 5ⁱ(5^p – 1) = n’·5^m2^m = n’·10^m.

Reversing the process above, we can show congruence mod 10^m:

5ⁱ(5^p – 1) = 5^i+p – 5ⁱ = n’·10^m.

This shows that for i≥m, , so p is a cycle mod 10^m.

The cycle mod 10^m is the period mod 2^m

We’ve shown that p is a cycle mod 10^m, but not that p is the period mod 10^m. In other words, it’s possible the period mod 10^m is less than p — we have to show it isn’t.

For i ≥ m, we’ve shown that . Using the argument above, congruence mod 10^m means that:

5^i+p – 5ⁱ = n·10^m = n·5^m2^m = (n·5^m)2^m.

This shows that for i≥m, , so p is the period mod 2^m and mod 10^m.

Empirical Evidence

This section is not part of the proof, but it shows some PARI/GP calculations that support it.

Compute the order of 5 mod 2 through 5 mod 2¹⁶ (you can see they’re powers of two):

? for(m=1,16,print(znorder(Mod(5,2^m))))
1
1
2
4
8
16
32
64
128
256
512
1024
2048
4096
8192
16384

Show the factors in 5¹⁶ – 1:

? factor(5^16-1)
%1 =
[2 6]
[3 1]
[13 1]
[17 1]
[313 1]
[11489 1]

? factor(4*(5^1+1)*(5^2+1)*(5^4+1)*(5^8+1))
%2 =
[2 6]
[3 1]
[13 1]
[17 1]
[313 1]
[11489 1]

? factor((5^1+1))
%3 =
[2 1]
[3 1]

? factor((5^2+1))
%4 =
[2 1]
[13 1]

? factor((5^4+1))
%5 =
[2 1]
[313 1]

? factor((5^8+1))
%6 =
[2 1]
[17 1]
[11489 1]

Show the factors of two in 5ⁿ – 1, n = 1 to 32 (you can see how the nested cycles correspond to the powers of two):

? for(i=1,32,print("5^",i,"-1 has factor 2^",factor(5^i-1)[1,2]))

5^1-1 has factor 2^2
5^2-1 has factor 2^3
5^3-1 has factor 2^2
5^4-1 has factor 2^4
5^5-1 has factor 2^2
5^6-1 has factor 2^3
5^7-1 has factor 2^2
5^8-1 has factor 2^5
5^9-1 has factor 2^2
5^10-1 has factor 2^3
5^11-1 has factor 2^2
5^12-1 has factor 2^4
5^13-1 has factor 2^2
5^14-1 has factor 2^3
5^15-1 has factor 2^2
5^16-1 has factor 2^6
5^17-1 has factor 2^2
5^18-1 has factor 2^3
5^19-1 has factor 2^2
5^20-1 has factor 2^4
5^21-1 has factor 2^2
5^22-1 has factor 2^3
5^23-1 has factor 2^2
5^24-1 has factor 2^5
5^25-1 has factor 2^2
5^26-1 has factor 2^3
5^27-1 has factor 2^2
5^28-1 has factor 2^4
5^29-1 has factor 2^2
5^30-1 has factor 2^3
5^31-1 has factor 2^2
5^32-1 has factor 2^7

(The pattern of the powers of two is reminiscent of the representation of powers of two in a binary counter.)

Discussion

Part 2 of the proof is identical to part 2 of the proof in my article “Cycle Length of Powers of Two Mod Powers of Ten”, except that 5 and 2 are interchanged. Part 1 is quite different though. I could not use the same approach, since powers of two beyond four don’t have primitive roots.

Coming up with an alternate approach took some time. I had the elements of it early on: knowing that the order must be a power of two, knowing how to factor differences of squares, and knowing that each factor was even. But I was missing the piece that let me show each factor had exactly one factor of two — until I realized I could apply the divisibility by four test.

The binary structure of powers of a number computed mod powers of two make this approach possible (it should work for powers of any odd number, although it may be more difficult to show the factors of two).

Summary

To prove the period formula for the positive powers of five mod powers of ten, I first proved it for mod powers of two. I showed that the period must be a power of two, and derived an expression that showed which power of two. I then showed that the period mod powers of two and mod powers of ten must be the same, after a certain starting point.

I’ve previously shown a connection between the positive powers of five and negative powers of two; now I’ve shown a connection between the positive powers of five and the positive powers of two.

By Rick Regan (Copyright © 2008-2012 Exploring Binary)

Cycle Length of Powers of Five Mod Powers of Ten

Cycles in the Ending Digits of the Powers of Five

I will show you why these cycles exist, how they are expressed mathematically, and how to visualize them.

(This article is the companion article to “Patterns in the Last Digits of the Positive Powers of Two”. It describes the digit patterns in the negative powers of two, indirectly, since positive powers of five look like negative powers of two.)

Cycles in the Last One to Four Digits

You can use modular arithmetic to show the repeating digit cycles of the positive powers of five. You can find the last m digits of a positive power by computing its common residue mod 10^m. You can find the last m digits of a sequence of powers by computing them incrementally, mod 10^m. The cycle restarts when you get a result you’ve already seen.

Last Digit

Starting with 5, the last digit repeats in a cycle of period one: 5. You show it with incremental calculations mod 10:

1. 
2. 
3. …

In other words, every positive power of five ends in 5.

Last Two Digits

Starting with 5², the last two digits repeat in a cycle of period one: 25. You show it with incremental calculations mod 10²:

1. 
2. 
3. 
4. …

Last Three Digits

Starting with 5³, the last three digits repeat in a cycle of period two: 125, 625. You show it with incremental calculations mod 10³:

1. 
2. 
3. 
4. 
5. 
6. …

Last Four Digits

Starting with 5⁴, the last four digits repeat in a cycle of period four: 0625, 3125, 5625, 8125. You show it with incremental calculations mod 10⁴:

1. 
2. 
3. 
4. 
5. 
6. 
7. 
8. 
9. …

Whenever the residue is less than m digits, it is implicitly padded out with leading zeros. For example, the last 4 digits of 5⁸ = 390625 are 0625.

Cycles in the Last m Digits

If you continue this process, you’ll see that the period doubles for each additional ending digit; I did the calculations through ten ending digits:

This table implies that the ending m digits of the positive powers of five cycle with period 2^m-2, m ≥ 2, starting at 5^m. (The proof is here.)

Powers of five with exponents that are congruent mod 2^m-2 are themselves congruent mod 10^m. That is, 5ⁱ and 5^i+(2m-2)·k, i ≥ m, k ≥ 0, end with the same m digits.

Visualizing the Nesting of Cycles

The cycles in m digit, m-1 digit, m-2 digit, …, 1-digit endings can be viewed as nested, even though their starting points are staggered. You just have shift the starting points of the lesser digit cycles to make them coincide. For example, each copy of the length 4 cycle of four digit endings has within it two copies of the length 2 cycle of three digit endings; each copy of the length 2 cycle of three digit endings has within it 2 copies of the length 1 cycle of two digit endings.

This diagram shows the nesting by shading every other occurrence of a cycle (the millions place is fully shaded, because only one cycle of the last seven digits is shown):

Nested 1-7 Digit Ending Patterns from 5⁷ to 5³⁸

Not surprisingly, the nested powers of two in this pattern make it look similar to the pattern in consecutive binary integers.

Binary Tree

Another way to show the nesting of powers of two is with a binary tree. Here is a tree showing the ending 1-5 digits:

Binary Tree Showing Nested 1-5 Digit Ending Patterns (Digits)

Each level contains the ending digits for a given cycle, from the last digit (the root, or first level) to the eight 5 digit endings (the leaves, or fifth level). Starting with the second level, each m digit ending is the suffix of two m+1 digit endings.

Here’s the same tree, except labeled with the smallest power of five corresponding to the ending digits:

Binary Tree Showing Nested 1-5 Digit Ending Patterns (Powers)

Each level from level 2 down includes the endings for 5^m through 5^(m+2m-2-1).

Notice two interesting things that become apparent with the binary tree representations:

• The starting digits of each pair of children, for m ≥ 3, differ by 5. For example, 3125 and 8125 on level 4 (digits 3 and 8).
• The exponents of each pair of children, for m ≥ 3, differ by 2^m-3. For example, 5⁵ and 5⁷ on level 4.

(For more details, see my article “Ending Digits of Powers of Five Form a Binary Tree”.)

Exploring Ending Digits with PARI/GP

You can use PARI/GP to explore the cycles in the ending digits; here are two examples:

• Print the first 20 positive powers of five:

  ? for(i=1,20,print("5^",i,": ",5^i))
  5^1: 5
  5^2: 25
  5^3: 125
  5^4: 625
  5^5: 3125
  5^6: 15625
  5^7: 78125
  5^8: 390625
  5^9: 1953125
  5^10: 9765625
  5^11: 48828125
  5^12: 244140625
  5^13: 1220703125
  5^14: 6103515625
  5^15: 30517578125
  5^16: 152587890625
  5^17: 762939453125
  5^18: 3814697265625
  5^19: 19073486328125
  5^20: 95367431640625
  

• Show the cycle in the last m (in this case 5) digits:

  ? m=5; for(i=1,2^(m-2)+m,print("5^",i," mod 10^",m": ",5^i%10^m))
  5^1 mod 10^5: 5
  5^2 mod 10^5: 25
  5^3 mod 10^5: 125
  5^4 mod 10^5: 625
  5^5 mod 10^5: 3125
  5^6 mod 10^5: 15625
  5^7 mod 10^5: 78125
  5^8 mod 10^5: 90625
  5^9 mod 10^5: 53125
  5^10 mod 10^5: 65625
  5^11 mod 10^5: 28125
  5^12 mod 10^5: 40625
  5^13 mod 10^5: 3125
  

  (The leading 0 of 3125 is not printed.)

By Rick Regan (Copyright © 2008-2012 Exploring Binary)

Patterns in the Last Digits of the Positive Powers of Five

Powers of Two and Powers of Five that Look Alike

This relationship is not coincidence; it’s a by-product of how fractions are represented as decimals. I’ll show you simple algebra that proves it, as well as algebra that proves similar properties — in products involving negative powers.

Positive Powers of Five in Negative Powers of Two

A negative power of two in decimal form looks like a positive power of five, except that it has a decimal point and possibly some leading zeros. Here are some examples, the first eight pairs of positive powers of five and negative powers of two:

The powers that are paired have exponents that are additive inverses of each other. It’s easy to show this relationship holds for all such pairs of powers. A negative power of two is a number of the form 2ⁿ, n and integer less than 0, or equivalently, a number of the form 2^-n, n > 0. By the laws of exponents, 2^-n = 1/2ⁿ = (1/2)ⁿ = (5/10)ⁿ = 5ⁿ/10ⁿ. Written as a decimal, 5ⁿ/10ⁿ looks like a positive power of five, except that it’s a fraction with n decimal places.

A different transformation gives another way to look at this: 2^-n = 1/2ⁿ = (1/2)ⁿ = (0.5)ⁿ. If you were to multiply this out by hand, you would ignore the decimal point, multiply all the factors of five, and then place the decimal point when you were done. You’d end up with a positive power of five, but preceded with a decimal point and zero or more leading 0s — to pad it out to n decimal places.

Positive Powers of Two in Negative Powers of Five

Not surprisingly, a similar relationship exists in reverse; that is, between the positive powers of two and the negative powers of five. This makes sense, given the role of the numbers two and five in the decimal system.

A negative power of five in decimal form looks like a positive power of two, except that it has a decimal point and possibly some leading zeros. Here are some examples, the first eight pairs of positive powers of two and negative powers of five:

Again, the powers that are paired have exponents that are additive inverses of each other. Using a similar transformation as above, we can show this relationship holds for all such pairs of powers: 5^-n = 1/5ⁿ = (1/5)ⁿ = (2/10)ⁿ = 2ⁿ/10ⁿ. Written as a decimal, 2ⁿ/10ⁿ looks like a positive power of two, except that it’s a fraction with n decimal places.

There’s also an alternate way to view this: 5^-n = 1/5ⁿ = (1/5)ⁿ = (0.2)ⁿ. As above, think of it as multiplying the factors to get a power of two and then prefixing a decimal point and leading zeros.

Products of Negative Powers of Two and Positive Powers of Five

Because a negative power of two looks like a positive power of five, multiplying it by a positive power of five makes it look like yet another positive power of five. The resulting power of five has an exponent that is the sum of the absolute values of the exponents of the multiplied powers. For example: 2^-3·5⁴ = 78.125, which looks like 5⁷ = 78125; 2^-7·5³ = 0.9765625, which looks like 5¹⁰ = 9765625.

Algebraically, the multiplication is expressed as 2^-t·5^f, with t, f > 0. The result looks like 5^(t+f); here’s why: 2^-t = 5^t/10^t, so 2^-t·5^f = 5^(t+f)/10^t. This expression describes a positive power of five, only shifted right t decimal places.

You can think of these as looking like negative powers of two as well. The product 2^-t·5^f can be written equivalently as 2^-t·5^f·(2^-f·2^f) = 2^-(t+f)·10^f, which is 2^-(t+f) shifted left f decimal places.

Products of Negative Powers of Five and Positive Powers of Two

Similarly, the product of a negative power of five and a positive power of two looks like a positive power of two. For example, 5^-3·2³ = 0.064, which looks like 2⁶ = 64, and 5^-4·2¹² = 6.5536, which looks like 2¹⁶ = 65536.

Algebraically, the multiplication is expressed as 5^-f·2^t, with f, t > 0. The result looks like 2^(f+t); here’s why: 5^-f = 2^f/10^f, so 5^-f·2^t = 2^(f+t)/10^f. This expression describes a positive power of two, only shifted right f decimal places.

You can think of these as looking like negative powers of five as well. The product 5^-f·2^t can be written equivalently as 5^-f·2^t·(5^-t·5^t) = 5^-(f+t)·10^t, which is 5^-(f+t) shifted left t decimal places.

Products of Negative Powers of Two and Negative Powers of Five

A negative power of two times a negative power of five results in a string of digits that looks like either a positive power of five or a positive power of two, depending on which exponent is smaller. For example, 2^-4·5^-2 = 0.0025, which looks like a positive power of five, and 2^-2·5^-4 = 0.0004, which looks like a positive power of two.

Algebraically, the multiplication is expressed as 2^-t·5^-f, with t, f > 0, and f ≠ t. There are two cases:

• f < t: the result looks like 5^(t-f), with t decimal places. Here’s why:

  Recall from above that 2^-t = 5^t/10^t and that 5^-f = 2^f/10^f. This gives 2^-t·5^-f = (5^t/10^t)(2^f/10^f) = (5^t·2^f)/10^(t+f). Now factor out 10^f from the numerator, which we can do since f < t: (5^t·2^f)/10^(t+f) = (10^f·5^(t-f))/10^(t+f) = 5^(t-f)/10^t.

• t < f: the result looks like 2^(f-t), with f decimal places. Here’s why:

  Starting from (5^t·2^f)/10^(t+f), factor out 10^t from the numerator, which we can do since t < f: (5^t·2^f)/10^(t+f) = (10^t·2^(f-t))/10^(t+f) = 2^(f-t)/10^f.

(If you allow f = t, the answer would be 2⁰ = 5⁰ = 1, a nonnegative power, with f = t decimal places.)

Discussion

I came up with the algebra for the multiplication cases after I played around with some numbers in PARI/GP. I noticed a pattern, and then sought to state it mathematically.

The patterns hold for quotients of powers as well, since divisions can be converted to multiplications by inverting exponents.

Some Practical Uses

If you know the positive powers of five, or if you learn them, you will recognize negative powers of two more readily — similarly for the positive powers of two and negative powers of five.

If you wanted to print a table of negative powers of two or negative powers of five using a computer program, you could do so easily, just by using integers: convert the positive powers to strings, and then prefix each with a decimal point and the appropriate number of leading zeros.

References

• learner.org article “Fractions to Decimals”.

By Rick Regan (Copyright © 2008-2012 Exploring Binary)

Seeing Powers of Five in Powers of Two and Vice Versa

• Part 1 shows that the powers of two mod 5^m cycle with period 4·5^m-1, starting at 2⁰.
• Part 2 shows that the powers of two mod 10^m cycle with the same period as the powers of two mod 5^m, starting at 2^m.

To understand the proof, you’ll need to know some elementary number theory.

Part 1: Period mod 5^m is 4·5^m-1

The main goal of part 1 is to show that two is a primitive root mod powers of five, meaning that the order of 2 mod 5^m is the value of Euler’s phi function for 5^m. I break this into three steps, showing that

• = 4·5^m-1, where is Euler’s phi function (also known as Euler’s totient function).
• 2 is a primitive root mod 5.
• 2 is a primitive root mod 5^m.

In addition, I show that the powers of two mod 5^m cycle starting at 2⁰; this is important because it will show that the cycle mod 10^m always starts after the cycle mod 5^m.

= 4·5^m-1

is the number of positive integers less than or equal to n that are relatively prime to n. For a power of a prime number p, we can use this formula to compute it: . For p=5, this gives = 4·5^m-1.

2 is a Primitive Root Mod 5

To prove this, we just have to show that the order (also known as multiplicative order) of 2 mod 5 is equal to .

= 4, by the formula above. The order of 2 mod 5 — the smallest exponent a such that — is four, as shown by the following sequence of four calculations:

Thus, 2 is a primitive root mod 5.

2 is a Primitive Root Mod 5^m

To show that 2 is a primitive root mod all positive powers of 5, we can use this result from page 26 of “A Course in Computational Algebraic Number Theory” by Henri Cohen:

“…to find a primitive root modulo p^a for p an odd prime and a ≥ 2, proceed as follows: first compute g a primitive root modulo p …, then compute g₁ = g^p-1 mod p². If g₁ ≠ 1, g is a primitive root modulo p^a for every a…” .

In our case g = 2 and p = 5. We’ve already shown 2 to be a primitive root mod 5, so now all we need to show is that , which is trivial: .

So, 2 is a primitive root mod 5^m, which means that the order of 2 mod 5^m is . This proves that the powers of two mod 5^m cycle with period 4·5^m-1.

The Powers of Two Mod 5^m Cycle Starting at 2⁰

Let the period p = 4·5^m-1. , since 2 is a primitive root mod powers of 5 (Euler’s theorem applies). We also know, trivially, that . The difference between the exponents p and 0 is p, showing a full cycle occurs starting at 2⁰.

Part 2: Period mod 10^m is 4·5^m-1

Part 2 shows, using the definition of modular arithmetic, the laws of exponents, and simple algebra, that the powers of two mod 10^m have the same period as the powers of two mod 5^m. It’s broken into two halves, showing that

• The period mod 5^m is a cycle mod 10^m.
• The cycle mod 10^m is the period mod 5^m.

The first half only shows that the period mod 5^m is a multiple of the period mod 10^m. The second half shows that the period mod 10^m is in fact the same as the period mod 5^m.

I’ll use the variable p for the period mod 5^m, as above, to simplify the notation in the proof that follows.

The period mod 5^m is a cycle mod 10^m

For i ≥ 0, what we’ve shown is that . That is, powers of two with exponents a period apart are congruent mod 5^m.

Congruence mod 5^m means that 2^i+p – 2ⁱ is a multiple of 5^m; that is, 2^i+p – 2ⁱ = n·5^m, for some nonnegative integer n.

Factoring out 2ⁱ we get 2ⁱ(2^p – 1) = n·5^m. When i≥m, n is a multiple of 2^m, so n = n’·2^m.

Substituting n’·2^m for n we get 2ⁱ(2^p – 1) = n’·2^m5^m = n’·10^m.

Reversing the process above, we can show congruence mod 10^m:

2ⁱ(2^p – 1) = 2^i+p – 2ⁱ = n’·10^m.

This shows that for i≥m, , so p is a cycle mod 10^m.

The cycle mod 10^m is the period mod 5^m

We’ve shown that p is a cycle mod 10^m, but not that p is the period mod 10^m. In other words, it’s possible the period mod 10^m is less than p — we have to show it isn’t.

For i ≥ m, we’ve shown that . Using the argument above, congruence mod 10^m means that:

2^i+p – 2ⁱ = n·10^m = n·2^m5^m = (n·2^m)5^m.

This shows that for i≥m, , so p is the period mod 5^m and mod 10^m.

Empirical Evidence that 2 is a Primitive Root mod 5^m

This section is not part of the proof, but it shows some PARI/GP calculations that support it.

These two commands show indirectly that 2 is a primitive root mod 5 through mod 5²⁰, by showing that the order equals Euler’s phi function:

? for(m=1,20,print(eulerphi(5^m)))
4
20
100
500
2500
12500
62500
312500
1562500
7812500
39062500
195312500
976562500
4882812500
24414062500
122070312500
610351562500
3051757812500
15258789062500
76293945312500

? for(m=1,20,print(znorder(Mod(2,5^m))))
4
20
100
500
2500
12500
62500
312500
1562500
7812500
39062500
195312500
976562500
4882812500
24414062500
122070312500
610351562500
3051757812500
15258789062500
76293945312500

This command shows directly that 2 is a primitive root mod powers of 5, using PARI/GP’s znprimroot() function (znprimroot only returns the lowest primitive root, which luckily for us is 2):

? for(m=1,20,print(znprimroot(5^m)))
Mod(2, 5)
Mod(2, 25)
Mod(2, 125)
Mod(2, 625)
Mod(2, 3125)
Mod(2, 15625)
Mod(2, 78125)
Mod(2, 390625)
Mod(2, 1953125)
Mod(2, 9765625)
Mod(2, 48828125)
Mod(2, 244140625)
Mod(2, 1220703125)
Mod(2, 6103515625)
Mod(2, 30517578125)
Mod(2, 152587890625)
Mod(2, 762939453125)
Mod(2, 3814697265625)
Mod(2, 19073486328125)
Mod(2, 95367431640625)

Summary

To prove the period formula for the positive powers of two mod powers of ten, I first proved it for mod powers of five. I showed that two is a primitive root mod powers of five, meaning that the period formula is Euler’s phi function for the powers of five. I then showed that the period mod powers of five and mod powers of ten must be the same, after a certain starting point.

References

These are the main influences behind my proof:

• Challenging Mathematical Problems with Elementary Solutions, Volume 1. Page 198: “It can be proved that the last k digits of the number 2ⁿ are repeated in groups of 4·5^k-1, starting with the number 2^k.”
• The USSR Olympiad Problem Book; Selected Problems and Theorems of Elementary Mathematics. Page 57, problem 243, and its answer, on page 354, discuss the case for powers of two mod 10¹⁰.
• Question I asked at physicsforums.com. User hamster143‘s proof that the period for powers of five and ten are equal.
• Discussion at physicsforums.com. User shmoe‘s comment: “a power of 2′s congruence class mod 10 is completely determined by its congruence class mod 5”.
• Hakmem Item 57. Two comments by Schroeppel:
  • “After the nth, they are all multiples of 2ⁿ.”
  • “They get into a loop of length 4·5^m-1. (Because 2 is a primitive root of powers of 5.)”
• Question I asked at mathoverflow.net.

By Rick Regan (Copyright © 2008-2012 Exploring Binary)

Cycle Length of Powers of Two Mod Powers of Ten

Modular arithmetic, and in particular, modular exponentiation, comes to the rescue. It provides an efficient way to find the last m digits of a power, by hand, with perhaps only a little help from a pocket calculator. All you need to do is compute the power incrementally, modulo 10^m.

In this article, I will discuss three methods — all based on modular exponentiation and the laws of exponents — for finding the ending digits of a positive power of two. The techniques I use are easily adapted to powers of any number.

1. Ad Hoc Exponentiation

In this method, you reduce a power of two modulo 10^m repeatedly until you get a congruent power, or product of powers, for which the end digits are known — or easily computed. You use your knowledge of smaller powers of two, in conjunction with the power of a power and product of powers rules, to set up easier sub problems to solve.

You start by dividing the exponent of 2ⁿ by the exponent of a known, smaller power of two, 2^a, getting a quotient q and a remainder r. You then rewrite 2ⁿ as (2^a)^q · 2^r.

Finding the Last Digit

I’ve categorized two sub methods of the ad hoc method that make it more systematic when dealing specifically with powers of two. I call them the powers of two method and the powers of six method.

Powers of Two Method

In the powers of two method, you reduce a power of two by using a power of two ending in 2. This reduces the problem at each stage to a smaller power of two, giving the method a recursive feel. The intermediate powers of two are in effect nested.

For example, let’s find the last digit of 2²⁰⁰⁹, using 2⁵ (32) to reduce the problem at each step:

So , showing that 2²⁰⁰⁹ ends in 2.

The intermediate results — 2⁴⁰⁵, 2⁸¹, 2¹⁷, and 2⁵ — are all congruent, ending in 2. If you recognize this along the way, you can stop. For example, if you happen to know that 2¹⁷ is 131,072, you can stop after the third step.

Any power of two ending in 2 works. Here’s how the process goes when using 2⁹ (512):

Using 2⁹, there is one less step, but the arithmetic is slightly harder (division by 9 instead of division by 5).

Powers of Six Method

In the powers of six method, you reduce a power of two using a power of two that ends in 6. This introduces powers of six, which have to be handled separately and combined with the “remainder” powers of two. For example, let’s do our example with 2⁴ = 16. The first step would give:

But wait! All powers of six end in 6 (6 times 6 mod 10 is 6, and around it goes…), so we just turned this into a very simple problem: .

So, while it doesn’t have the elegance of the powers of two method, the powers of six method is simpler.

Finding the Last Two Digits

The powers of six method does not apply to mod 100, but the powers of two method does — indirectly. Although there is no power of two that ends in 02, we can use any two-digit ending that is a power of two; we just convert it to 2^y using the laws of exponents.

Let’s go back to our example, 2²⁰⁰⁹. Consulting a table of positive powers of two, find a power of two that ends in 04; for example, 2²² (4,194,304). Let’s use this to reduce our problem at each step:

Finding the Last m Digits

For the last m digits, find an m-digit power of two greater than or equal to 2^m and convert it to 2^y as above. Of course, the remainders will become larger and larger as the modulus increases.

2. Successive Squaring

The method of successive squaring, also called repeated squaring or binary exponentiation, is a very systematic way to do modular exponentiation. It’s a generic four step process applicable to any base and modulus, but here’s how to use it to compute 2ⁿ mod 10^m specifically:

1. Rewrite 2ⁿ so that n is a sum of powers of two (essentially, convert n to binary).
2. Rewrite 2ⁿ using the product of powers rule.
3. Create a list of powers 2²ⁱ mod 10^m, by repeatedly squaring the prior result.
4. Combine, with multiplication mod 10^m, the powers in the list that make up 2ⁿ.

This process is independent of the number of ending digits m, although you have to deal with bigger and bigger numbers as m increases.

Example: Find the Last Digit of 2²⁰⁰⁹

Let’s use this method to find the last digit of 2²⁰⁰⁹:

1. 2²⁰⁰⁹ = 2^{1024 + 512 + 256 + 128 + 64 + 16 + 8 + 1}
2. 2²⁰⁰⁹ = 2¹⁰²⁴ · 2⁵¹² · 2²⁵⁶ · 2¹²⁸ · 2⁶⁴ · 2¹⁶ · 2⁸ · 2¹
3. Create a list of powers of two raised to powers of two, mod 10:
   • 
   • 
   • 
   • 
   • 
   • 
   • 
   • 
   • 
   • 
   • 

   (I wrote out the whole list for completeness, but it was unnecessary to go beyond 2⁴. Again, that’s because all powers of six end in 6.)

4. Combine the required powers:
   • 
   • 
   • 

Example: Find the Last Two Digits of 2²⁰⁰⁹

Let’s use this method to find the last two digits of 2²⁰⁰⁹:

1. 2²⁰⁰⁹ = 2^{1024 + 512 + 256 + 128 + 64 + 16 + 8 + 1}
2. 2²⁰⁰⁹ = 2¹⁰²⁴ · 2⁵¹² · 2²⁵⁶ · 2¹²⁸ · 2⁶⁴ · 2¹⁶ · 2⁸ · 2¹
3. Create a list of powers of two raised to powers of two, mod 100:
   • 
   • 
   • 
   • 
   • 
   • 
   • 
   • 
   • 
   • 
   • 

   (Notice that the powers in this list cycle after a point, so it is not necessary to compute them all.)

4. Combine the required powers:
   • 
   • 
   • 
   • 
   • 
   • 
   • 

(I could have used negative numbers in the intermediate steps to make the math easier; for example, -4 instead of 96. Both are congruent mod 100.)

3. Cyclic Powers

In this method, you exploit the fact that the ending m digits of the positive powers of two repeat in cycles; specifically, cycles of length 4·5^m-1, starting at 2^m. Powers of two that differ in their exponents by 4·5^m-1 have the same ending m digits.

There are two ways to use the cycle information, in techniques I call the table method and the base power method.

Table Method

In the table method, you compute the powers of two mod 10^m in sequence, until the ending digits cycle. You label entries sequentially starting at m, wrapping around 0 to end at m – 1. I’ve created tables for the last one, two, and three ending digits; in other words, tables for powers of two mod 10, mod 100, and mod 1000.

To find where in the cycle your power 2ⁿ falls, compute n mod 4·5^m-1, or equivalently, find the remainder of n/(4·5^m-1). The last m digits of 2ⁿ are the digits in the table with the label corresponding to that remainder.

For example, let’s find the last digit of 2²⁰⁰⁹. The last digit of the positive powers of two cycles with length 4, and . According to the table, a remainder of 1 corresponds to a last digit of 2.

Almost as simply, we can find the last two digits of 2²⁰⁰⁹. The last two digits of the positive powers of two cycle with length 20, and . According to the table, a remainder of 9 corresponds to the ending digits 12.

Finding the Last m Digits

The table method works for any number of ending digits, but beyond two or three, is impractical. The tables grow large, by a factor of five for each additional ending digit.

Base Power Method

Every power of two 2ⁿ, mod 10^m, is congruent to some power of two in the first instance of the cycle; that is, a power of two between 2^m and 2^{(m + 4·5m – 1 – 1)}. You need to determine which power of two in this range it is — what I call the base power of two — and then use another method to find its ending digits.

You can find the base power of two directly: it is 2^{m + j}, where j is an offset given by the expression n-m (mod 4·5^m-1).

For example, let’s find the last digit of 2²⁰⁰⁹. , so the base power of two is 2¹⁺⁰ = 2¹ = 2. Trivially, we can see the ending digit is 2.

For the last two digits of 2²⁰⁰⁹, compute . The base power of two is 2²⁺⁷ = 2⁹, which is small enough to compute directly: 512. The last two digits are 12.

Finding the Last m Digits

The base power method works well for any number of ending digits, assuming n is greater than 4·5^m-1.

Cheating

I said find the last digits without using a computer, right? I wanted to verify my work, so I used PARI/GP to compute 2²⁰⁰⁹. It took an instant — here it is:

58,784,291,598,041,831,640,721,059,900,297,317,581,942,666,346,941,194,264,
455,308,125,479,232,583,289,360,069,460,965,699,405,121,019,824,433,389,516,
158,094,000,492,490,796,188,432,969,007,685,435,732,643,092,034,554,442,399,
887,360,352,654,923,898,902,974,171,610,618,912,504,957,328,187,117,386,950,
842,341,026,317,332,718,773,233,103,358,237,779,148,190,179,650,358,079,135,
564,562,516,081,648,810,332,848,214,481,400,042,754,868,418,296,221,651,998,
157,278,605,568,219,649,390,953,792,425,227,268,163,704,976,021,381,769,156,
258,409,778,685,642,966,081,035,151,287,502,869,585,844,829,824,788,935,390,
157,871,063,324,138,385,197,912,084,049,961,962,094,914,858,370,754,777,898,
867,719,950,514,578,646,749,211,908,564,621,201,347,904,089,822,990,746,021,
295,498,658,798,312,326,238,643,788,303,040,512

Summary

The three methods I’ve shown are efficient ways to do modular exponentiation, unlike the straightforward method, which requires n-1 multiplications to compute 2ⁿ mod 10^m. The three methods provide shortcuts to the answer, exploiting knowledge of the laws of exponents and the cycling of ending digits.

Which method should you use? The ad hoc method is the least systematic but allows for case-by-case optimization. The method of successive squaring is the most systematic but may be overkill for certain problems. Learn all three methods to get a deeper understanding, then decide which one you like.

For finding the last digit, I like the ad hoc powers of six method. It is the quickest.

Exercises

For these exercises, use any method you like — or all three if you’re feeling ambitious:

1. Find the last digit of 2⁴⁹⁷
2. Find the last digit of 2²⁰⁰⁰⁰
3. Find the last two digits of 2⁶¹³
4. Find the last two digits of 2⁵¹²
5. Find the last three digits of 2¹²⁹
6. Find the last three digits of 2²⁰⁰⁹

Answers

1. 2
2. 6
3. 92
4. 96
5. 912
6. 512

By Rick Regan (Copyright © 2008-2012 Exploring Binary)

How to Find the Last Digits of a Positive Power of Two

In this article, I will show you why these cycles exist, how long they are, how they are expressed mathematically, and how to visualize them.

Cycle in the Last Digit

The last digit — the ones place — of a decimal integer d is the remainder of the division d/10. Equivalently, the last digit is the result of the operation d mod 10, when following the convention that the least nonnegative value — the common residue — is returned. Modular arithmetic, combined with iterative generation of the positive powers of two, allows us to show the cycle in the last digit:

1. 
2. 
3. 
4. 
5. 
6. 
7. …

We start with 2, compute it mod 10, multiply that result by 2, compute it mod 10, etc. From this it’s clear the pattern will repeat, once a previous result — 2 in this case, at step 5 — is obtained. This shows that the numbers 2ⁿ, n ≥ 1, cycle through the four ending digits 2, 4, 8, and 6.

The cycle implies that powers of two with the same ending digit are related, their exponents differing by a multiple of four:

• Ends in 2: 2¹, 2⁵, 2⁹, 2¹³, 2¹⁷, … .
• Ends in 4: 2², 2⁶, 2¹⁰, 2¹⁴, 2¹⁸, … .
• Ends in 8: 2³, 2⁷, 2¹¹, 2¹⁵, 2¹⁹, … .
• Ends in 6: 2⁴, 2⁸, 2¹², 2¹⁶, 2²⁰, … .

You can express these relationships more succinctly using the laws of exponents, showing explicitly that the ending digit of the first four positive powers of two determine the ending digit of all positive powers of two:

• Ends in 2: 2¹·2^4k, or 2^1+4k, k ≥ 0.
• Ends in 4: 2²·2^4k, or 2^2+4k, k ≥ 0.
• Ends in 8: 2³·2^4k, or 2^3+4k, k ≥ 0.
• Ends in 6: 2⁴·2^4k, or 2^4+4k, k ≥ 0.

You can also relate the positive powers of two to their ending digits based on their exponents (n) mod 4:

• Ends in 2:
• Ends in 4:
• Ends in 8:
• Ends in 6:

This gives an easy way to determine the last digit of any positive power of two. For example, the last digit of 2³¹⁹ is 8, since .

In summary, the table says that if , .

Cycle in the Last Two Digits

A similar analysis, but using mod 100, shows that the last two digits of the powers of two, starting at 2², cycle with a period of 20:

Cycle in the Last Three Digits

To determine the cycle in the last three digits, repeat the same process using mod 1000. It shows that the powers of two, starting at 2³, cycle with a period of 100:

Cycle in the Last m Digits

The last m digits of a positive power of two are taken mod 10^m, and have a cycle of period 4·5^m-1, starting at 2^m. (The proof involves techniques of number theory, and is beyond the scope of this article.)

The periods grow very fast — exponentially — so it is impractical to list tables of ending digits for m greater than three.

Nesting of Cycles

The cycles in m digit, m-1 digit, m-2 digit, …, 1 digit endings can be viewed as nested, even though their starting points are staggered. You just have shift the starting points of the lesser digit cycles to make them coincide.

For example, one copy of the length 100 cycle of three digit endings has within it five copies of the length 20 cycle of two digit endings; one copy of the length 20 cycle of two digit endings has within it five copies of the length 4 cycle of one digit endings. This works when you view the one digit ending cycle as starting at 8 (shifted two positions), the two digit ending cycle as starting at 08 (shifted one position), and the three digit ending cycle as starting at 008 (not shifted).

The diagram below shows the nesting by shading every other occurrence of a cycle (the 100s place is fully shaded, because only 100 powers of two — one cycle of the last three ending digits — are shown).

Nested 1-3 Digit Ending Patterns From 2³ to 2¹⁰²

Exploring Ending Digits with PARI/GP

I used PARI/GP to make some of the calculations above and to check my work. Here are three examples:

• Print the first 20 powers of two that end in 2:

  ? for (i=0,19,print("2^",1+4*i,": ",2^(1+4*i)))
  2^1: 2
  2^5: 32
  2^9: 512
  2^13: 8192
  2^17: 131072
  2^21: 2097152
  2^25: 33554432
  2^29: 536870912
  2^33: 8589934592
  2^37: 137438953472
  2^41: 2199023255552
  2^45: 35184372088832
  2^49: 562949953421312
  2^53: 9007199254740992
  2^57: 144115188075855872
  2^61: 2305843009213693952
  2^65: 36893488147419103232
  2^69: 590295810358705651712
  2^73: 9444732965739290427392
  2^77: 151115727451828646838272
  

• Print the cycle of the last two digits (the ‘%’ operator returns the remainder, which in effect is how we’re using modular arithmetic):

  ? for (i=2,21,print("2^",i," mod(100): ",2^i % 100))
  2^2 mod(100): 4
  2^3 mod(100): 8
  2^4 mod(100): 16
  2^5 mod(100): 32
  2^6 mod(100): 64
  2^7 mod(100): 28
  2^8 mod(100): 56
  2^9 mod(100): 12
  2^10 mod(100): 24
  2^11 mod(100): 48
  2^12 mod(100): 96
  2^13 mod(100): 92
  2^14 mod(100): 84
  2^15 mod(100): 68
  2^16 mod(100): 36
  2^17 mod(100): 72
  2^18 mod(100): 44
  2^19 mod(100): 88
  2^20 mod(100): 76
  2^21 mod(100): 52
  

  (Single-digit results have an implicit leading 0.)

• Print the period of the last 1 to 10 digits:

  ? for (i=1,10,print(i,": ",4*5^(i-1)))
  1: 4
  2: 20
  3: 100
  4: 500
  5: 2500
  6: 12500
  7: 62500
  8: 312500
  9: 1562500
  10: 7812500
  

By Rick Regan (Copyright © 2008-2012 Exploring Binary)

Patterns in the Last Digits of the Positive Powers of Two

PARI/GP is a sophisticated tool, with several components — yet it’s easy to install and use. I use its command shell in particular, the PARI/GP calculator, or gp for short. I will show you how to use simple gp commands to explore binary numbers.

PARI/GP Calculator (Sample of Calculations Used to Explore Binary Numbers)

In particular, I’ll show you how to:

• Generate arbitrarily large or small powers of two
• Check if an integer is a positive power of two
• Sum powers of two
• Find the largest power of two contained in a number
• Convert dyadic decimals to dyadic fractions
• Discover properties of binary representations
  • Convert decimal to binary
  • Count number of trailing zeros
  • Count number of bits
  • Count number of 1 bits

In the sections that follow, I’ve chosen to show the text log of commands — formatted for display — rather than screenshots of the command shell.

Generate Arbitrarily Large or Small Powers of Two

Nonnegative Powers of Two

gp can generate any arbitrary nonnegative power of two:

? 2^8
%1 = 256

? 2^64
%2 = 18446744073709551616

? 2^200
%3 = 1606938044258990275541962092341162602522202993782792835301376

gp can also generate sequences of nonnegative powers of two, from 2⁰ to 2ⁿ:

? for(i=0,16,print(2^i))
1
2
4
8
16
32
64
128
256
512
1024
2048
4096
8192
16384
32768
65536

There are also a few “fun” ways to generate sequences of nonnegative powers of two:

• By computing the divisors of 2ⁿ:

  ? divisors(2^16)
  %1 = [1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096,
  8192, 16384, 32768, 65536]
  

• By computing the sum of each row, 0 through n, of Pascal’s Triangle:

  ? for(i=0,16,print(sum(j=0,i,binomial(i,j))))
  1
  2
  4
  8
  16
  32
  64
  128
  256
  512
  1024
  2048
  4096
  8192
  16384
  32768
  65536
  

• By computing the value of Euler’s phi function for 2¹ through 2ⁿ⁺¹:

  ? for(i=0,16,print(eulerphi(2^(i+1))))
  1
  2
  4
  8
  16
  32
  64
  128
  256
  512
  1024
  2048
  4096
  8192
  16384
  32768
  65536
  

• By computing the order of 5 mod powers of two for 2² through 2ⁿ⁺²:

  ? for(i=0,16,print(znorder(Mod(5,2^(i+2)))))
  1
  2
  4
  8
  16
  32
  64
  128
  256
  512
  1024
  2048
  4096
  8192
  16384
  32768
  65536
  

Negative Powers of Two

gp can generate any arbitrary negative power of two:

• In fraction form:

  ? 2^-16
  %1 = 1/65536
  
  ? 1/2^16
  %2 = 1/65536
  

• In decimal form:

  ? 2.^-16
  %1 = 0.00001525878906250000000000000000
  
  ? 1/2.^16
  %2 = 0.00001525878906250000000000000000
  
  ? 0.5^16
  %3 = 0.00001525878906250000000000000000
  

gp can also generate sequences of negative powers of two, from 2^-1 to 2^-n, in fraction or decimal form:

? for(i=1,8,print(2^-i))
1/2
1/4
1/8
1/16
1/32
1/64
1/128
1/256

? for(i=1,8,print(2.^-i))
0.500000000000000000000000000000
0.250000000000000000000000000000
0.125000000000000000000000000000
0.0625000000000000000000000000000
0.0312500000000000000000000000000
0.0156250000000000000000000000000
0.00781250000000000000000000000000
0.00390625000000000000000000000000

A Note on Precision

When dealing with decimals, you have to make sure you are using the proper precision. The default precision, 28 significant decimal digits, is enough for the examples above — in fact, more than enough, as judged by the trailing 0s.

Here’s an example where the default precision is not enough:

? 2.^-42
%1 = 0.0000000000002273736754432320594787597656

How do I know there’s not enough precision? First of all, the displayed answer has only 40 decimal places, and 2^-42 has 42 decimal places (2^-n has n decimal places). Also, the displayed answer does not end with the digit ‘5’; all negative powers of two must end in ‘5’.

Here is 2^-42 again, printed with 42 significant digits using the ‘\p’ command:

? \p 42
   realprecision = 48 significant digits (42 digits displayed)

? 2.^-42
%1 = 0.000000000000227373675443232059478759765625000000000000

(PARI/GP sets its precision based on machine word boundaries, so the number of significant digits used for computation may be greater than the number of significant digits displayed.)

As you can see, we overshot — trailing 0s were added. Even though 2^-42 has 42 decimal places, it only has 30 significant digits; that is, 30 digits after the 12 leading 0s.

Here’s how to display 2^-42 exactly, without the trailing 0s:

? \p 30
   realprecision = 38 significant digits (30 digits displayed)

? 2.^-42
%1 = 0.000000000000227373675443232059478759765625

Check if an Integer is a Positive Power of Two

An integer is a positive power of two if its only prime factor is 2, as demonstrated by these examples:

? 2^64
%1 = 18446744073709551616

? factor(18446744073709551616)
%2 =
[2 64]

? 2^64-1
%3 = 18446744073709551615

? factor(18446744073709551615)
%4 =
[3 1]

[5 1]

[17 1]

[257 1]

[641 1]

[65537 1]

[6700417 1]

Sum Powers of Two

A binary number represents a sum of powers of two, so you can convert it to decimal simply by performing addition.

Sum Nonnegative Powers of Two

Converting binary integers to decimal is straightforward. For example, 1100100₂, which represents 2⁶ + 2⁵ + 2², is 100₁₀:

? 2^6 + 2^5 + 2^2
%1 = 100

Sum Negative Powers of Two

Binary fractionals can be finite or infinite; converting finite binary fractionals to decimal is straightforward. For example, 0.101₂ is 5/8, or 0.625, in decimal:

? 2^-1 + 2^-3
%1 = 5/8

? 2.^-1 + 2^-3
%2 = 0.6250000000000000000000000000

0.111111₂ is 63/64, or 0.984375, in decimal:

? sum(i=1,6,2^-i)
%1 = 63/64

? sum(i=1,6,2.^-i)
%2 = 0.9843750000000000000000000000

Infinite binary fractionals are more involved; they require infinite sums of negative powers of two. To show a simple example, consider 0.01₂. This represents an infinite geometric series of all negative powers of two except 2^-1. It converges to 1/2:

(This is the same summation shown in this site’s header image.)

? suminf(i=2,2^-i)
%1 = 0.5000000000000000000000000000

You can see the series converge by using finite sums, adding more terms each time:

? \p 50
? sum(i=2,10,2.^-i)
%1 = 0.49902343750000000000000000000000000000000000000000

? sum(i=2,20,2.^-i)
%2 = 0.49999904632568359375000000000000000000000000000000

? sum(i=2,30,2.^-i)
%3 = 0.49999999906867742538452148437500000000000000000000

? sum(i=2,40,2.^-i)
%4 = 0.49999999999909050529822707176208496093750000000000

? sum(i=2,50,2.^-i)
%5 = 0.49999999999999911182158029987476766109466552734375

Find the Largest Power of Two Contained in a Number

The largest power of two less than or equal to a positive integer n is 2 raised to the integer part of the base 2 logarithm of n; that is, 2^{floor(log₂(n))}, or . This is used, for example, in the “subtract largest power of two” method of decimal to binary conversion, or in the formula for the solution to the Josephus Problem.

PARI/GP doesn’t compute base 2 logarithms directly, so you have to do a change of base. For example, the largest power of two in 100 is:

? 2^(floor(log(100)/log(2)))
%1 = 64

You have to be careful with larger numbers, especially those that are powers of two or close to one. The logarithm function will require greater precision to give the correct result:

? n=2^50
%1 = 1125899906842624

? 2^(floor(log(n)/log(2)))
%2 = 562949953421312

? \p 100
   realprecision = 105 significant digits (100 digits displayed)

? 2^(floor(log(n)/log(2)))
%3 = 1125899906842624

(I don’t know if there’s a systematic way to pick the exact precision required — I picked 100 digits arbitrarily, assuming it would be large enough.)

Convert Dyadic Decimals to Dyadic Fractions

A dyadic fraction is a number that can be written in the form a/2ⁿ. For our purposes, a < 2ⁿ, so it is simply a sum of negative powers of two. For example, 3/8 = 1/4 + 1/8.

A dyadic decimal is what I call a dyadic fraction in decimal form; for example, 0.375. A dyadic decimal can be converted to a dyadic fraction easily: just write it in the form b/10^m and reduce to lowest terms. For example:

? 375/1000
%1 = 3/8

? 375/10^3
%2 = 3/8

PARI/GP reduces fractions to lowest terms automatically, so there’s nothing further to do.

For longer decimals, like those that represent double-precision floating-point numbers, it’s cumbersome — and error prone — to count decimal places by hand. I wrote this gp script to automate the process:

dyadic(decimal_num_string)=
{
 decimal_num = eval(decimal_num_string);
 decimal_den_exponent = length(decimal_num_string);
 decimal_den = 10^decimal_den_exponent;

 dyadic_num = numerator(decimal_num/decimal_den);
 dyadic_den = denominator(decimal_num/decimal_den);
 dyadic_den_exponent = valuation(dyadic_den,2);

 print("Decimal: 0.",decimal_num_string);
 print("Dyadic fraction: ",dyadic_num,"/2^",dyadic_den_exponent);
}

It takes as argument the string representation of the numerator of the dyadic fraction, which is just the dyadic decimal without the “0.” prefix (a string preserves leading 0s, which must be counted to determine the proper power of 10 denominator!). It computes the denominator that matches the length of the numerator and then divides the two.

The script has the added bonus of displaying the denominator in the form 2ⁿ, since large denominators are otherwise not readily identified as powers of two.

Here’s the script in action:

? \r dyadic

? dyadic("375")
Decimal: 0.375
Dyadic fraction: 3/2^3

? 2.^-8
%1 = 0.003906250000000000000000000000

? dyadic("00390625")
Decimal: 0.00390625
Dyadic fraction: 1/2^8

? dyadic("1000000000000000055511151231257827021181583404541015625")
Decimal: 0.1000000000000000055511151231257827021181583404541015625
Dyadic fraction: 3602879701896397/2^55

? dyadic("59999999999999997779553950749686919152736663818359375")
Decimal: 0.59999999999999997779553950749686919152736663818359375
Dyadic fraction: 5404319552844595/2^53

(You might recognize those last two examples as the double-precision floating-point approximations of the decimal values 0.1 and 0.6, respectively.)

Discover Properties of Binary Representations

You can discover properties of the binary representation of a number in two ways: directly, by converting it to binary and viewing the result, or indirectly, by doing calculations related to its underlying binary representation. PARI/GP lets you do both.

Convert Decimal to Binary

PARI/GP can convert integers to binary, although to an awkward format:

? binary(7)
%1 = [1, 1, 1]

? binary(2^50-1)
%2 = [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1]

PARI/GP can also convert real numbers to binary, to an even more awkward format:

? binary(0.1)
%1 = [[0], [0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0,
0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1,
1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0,
0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1,
1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 1]]

? binary(0.6)
%2 = [[0], [1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0,
0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1,
1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0,
0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1,
1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0]]

You can’t tell from the output whether these are finite or infinite expansions, but we know in this case they are infinite: 0.1₁₀ is 0.00011₂, and 0.6₁₀ is 0.1001₂. They are rounded to the nearest dyadic allowed with the given precision.

More bits are printed if you increase the precision with the ‘\p’ command.

Count Number of Trailing Zeros

You can determine the number of trailing zeros in the binary representation of an integer greater than one without first converting it to binary — all you need to do is factor it. If it has no factors of 2, it has no trailing zeros; if it has a factor of 2, it has a number of trailing zeros equal to the exponent to which 2 is raised. For example, 100₁₀ = 1100100₂ has two trailing zeros:

? factor(100)
%1 =
[2 2]
[5 2]

The valuation function gives the same answer. It prints the exponent of 2 directly:

? valuation(100,2)
%1 = 2

Count Number of Bits

The number of bits in the binary representation of an integer n > 0 is :

? n=2^32-1
%1 = 4294967295

? floor(log(n)/log(2))+1
%2 = 32

? n=2^32
%3 = 4294967296

? floor(log(n)/log(2))+1
%4 = 33

? n=100
%5 = 100

? floor(log(n)/log(2))+1
%6 = 7

(As mentioned above, for large numbers, be sure to set the precision high enough so that the logarithm gives the correct result.)

Count Number of 1 Bits

The number of 1 bits in the binary representation of an integer n > 0 is computed by checking each bit of the binary representation with the bittest() function:

? n=7
%1 = 7

? sum(i=1,floor(log(n)/log(2))+1,bittest(n,i-1))
%2 = 3

? n=2^32-1
%3 = 4294967295

? sum(i=1,floor(log(n)/log(2))+1,bittest(n,i-1))
%4 = 32

? n=2^32
%5 = 4294967296

? sum(i=1,floor(log(n)/log(2))+1,bittest(n,i-1))
%6 = 1

(As mentioned above, for large numbers, be sure to set the precision high enough so that the logarithm gives the correct result.)

bittest() works on the underlying binary representation of a number. It numbers bits from right to left, starting at bit 0.

Counting the number of 1 bits is another way to check whether an integer is a power of two — it is a power of two if it has one, and only one, 1 bit.

You can also compute the distribution of 1 bits for the numbers 0 to 2ⁿ – 1 by computing row n of Pascal’s Triangle. For example, here are the binary representations of the numbers 0 through 7:

? for(i=0,7,print(binary(i)))
[0]
[1]
[1, 0]
[1, 1]
[1, 0, 0]
[1, 0, 1]
[1, 1, 0]
[1, 1, 1]

Here is row 3 of Pascal’s Triangle:

? n=3
%1 = 3

? for(i=0,n,print(binomial(n,i)))
1
3
3
1

This tells us that

• 1 number has 0 1 bits (0)
• 3 numbers have 1 1 bit (1, 10, 100)
• 3 numbers have 2 1 bits (11, 101, 110)
• 1 number has 3 1 bits (111)

Acknowledgements

Thanks to the PARI/GP mailing list, especially Kurt Foster, John Cremona, Bill Allombert, and Karim Belabas, for their ideas and for answering my questions.

By Rick Regan (Copyright © 2008-2012 Exploring Binary)

Exploring Binary Numbers With PARI/GP Calculator

Here’s an example with 13 participants:

Alternating Elimination with 13 people, order of elimination shown in red (winner is person 11)

As Pradeep and his readers point out, there’s no need to work through the elimination process — a simple formula will give the answer. This formula, you won’t be surprised to hear, has connections to the powers of two and binary numbers. I will discuss my favorite solution, one based on the powers of two.

When the Number of Participants is a Power of Two

Alternating elimination means one of every two participants is eliminated. This is halving, and suggests powers of two are involved. Let’s first explore this with the special case where the number of participants is a power of two, since powers of two halve neatly into powers of two.

Here is an example circle with eight people:

Alternating Elimination (8 people)

The elimination process works like this: the first pass starts at person 1 and proceeds clockwise, and each new pass starts every time person 1 is reached. The people eliminated on a pass are crossed out, and are marked to indicate the order in which they were eliminated. Eliminated people are then omitted in subsequent diagrams.

Three passes are required to determine the winner:

• Pass 1 eliminates four people: 2, 4, 6, 8.
• Pass 2 eliminates two people: 3, 7.
• Pass 3 eliminates one person: 5.

This leaves person 1 the winner.

The number of people eliminated (and equivalently, remaining) per pass follows the same pattern as in a single-elimination tournament with a power of two number of participants.

Analysis

Regardless of the number of participants n, person 1 survives the first pass. Since n is even, as every positive power of two is, person 1 survives the second pass as well. In the first pass, the process goes like this: person 1 is skipped, person 2 is eliminated, person 3 is skipped, person 4 is eliminated, … , person n-1 is skipped, person n is eliminated. The second pass starts by skipping person 1.

As long as the number of participants per pass is even, as it will be for a power of two starting point, the same pattern is followed: person 1 is skipped each time. Therefore, for any power of two, person 1 always wins.

Proof by Induction

You can also show that person 1 is the winner using an inductive proof (for details see Miguel Lerma’s proof of the Josephus problem). Compared to the argument above, induction works in the opposite direction; that is, it builds up to a more complicated problem from a simpler one.

For n = 2¹ = 2 participants, the base case, it’s easy to see that person 1 is the winner. For the induction hypothesis, assume person 1 is the winner for n = 2^m. Show person 1 is the winner for n = 2^m+1.

When n = 2^m+1, 2^m people — all the even numbered people — are eliminated in the first pass, leaving 2^m people — all the odd numbered people — remaining. By the induction hypothesis, person 1 is the winner of the n = 2^m remaining people, and thus the winner among all n = 2^m+1 people. Therefore, for any power of two, person 1 always wins.

When the Number of Participants is NOT a Power of Two

When the number of participants is not a power of two, we know this much: person 1 can’t be the winner. This is because at least one pass will have an odd number of participants. Once the first odd participant pass is complete, person 1 will be eliminated at the start of the next pass.

So is there an easy way to determine who is the winner? Let’s step back and take a closer look at the elimination process in the 13-person example:

Alternating Elimination (13 people)

Four passes are required to determine the winner:

• Pass 1 eliminates six people: 2, 4, 6, 8, 10, 12.
• Pass 2 eliminates four people: 1, 5, 9, 13.
• Pass 3 eliminates one person: 7.
• Pass 4 eliminates one person: 3.

This leaves person 11 the winner.

Analysis

Powers of two come into play here, but you have to change your perspective to see them. They don’t occur on pass boundaries — they span them. At some point, during pass 1, the number of participants remaining becomes a power of two. In this example, that occurs when 5 of the 13 people are eliminated, leaving an 8 person problem: 11, 12, 13, 1, 3, 5, 7, 9. This means person 11, the first person in the new power of two circle, wins.

Here’s the 13-person example again, with the 8-person power of two sub case shown explicitly with passes realigned:

8-Person Power of Two Subset of 13-Person Problem

(It occurred to me that the alternating elimination process is an indirect way to check whether a number is a positive power of two. A number is a positive power of two if and only if, when halved repeatedly, becomes 1.)

The Equations

We can solve both cases — in other words, for an arbitrary number of participants — using a little math.

Write n as n = 2^m + k, where 2^m is the largest power of two less than or equal to n. k people need to be eliminated to reduce the problem to a power of two, which means 2k people must be passed over. The next person in the circle, person 2k + 1, will be the winner. In other words, the winner w is w = 2k + 1.

Let’s apply these equations to a few examples:

• n = 8: The equations still apply, although using them is unnecessary: n = 8 + 0, so k = 0 and w = 0 + 1 = 1.
• n = 13: n = 8 + 5, so k = 5 and w = 2*5 + 1 = 11.
• n = 1000: This is the example in the New York Times: n = 1000 = 512 + 488, so k = 488 and w = 2*488 + 1 = 977.

The Formula

We can combine the equations n = 2^m + k and w = 2k + 1 to get a single formula for w:

• Rearrange n = 2^m + k to isolate k: k = n – 2^m.
• Substitute this expression for k into w = 2k + 1:

  w = 2(n – 2^m) + 1

The Formula as a Function of One Variable

As written, the formula is a function of two variables, n and m. This works fine, but ideally it should be a function of only one variable — n. This means we have to eliminate m, or more precisely, make m itself a function of n.

We described 2^m loosely as “the largest power of two less than or equal to n,” but that is an algorithmic description. Described mathematically, m is the integer part of the base 2 logarithm of n; that is, m = floor(log₂(n)), or .

So now we have a formula in terms of the number of participants n:

Summary

An alternating elimination Josephus problem has a deep connection to the powers of two, a connection reflected in the formula we derived to find the winning spot. The formula requires a few simple calculations, and is a function of the number of participants n: find the largest power of two in n, subtract it from n, double the result, and add 1. The person in that spot will be the winner.

By Rick Regan (Copyright © 2008-2012 Exploring Binary)

Powers Of Two In The Josephus Problem

Basketball or Binary?

If you’re like me, you also think of powers of two, binary trees, logarithms, laws of exponents, geometric sequences, geometric series, and Bernoulli trials — in short, the elements of binary numbers, binary code, and binary logic.

I hope you’re thinking: “wow — all that math stuff I hated actually has a practical use!”

(As of 2011, the format of the tournament has changed. Four play-in games, known as the “First Four,” determine the last four entrants to the 64-team field. My description below applies to the six-round, 64-team part of the tournament, although the NCAA has renumbered the rounds — the old first round is now the second round, etc.)

Powers of Two

The NCAA Basketball Tournament, depicted in the diagram above, is a 64-team, single-elimination tournament. Why 64 teams? Because 64 is a power of two. Powers of two halve into powers of two, which gives the tournament symmetry: all 64 teams play in the first round, and teams that meet in rounds thereafter have won the same number of games.

Powers of two that pop up in the tournament are: the nonnegative powers of two 1, 2, 4, 8, 16, 32, 64, and 2⁶³, and the negative powers of two 1/2, 1/4, 1/8, 1/16, 1/32, 1/64, and 1/2⁶³.

Binary Trees

A binary tree is a model of things that have binary structure. The tournament has binary structure because its games are binary; each has one of two outcomes. You can see binary trees in the brackets. Although they are laid out as four separate trees, they are actually part of one big binary tree. This diagram connects them:

Bracketless Representation of the NCAA Basketball Tournament.

You can see the nested structure in the tree. It is made up of subtrees, each of which is effectively a power of two sized sub-tournament. One 64-team tournament is comprised of two 32-team sub-tournaments, each of which is comprised of two 16-team sub-tournaments, etc. This symmetry, due to the initial power of two number of teams, makes the tree a perfect binary tree.

A computer scientist would draw the tree like this:

The NCAA Tournament Represented As a Binary Tree (click for higher resolution).

Each circle, called a node, represents a team. The 64 nodes at the bottom, called leaf nodes, represent the 64 teams selected. The 63 nodes above that, called internal nodes, represent teams that have won one or more games. The 63 internal nodes correspond to the 63 games played in the tournament. Each level of internal nodes represents winners of one round. Winners advance up the tree until one team remains at the top node, called the root node.

The internal nodes are labeled by level, from round 1 to round 6. These represent the status of the tournament after the given round is complete. I’ve labeled the leaf nodes round 0, but they don’t really represent a round; the 64 teams get there by selection. Round 6 marks the end of the tournament — the lone team remaining has won all 6 of its games.

Logarithms

A logarithm is a fancy word for power, or exponent. You’re probably familiar with logarithms in base 10, but you can compute logarithms in any base. Logarithms in base 2 are useful when dealing with powers of two. The logarithm in base 2 of a power of two is the exponent corresponding to that power of two. For example, the logarithm in base 2 of 64, or log₂(64), is 6, which means 64 is 2⁶. In other words, it takes 6 rounds to halve 64 teams down to 1.

Laws of Exponents

You can use the laws of exponents to determine how many games are played in any round k:

That is a compact way of expressing the halving of 64 teams k times.

Applying the formula, 2^(6-3) = 2³ = 8 games are played in round 3. How many games are played in round 6? That’s easy: 2^(6-6) = 2⁰ = 1.

Geometric Sequences

A geometric sequence, also known as a geometric progression, is a sequence of numbers that are related by a constant multiple, like sequences of powers of two. In sequences of powers of two, the multiple is either 2 or 1/2. For example, in the sequence 1, 2, 4, 8, 16, 32, 64, each number is double the previous; in the sequence 64, 32, 16, 8, 4, 2, 1, each number is half the previous.

Here are sequences of powers of two that arise in the tournament:

• 64, 32, 16, 8, 4, 2. The number of teams that play, by round.
• 32, 16, 8, 4, 2, 1. The number of games played, by round. Equivalently, it’s the number of winning teams per round, or the number of losing teams per round.
• 1/2, 1/4, 1/8, 1/16, 1/32, 1/64. The fraction of teams that remain, by round.

Geometric Series

A series is a sum, and a geometric series is a sum of numbers from a geometric sequence. The total number of games played in the tournament is a geometric series: 32 + 16 + 8 + 4 + 2 + 1 = 63. Writing it in reverse order — 1 + 2 + 4 + 8 + 16 + 32 = 63 — makes it clearer that this notational shortcut applies:

The symbol means sum, and the way you read it is the sum from i = 0 to 5 of 2ⁱ is 63. In other words, the sum of the first 6 nonnegative powers of two is 63.

There is a nice formula that gives the sum without any adding:

It says the sum of the first n+1 nonnegative powers of two is one less than the n+2nd nonnegative power of two. For us, n is 5, so the sum is 2⁶ – 1, or 63.

(There’s another — some say simpler — way to count the number of games. Since there are 64 teams, and all but one team loses exactly one game, there must be 63 games. It works, but I find the geometric series explanation more satisfying.)

The formula might be overkill for a basketball tournament, but it comes in handy in other contexts. What’s the largest value that can fit in a 32-bit binary number? The largest value is represented in binary as 32 1 bits, which represents the first 32 nonnegative powers of two. Applying the formula, we know the sum is 2³² – 1, or 4,294,967,295.

Games Played Through Round k

There’s a more general formula to compute the number of games played through any round, not just the last. Doing it by hand, you’ll see that the total number of games played through each of the 6 rounds is 32, 48, 56, 60, 62, and 63, respectively. That’s 32, 32 + 16, 32 + 16 + 8, 32 + 16 + 8 + 4, 32 + 16 + 8 + 4 + 2, and 32 + 16 + 8 + 4 + 2 + 1. Except for the last one, those series don’t start at 1, so the summation formula above does not apply.

That gives me a good excuse to introduce series of negative powers of two. Look at the sum 32 + 16, for instance. That’s equal to 64 * (1/2 + 1/4). The sum 32 + 16 + 8 + 4 is equal to 64 * (1/2 + 1/4 + 1/8 + 1/16). Those sums all include a series of negative powers of two starting at 2^-1, and conveniently, there’s a formula for them:

(For those of you that care for such things, that represents the value of a binary fraction consisting of k 1s following the radix point — but I digress.)

To use that formula, just plug in your value for round k and then multiply by 64, the original number of teams. For example, the number of games played through round 3 is 64 * (2³ – 1)/2³, or 64 * (7/8), or 56.

Bernoulli Trials, Outcomes, and Probability

A Bernoulli trial is an event that has one of two possible random outcomes — flipping a coin is a good example. Each game can be considered a Bernoulli trial; either team A wins or team B wins. From that you can derive the number of different outcomes for the entire tournament. That is, the number of ways the brackets can be filled out: 9,223,372,036,854,775,808.

That’s nine quintillion, two hundred twenty-three quadrillion, three hundred seventy-two trillion, thirty-six billion, eight hundred fifty-four million, seven hundred seventy-five thousand, eight hundred and eight. Or, if you prefer, 2⁶³.

That’s a big number, but it’s arrived at fairly simply. One way is to notice that there are 63 games with two outcomes each. By the rules of counting, that’s 2⁶³. But there’s a longer explanation, one I find more intuitive, and it’s based on the number of outcomes per round.

There are 2³² outcomes in round 1, since there are 32 games; there are 2¹⁶ outcomes in round 2, since there are 16 games; etc. But those outcomes are based on only one selection of teams entering that round; you must account for all possible selections. You do that by multiplying. For example, for each of the 2³² round 1 outcomes, there are 2¹⁶ round 2 outcomes. This means there are 2³² * 2¹⁶ = 2^(32+16) = 2⁴⁸ outcomes (by the laws of exponents) through round 2. Putting all 6 rounds together, you get:

2³² * 2¹⁶ * 2⁸ * 2⁴ * 2² * 2¹ = 2^{(32+16+8+4+2+1)} = 2⁶³ .

(I resisted putting a summation symbol in the exponent .)

The Office Pool

If you’re filling in brackets for your office pool, your odds of guessing all outcomes correctly, in theory, are 1 in 2⁶³. In reality, your odds are much better than that, because not all teams are evenly matched.

How about the odds of picking just the winner of the tournament? Those are about 18 orders of magnitude better, at 1 in 64. Again, that’s theoretical, since it assumes all teams are evenly matched.

What if the lowest numbered seeds were guaranteed to win all their games? Well, it wouldn’t be much of a tournament, because you might as well skip the first four rounds. The four number one seeds would face each other in the final four, and at that point you’d have a 1 in 4 chance of picking the winner, and a 1 in 8 chance of filling out the brackets correctly (4 round 5 outcomes multiplied by 2 round 6 outcomes).

Application to Other Tournaments

The analysis above applies directly to other 64-competitor tournaments, like the World Golf Championship match play tournament and the Wimbledon doubles tennis tournaments. The analysis also generalizes to other power of two sized tournaments, like the NCAA National Invitational Tournament (32 teams, 31 games, and 5 rounds) and the Wimbledon singles tennis tournaments (128 players, 127 matches, and 7 rounds).

(Caution: just because a tournament starts with a power of two number of competitors doesn’t mean it has to be laid out symmetrically. For example, in the 16-team Big East tournament, there are two rounds of byes before it becomes a symmetric, 8-team tournament.)

Elements of the above analysis apply to many other single-elimination tournaments, even when they don’t start with a power of two number of competitors. The Big Ten Conference Men’s Basketball Tournament, for example, starts with eleven teams; it becomes a symmetric 8-team, 7 game, 3 round tournament after the first round.

Elements of the above analysis also apply to other tournament formats, like the winner’s bracket in double-elimination tournaments, and the upper bracket in triple-elimination tournaments.

Tournaments in just about every sport have elements of binary.

By Rick Regan (Copyright © 2008-2012 Exploring Binary)

Elements of Binary in the NCAA Basketball Tournament

.

Why are they called powers of two? What is the pattern you see? How is the set described mathematically? What are the set’s components? We will answer those questions in this article.

Let’s start by naming its three basic components, or subsets: the positive powers of two, the negative powers of two, and two to the zeroth power. These will serve as a guide through our explanation.

The Positive Powers of Two

Consider the set of numbers you get when you start with the number two and keep doubling: {2, 4, 8, 16, 32, 64, …}. This is the infinite set of positive integers known as the positive powers of two.

Why are numbers in this set called positive powers of two? First, what is a “positive power”? A positive power is another term for “positive exponent,” which is just shorthand for multiplication. Instead of 2·2·2·2·2, for example, you could write 2⁵. This reads “two raised to the fifth power” or “two to the fifth power” or simply “two to the fifth.” However you choose to say it, it means five twos multiplied together — 32. Since it is two we are raising to the power, we call 2⁵ a power of two; and more specifically in this case, a positive power of two.

In the order in which we’ve listed them, each positive power of two is two times the previous. You can show the factors of two explicitly by writing

{2, 2·2, 2·2·2, 2·2·2·2, 2·2·2·2·2, 2·2·2·2·2·2, …}.

Using exponents, you can rewrite it as {2¹, 2², 2³, 2⁴, 2⁵, 2⁶, …}. (Recall that 2¹ simply means 2 — no multiplication is involved.) Recognizing the pattern, we can write it more concisely as {2ⁿ | n is an integer > 0}, which reads “the set of all elements 2ⁿ such that n is a positive integer.”

The Negative Powers of Two

Consider the set of numbers you get when you start with the fraction 1/2 and keep halving: . Equivalently, consider the set of numbers you get when you start with the decimal fraction 0.5 and keep halving: {0.5, 0.25, 0.125, 0.0625, 0.03125, 0.015625, …}. In either form, this is the infinite set of positive fractions known as the negative powers of two.

Why are numbers in this set called negative powers of two? Well first of all, what is a negative power, or negative exponent? Surely it isn’t shorthand for multiplying a negative number of factors — that wouldn’t make sense. Then what, for example, does 2^-5 (two to the negative fifth or two to the minus fifth) mean? It means raise two to the fifth power and then invert it, giving 1/2⁵ = 1/32 = 0.03125. This inverse relationship is clear when you compare the sets {2, 4, 8, 16, 32, 64, …} and .

In the order in which we’ve listed them, each negative power of two is one-half the previous. You can show the factors of one-half explicitly by writing

.

If you combine the denominators you get

.

This can be expressed using positive exponents as

.

Recognizing these as negative exponents in disguise, you can rewrite it as {2^-1, 2^-2, 2^-3, 2^-4, 2^-5, 2^-6, …}. Written more concisely, this is {2ⁿ | n is an integer < 0}, which reads “the set of all elements 2ⁿ such that n is a negative integer.”

It’s worth emphasizing — the word “negative” qualifies the word “powers” and not the phrase “powers of two.” It is the power that is negative, not the power of two. We wouldn’t say, for example, that -8 is a negative power of two (although it is in a different sense — it’s a power of two that’s negative). Just remember — negative powers of two are positive numbers!

Two to The Zeroth Power

To this point we’ve discussed raising two to all integer powers except zero. So what is two to the zeroth power, or 2⁰? An exponent of zero does not have a neat interpretation like positive or negative exponents. It’s just defined to be one (so that exponent arithmetic works out). So like any number (except zero) raised to the zeroth power, 2⁰ = 1. So one is a power of two, and for our purposes we’ll refer to it as the one-element set {1} or {2⁰}.

The Powers of Two

If you combine the positive powers of two, the negative powers of two, and {2⁰} you get the set known collectively as the powers of two. This is the set of all elements 2ⁿ such that n is an integer, or {2ⁿ | n is an integer}. These are the values

,

or equivalently, with fractions in decimal form,

{…, 0.015625, 0.03125, 0.0625, 0.125, 0.25, 0.5, 1, 2, 4, 8, 16, 32, 64, …}.

You’ll notice that in this set we’ve written the negative powers of two in reverse. This is the standard way to display a set that is infinite in “both directions.” It corresponds to the orientation of the number line; the numbers get infinitely small as you go left and infinitely large as you go right. Also, it shows a simple relationship when written this way: for any two adjacent numbers, the one to the right is double the one to the left, or equivalently, the one to the left is half the one to the right.

Other partitions

It is sometimes convenient to partition the powers of two in other ways. For example, you can think of the set as being composed of just two subsets: the negative powers of two and the nonnegative powers of two. The nonnegative powers of two combine {2⁰} and the positive powers of two: {1, 2, 4, 8, 16, 32, 64, …} or {2ⁿ | n is an integer ≥ 0}. This breakdown is commonly used in the context of binary numbers, for example.

Similarly, you can think of the powers of two as being composed of the positive powers of two and the nonpositive powers of two. The nonpositive powers of two combine {2⁰} and the negative powers of two, and is written or {2ⁿ | n is an integer ≤ 0}. This breakdown is not used commonly, however.

Summary

The following diagram summarizes what we’ve discussed:

The powers of two and its major subsets.

Here is the same, in symbols:

• The powers of two = {2ⁿ | n is an integer}.
• The positive powers of two = {2ⁿ | n is an integer > 0}.
• The negative powers of two = {2ⁿ | n is an integer < 0}.
• The nonnegative powers of two = {2ⁿ | n is an integer ≥ 0}.
• The nonpositive powers of two = {2ⁿ | n is an integer ≤ 0}.

And now the same, in words:

A power of two is the number two raised to an integer power. A power of two is classified according to the sign of its exponent: positive, negative, nonnegative, or nonpositive. Powers of two are positive numbers. Nonnegative powers of two are integers, and negative powers of two are fractions. The negative powers of two are just the reciprocals of the positive powers of two, and are expressed as either proper fractions or decimals.

Endnotes

Competing Definitions

If this definition of the powers of two is broader than one you’ve seen it’s because there are other definitions in common use (see “A Standard Definition of The Powers of Two”).

Set vs. Sequence

Technically, we’re being loose about the distinction between a set and a sequence. A set describes its elements without specifying order, whereas a sequence describes a particular ordering of elements. {1, 2, 4} and {4, 1, 2} are the same set, but (1, 2, 4) and (4, 1, 2) are different sequences.

We’ve listed powers of two in a particular order and described properties of that ordering, which means we’ve implicitly defined a sequence as well — a geometric sequence in fact. Where this does not cause confusion, we will exploit this fact, as in phrases like “add the first six nonnegative powers of two.”

By Rick Regan (Copyright © 2008-2012 Exploring Binary)

The Powers of Two