Special Case Laws of Exponents for Powers of Two

In my article “Composing Powers of Two Using The Laws of Exponents” I showed how to combine powers of two using the standard laws of exponents. There are two other rules I use when combining powers of two; I call them the add duplicate power of two rule and the subtract half power of two rule. These are nonstandard rules, applying only to powers of two. Although these are special cases of the existing multiplication and division rules, I’ve found value in recognizing them in addition and subtraction form. I’ll state these rules and show examples of their usage.

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Ending Digits of Powers of Five Form a Binary Tree

In my article “Patterns in the Last Digits of the Positive Powers of Five” I showed that the cycles of ending digits of the positive powers of five could be represented with a binary tree:

Binary Tree Showing Nested Ending Digit Patterns of the Positive Powers of 5
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.

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Counting Binary/Hexadecimal Palindromes

In my article “Counting Binary and Hexadecimal Palindromes” I derived formulas for counting binary palindromes and hexadecimal palindromes. For each type of palindrome, I derived two pairs of formulas: one pair to count n-digit palindromes, and one pair to count palindromes of n digits or less.

In this article, I will derive similar formulas to count binary/hexadecimal palindromes — multi-base palindromes I’ve shown to have an algorithmically defined structure.

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Counting Binary and Hexadecimal Palindromes

How many nonzero, n-digit, decimal number palindromes are there? These two formulas give the answer:

  • When n is even: 9·10n/2-1
  • When n is odd: 9·10(n+1)/2-1

How many nonzero, decimal number palindromes are there, consisting of n-digits or less? These two formulas give the answer:

  • When n is even: 2(10n/2 – 1)
  • When n is odd: 11·10(n-1)/2 – 2

So for example, there are 900 5-digit decimal palindromes, 9,000 8-digit decimal palindromes, 1,098 decimal palindromes of 5 digits or less, and 19,998 decimal palindromes of 8 digits or less.

In this article, I will derive similar formulas to count binary and hexadecimal number palindromes.

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Cycle Length of Powers of Five Mod Powers of Ten

In my article “Patterns in the Last Digits of the Positive Powers of Five” I noted that the positive powers of five modulo 10m cycle with period 2m-2, m ≥ 2, starting at 5m. In this article, I’ll present my proof, which has two parts:

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

The highlight of my proof is in part 1, where I derive a formula to show that the period, or order, of 5 mod 2m is 2m-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 2mdue to a hidden, binary structure I’ve uncovered.

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Seeing Powers of Five in Powers of Two and Vice Versa

The decimal representations of oppositely signed powers of two and powers of five look alike, as seen in these examples: 2-3 = 0.125 and 53 = 125; 5-5 = 0.00032 and 25 = 32. The significant digits in each pair of powers is the same, even though one is a fraction and one is an integer. In other words, a negative power of one base looks like a positive power of the other.

Powers of Two and Powers of Five that Look Alike
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.

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Cycle Length of Powers of Two Mod Powers of Ten

In my article “Patterns in the Last Digits of the Positive Powers of Two” I noted that the positive powers of two modulo 10m cycle with period 4·5m-1, starting at 2m. For example, the powers of two mod 10 cycle with period four: 2, 4, 8, 6, 2, 4, 8, 6, … . In this article, I’ll present my proof, which has two parts:

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

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Nines in Quinary

In my article “Nines in Binary”, I proved the following: positive integers of the form 10n-1, that is, integers made up of n digits of 9s, have binary representations with exactly n digits of trailing 1s. Pat Ballew made a clever observation, adapting my result to prove an equivalent statement for base 5 (quinary): positive integers of the form 10n-1 have quinary representations that have exactly n digits of trailing 4s. For example, 9999 in decimal is 304444 in quinary.

In “Nines in Binary”, I derived an expression for 10n – 1 that shows its structure as a binary number:

10n – 1 = (5n – 1) 2n + (2n – 1)

Pat derived a similar expression for 10n – 1 that shows its structure as a quinary number:

10n – 1 = (2n – 1) 5n + (5n – 1)

In essence, he swapped the 2s and 5s, making it the “dual” of my formula, if you will.

I’ll show the details of the derivation and prove why the formula works.

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Powers Of Two In The Josephus Problem

Pradeep Mutalik of The New York Times recently blogged about a puzzle that is an instance of the Josephus Problem. The problem, restated simply, is this: there are n people standing in a circle, of which you are one. Someone outside the circle goes around clockwise and repeatedly eliminates every other person in the circle, until one person — the winner — remains. Where should you stand so you become the winner?

Here’s an example with 13 participants:

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

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Nines in Binary

I discovered a cool property of positive integers of the form 10n-1, that is, integers made up of n digits of 9s: they have binary representations that have exactly n digits of trailing 1s. For example, 9,999,999 in decimal is 100110001001011001111111 in binary.

The property is interesting in and of itself, but what is more interesting is the process I went through to discover it. It’s a small-scale example of experimental mathematics: I observed something interesting, experimented to collect more data, developed a hypothesis, and constructed a proof.

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Composing Powers of Two Using The Laws of Exponents

Powers of two can be combined, under the laws of exponents, to create other powers of two. Under these rules, you can multiply powers of two, divide powers of two, or raise a power of two to a power and still get another power of two. You can combine these rules to create complicated expressions, expressions that result in a single power of two. For example,

\mbox{\footnotesize{\displaystyle {{\left(\frac{2^4 \cdot 2^3}{2^5}\right)}^3 = \: 2^6}}} .

The laws of exponents apply generally to any base; two is no different. But since we’re interested in powers of two, we’ll couch them in terms of powers of two. Once we explain the laws in this way, you’ll understand the math behind the example above.

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