# Cycle Length of Powers of Two Mod Powers of Ten

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In my article “Patterns in the Last Digits of the Positive Powers of Two” I noted that the positive powers of two modulo 10^{m} cycle with period 4·5^{m-1}, starting at 2^{m}. 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 5**cycle with period 4·5^{m}^{m-1},**starting at 2**.^{0} - Part 2 shows that the powers of two
**mod 10**cycle with the same period as the powers of two mod 5^{m}^{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 2^{0}; 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 g_{1} = g^{p-1} mod p^{2}. If g_{1} ≠ 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 2^{0}

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 2^{0}.

## 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} – 2^{i} is a multiple of 5^{m}; that is, 2^{i+p} – 2^{i} = n·5^{m}, for some nonnegative integer n.

Factoring out 2^{i} we get 2^{i}(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 2^{i}(2^{p} – 1) = n’·2^{m}5^{m} = n’·10^{m}.

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

2^{i}(2^{p} – 1) = 2^{i+p} – 2^{i} = 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} – 2^{i} = n·10^{m} = n·2^{m}5^{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 5^{20}, 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 2
^{n}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 10
^{10}. - 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 2
^{n}.” - “They get into a loop of length 4·5
^{m-1}. (Because 2 is a primitive root of powers of 5.)”

- “After the nth, they are all multiples of 2
- Question I asked at mathoverflow.net.

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