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The first few powers of $5$ are given by:

\begin{array}{r}

5 \\

25 \\

125 \\

625 \\

3125 \\

15625\\

78125\\

390625\\

1953125\\

9765625\\

48828125\\

244140625\\

1220703125\\

6103515625\\

\end{array}

We can see that the last digit is always $5$ and the second to last digit is always $2$. The preceding digit cycles between $1$ and $6$, and the one before that between $3,5,8$ and $0$. We can continue:

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\begin{array}{ll}

\text{digit} & \text{period} \\

1 & 5\\

2 & 2\\

3 & 16\\

4 & 3580\\

5 & 17956240\\

6 & 3978175584236200

\end{array}

Apart from the first 2 $(5$ and $2)$ we see that all these periods appear to be congruent to 7 modulo 9.

Can this be proven?

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Write $a\%n$ for the least non-negative integer congruent to $a$ modulo $n$, i.e. $a\%n=r$ iff $a\equiv r\pmod{n}$ and $0\le r<n$.

In particular, $a\% 10^k$ is the number represented by the last $k$ digits of $a$.

Note that the cycle of the last four digits contains $625,3125,5625,8125$, which are $1,5,9,13$ times $5^4$. Similarly the cycle of the last five digits contains

$$

\begin{array}{ll}

3125 = 1\times 5^5 & 53125 = 17\times 5^5 \\

15625 = 5\times 5^5 & 65625 = 21\times 5^5 \\

28125 = 9\times 5^5 & 78125 = 25\times 5^5 \\

40625 = 13\times 5^5 & 90625 = 29\times 5^5

\end{array}

$$

In general, these cycles have the following properties for $k>1$:

**P1**: $5^n\%10^k$ repeats in a cycle of length $2^{k-2}$ for $n\ge k$.

**P2**: These sets, which we will denote by $S_k$, are the same, formalizing the pattern alluded to above:

$$S_k = \left\{5^n\% 10^k\right\}_{n=k}^{k+2^{k-2}-1}

= \left\{(4n+1)5^k\right\}_{n=0}^{2^{k-2}-1}$$

Let $$A_k = \sum_{x\in S_k} x \\

B_k = \sum_{x\in S_k} \left\lfloor \frac{x}{10^{k-1}}\right\rfloor = \sum_{x\in S_k} \frac{x – (x\%10^{k-1})}{10^{k-1}}

$$

Then $B_k$ is the sum of the digits in the number called the “period” in the question, which is congruent to the number modulo $9$.

From **P1**, for $k>2$ one cycle of the last $k$ digits contains two cycles of the last $k-1$ digits, so

$$

A_k = 10^{k-1} B_k + 2 A_{k-1} \\

B_k \equiv A_k-2A_{k-1} \pmod {9}

$$

that is, we can break down $A_k$ into two sums of the last $k-1$ digits and a sum of the first digits times $10^{k-1}$.

From **P2** we can evaluate $A_k$:

$$

\begin{align}

A_k & = \sum_{n=0}^{2^{k-2}-1}(4n+1)5^k \\

& = 5^k\left[ \left(4\sum_{n=1}^{2^{k-2}-1} n\right) + \sum_{n=0}^{2^{k-2}-1} 1 \right] \\

& = 5^k\left(4 \frac{2^{k-2}(2^{k-2}-1)}{2} + 2^{k-2}\right) \\

& = 25\cdot 10^{k-2} \left(2^{k-1}-1\right) \\

& \equiv 7 \left(2^{k-1}-1\right) \pmod{9}

\end{align}

$$

from which it follows that

$$

\begin{align}

B_k & \equiv A_k-2A_{k-1} \\

& \equiv 7 (2^{k-1}-1) – 2\cdot 7(2^{k-2}-1) \\

& \equiv 7 \pmod {9}

\end{align}

$$

establishing the desired result.

Now to fill in the blanks we’ll prove **P1** and **P2**.

First $2^n\not\mid n!$. The power of $2$ in $n!$ is $v_2(n!)=\lfloor n/2\rfloor + \lfloor n/4 \rfloor + \lfloor n/8 \rfloor + \cdots<n$ (see e.g. this).

Hence $$2^n \left\vert \binom{2^n}{i} 2^i \right. = \frac{2^n C}{i!} 2^{i} $$

for some $C\in \mathbb{Z}$.

Then using the binomial theorem

$$

5^{2^{k-3}} = (1+4)^{2^{k-3}} = 1+2^{k-3}\cdot 4 + \sum_{i=2}^{2^{k-3}}\binom{2^{k-3}}{i} 4^i \equiv 2^{k-1}+1 \pmod {2^k} \\

5^{2^{k-2}} = (1+4)^{2^{k-2}} = 1 + \sum_{i=1}^{2^{k-2}}\binom{2^{k-2}}{i} 4^i \equiv 1 \pmod {2^k}

$$

which establishes that the order of $5$ in $\mathbb{Z}/2^k\mathbb{Z}^{\times}$ is $2^{k-2}$ (since it must also divide $2^{k-1}$).

Hence for $1<k\le n < k+2^{k-2}$, we always have $5^n\%5^k=0$, and $5^n\%2^k$ takes on exactly $2^{k-2}$ distinct values; by the Chinese Remainder Theorem $5^n\%10^k$ also takes on exactly $2^{k-2}$ distinct values. This establishes **P1**.

Furthermore for $1<k\le n < k+2^{k-2}$ every $5^n\equiv (4j+1)5^k\pmod{10^k}$ for some $j$, and there are exactly $2^{k-2}$ such possibilities, so every one must occur for exactly one choice of $n$ in this range. This establishes **P2**.

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