25 how many times?

From $25^1 = 25$ , $25^2 = 625$ , and $25^3 = 15625$ , it would appear that the last two digits $25^n$ are $25$ for any positive integer $n$ .

This can be proved inductively:

The last two digits of the first case, $25^1 = 25$ , are $25$ .

Assuming the last two digits of $25^n$ are $25$ , then:

$25^n = 100x + 25$ (for some integer $x$ )

$25 \cdot (25^n) = 25(100x + 25)$

$25^{n + 1} = 25(100x + 25)$

$25^{n + 1} = 2500x + 625$

$25^{n + 1} = 2500x + 600 + 25$

$25^{n + 1} = 100(25x + 6) + 25$

$25^{n + 1} = 100x' + 25$ (for some integer $x' = 25x + 6$ )

the last two digits of $25^{n + 1}$ are also $25$ .

Therefore, the last two digits of the given number are $\boxed{25}$ .

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Bonus: $25$ is another automorphic number .

The number $25$ is automorphic that is its powers $25^n$ always end with itself. That is $25^n \equiv 25 \text{ (mod 100)}$ . We can prove this using Chinese remainder theorem as follows:

$25^n \equiv (24+1)^n \equiv 1^n \equiv 1 \text{ (mod 4)}$ and $25^n \equiv 0 \text{ (mod 25)}$ . Therefore, $25^n = 25m$ , where $m$ is an integer. Then $25m \equiv 1 \text{ (mod 4)} \implies m \equiv 1 \implies 25^n \equiv \boxed{25} \text{ (mod 100)}$ .

Quick explanation 1 sentence

5 to the power of any integer that is bigger than 2 ends in 25 so 25 to the power of any integer will end in 25

25 has a two-digit power cycle of 1. That means $25^1, 25^2, 25^3 \ldots 25^k$ all end with the last two digits as 25. So regardless of the exponent in 25, it will end with the last two digits as $\boxed{25}$