The Power of 5

Lets take a look at the first few powers of $5\mod1000$ to establish its pattern.

$005, 025, 125, 625, \dots125, \dots625, \dots125, \dots625, \dots$

We are only looking at the last 3 digits because the smallest 4 digit number $1000*5=5000\mod1000=0$ . Meaning no digit 4 or more can affect the last 3 digits.

Checking what happens after $625$ , $625*5=3125\mod1000=125$ . Meaning every number following a $\dots625$ will always end in $125$ . And we can see that $125*5=625$ follows the same logic.

Since the ending digits always alternate between $125$ and $625$ , we just need to establish the pattern. Since $125$ happens on the 3rd power, and $625$ appears on the 4th power, and these endings always alternate, ignoring $5^1$ and $5^2$ , the pattern is odd powers end in $125$ and even powers end in $625$ .

So, since all powers of $5$ are odd, this "simplifies" to $5^{odd}$ meaning it ends in $125$ .

We have that $5^5 = 3125$ , which is congruent to $125 \mod 1000$ . Now we need $125^{3125} \mod 1000$ , which is congruent to $(125^4)^5 * 125 \mod 1000$ . $125^4 = 244140625$ , which is then congruent to $625 \mod 1000$ . Then we have $625^5 *125 \mod 1000$ , again is congruent to $(625^2)^2 * 625 *125 \mod 1000$ . $625^2$ is congruent to $625 \mod 1000$ . Then we finally need to solve $625 *125 \mod 1000$ , which is $78125 \mod 1000$ . It is then evident that the last three digits is "125". This method uses the fact that representatives of a single equivalence class of the integers modulo n are equivalent when doing computations in the quotient group Z/nZ.

5^5^5^5 = 10^10^2184.125722088846 = ....203125