Question

What are the last `3` digits of `7^2175` ?

Answer 1

`943`

Explanation:

First let's look at how the last digit changes with ascending powers of `7`. That is effectively modulo `10` arithmetic and we find successive powers of `7` starting from `0` modulo `10` are:

`1, 7, 9, 3, 1, 7, 9, 3, 1, 7, 9, 3,...`

In particular note that:

`7^4 -= 1" "` modulo `10`

We find:

`7^4 = 2401 -= 1" "` modulo `100`

So now let's look at powers of `401` modulo `1000`:

`1, 401, 801, 201, 601, 1`

So:

`7^20 -= 401^5 -= 1" "` modulo `1000`

So:

`7^2175 = 7^(108*20+15) -= 7^15" "` modulo `1000`

`-= 401^3 * 7^3 -= 201 * 343 -= 943" "` modulo `1000`