what are the last three digits of the number 7^9999?

Answer 1

One way to figure this out is to write the powers of 7, but only the last 3 digits.

7^0 = 001
7^1 = 007
7^2 = 049
7^3 = 343
7^4 = 401
7^5 = 807
...
7^19 = 143
7^20 = 001

So it repeats every 20 powers.

If you take 9999 mod 20 (remainder after dividing by 20) you get 19.

That tells us 7^9999 will have the same last 3 digits as 7^19, namely 143.

Answer 2

7^phi(1000) = 7^400 = 1mod 1000. Then,
7^10000=(7^400)^25 = 1 mod 1000, 7^9999 = 7^(-1)mod1000. Note 1001=7*143 so 7^(-1) = 143. Therefore,
last three digits of 7^9999 are 143.

Answer 3

143