Final answer:
To find the last three digits of 7^9999, we use modular arithmetic and understand that 7 * 11 * 13 equals 1001. Simplifying (7 * (1000 + 1))^3333 * 7^3 modulo 1000 and recognizing that any multiple of 1000 will vanish, we calculate 7^3 to get 343. Therefore, the last three digits of 7^9999 are 343.
Step-by-step explanation:
To find the last three digits of 79999, we can use modular arithmetic, particularly focusing on the module 1000 as the last three digits of any number are its remainder when divided by 1000. First, let's consider 7 · 11 · 13 = 1001, which is one more than 1000.
Now we can express 79999 as (7 · 1001)3333 · 73, since 9999 = 3333 · 3 + 3. This expression can be simplified since 1001 = 1000 + 1 and we only care about the remainder modulo 1000:
(7 · (1000 + 1))3333 · 73 · 13333 · 73 · 343 (mod 1000)
Because 1000 is congruent to 0 (mod 1000), all the 1000 terms disappear, and we only need to compute 73, which is 343. So the last three digits of 79999 are 343.