[Guest]

(I) Determine the last three digits in the base ten expansion of 2004²⁰⁰⁴.

(II) Determine the last three digits in the base ten expansion of 1996¹⁹⁹⁶.

For an extra challenge, solve these without the aid of a computer program.

[superprismatic]

The answer to (I) is 256.

Method:

We can use the binomial expansion of (2000+4)²⁰⁰⁴ which equals 2004²⁰⁰⁴ to get the answer.

All we need do is realize that any term in the expansion which has a non-zero power of 2000 has

no effect on the last three digits. The only other term is just 4²⁰⁰⁴. So this term

alone will determine the last three digits. Now, 4²⁰⁰⁴=256⁵⁰¹. Using

a calculator, we can easily see that the last three digits of powers of 256 have a 25-long cycle.

Here's the cycle beginning 256¹, 256², 256³, ... :

256 536 216 296 776 656 936 616 696 176 56 336 16 96 576 456 736 416 496 976 856 136 816 896 376.

Since this is a 25-long cycle, 256¹=256²⁶=256⁵¹=...=256⁵⁰¹. Voila! 256 is our answer.

The answer to (II) is 896.

Method:

Similar to the above using the binomial expansion of (2000+(-4))¹⁹⁹⁶. Here, we use the fact that

(-4)¹⁹⁹⁶=256⁴⁹⁹. Then, using the cycle above, we find that 256⁴⁹⁹ has last three

digits 896.