what

There are infinitively many powers of 2 (pigeons), and only 2001 remainders when these numbers are divided by 2001 (pigeon holes). That’s why there are two powers of 2 that give the same remainder when divided by 2001. The difference of these powers is a multiple of 2001.
We are looking for 3k which ends with 001. This is the same as 3k - 1 being a multiple of 1000. Similarly to problem 1 we can prove that there are two powers of 3 that differ by a multiple of 1000. Let’s call these powers 3n and 3m. Factor their difference into diet pill
3m is relatively prime with 1000 (it is neither a multiple of 2 nor a multiple of 5), so 3n-m - 1 must be a multiple of 1000.
This problem is simular