Question

What is the largest value of $n$ less than 100,000 for which the expression $8(n-2)^5-n^2+14n-24$ is a multiple of 5?

Answer 1

The largest value is 99,997

Explanation:

We can see the congruence of the expression module 5 according to the congruence of n module 5. First, we simplify the expression module 5. 8 is replaced by 3, 14 n can be replaced by 4n (10n is 0 module 5), and -24 is replaced by 1. With this, the expression is congruent module 5 to the following one

`3(n-2)^5-n^2+4n+1`

Also, due to Fermat Theorem, if n-2 is not a multiple of 5, then (n-2)⁴ is congruent to 1 module 5. In any case, (n-2)⁵ is congruent to (n-2) module 5 (even if n-2 is a multiple of 5). So we can replace (n-2)⁵ with (n-2) in the expression.

`3(n-2)-n^2+4n+1 \equiv -n^2+7n -5 \equiv -n^2 +2n \mod (5)`

Now, lets see the congruences

• n congruent to 0 module 5:
  `-n^2 +2n \equiv 0 \mod (5)`
• n congruent to 1 module 5:
  `-n^2 +2n \equiv -1+2 \equiv 1 \mod(5)`
• n congruent to 2 module 5:
  `-n^2 +2n \equiv -4+4 \equiv 0 \mod (5)`
• n congruent to 3 module 5:
  `-n^2 + 2n \equiv -9+6 \equiv 2 \mod (5)`
• n congruent to 4 module 5:
  `-n^2 + 2n \equiv -16+8 \equiv 2 \mod (5)`

So, only for 0 and 2 the expression is a multiple of 5. As a result, the largest value less than 100,000 that is a multiple of 5 is 99,997, obtained by picking the largest number less than 100,000 that is either congruent to 0 or congruent to 2 module 5.