• #### For what integers N is ( N^3 + 100) divisible by (N+10) ?

From henhanna@gmail.com@21:1/5 to All on Sat Jul 2 08:15:40 2022
For what integers N is ( N^3 + 100) divisible by (N+10) ?

---------- pls wait a few days before posting answers or hints.

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• From Edward Murphy@21:1/5 to henh...@gmail.com on Sun Jul 3 13:52:54 2022
On 7/2/2022 8:15 AM, henh...@gmail.com wrote:
For what integers N is ( N^3 + 100) divisible by (N+10) ?

---------- pls wait a few days before posting answers or hints.

Note that his "please wait a few days" business is going against
favors "please use spoiler space and/or rot13"; the assumptions being
that (a) people who just don't want to accidentally see the solution
still have some protection, and (b) few if any people will bother to
read someone else's answer and then try to pass it off as their own.

Here's some spoiler space before my derivation of the answer.

Note that
N^3 + 100 = (N + 10) * (N^2 - 10N + 100) - 900
and let C = N^2 - 10N + 100

If N^3 + 100 is divisible by N + 10, then
N^3 + 100 = (N + 10) * K for some integer K

Now we can do the following:
(N + 10) * C - 900 = (N + 10) * K
(N + 10) * C = (N + 10) * K + 900
(N + 10) * (C - K) = 900
and note that C - K is an integer, so 900 is divisible by N + 10.

900's prime factorization is 2^2 * 3^2 * 5^2, so its factors are
1, 2, 4, 3, 6, 12, 9, 18, 36,
5, 10, 20, 15, 30, 60, 45, 90, 180,
25, 50, 100, 75, 150, 300, 225, 450, 900
and their additive inverses. Any value of N that is 10 less than
one of those factors should work, so that's 54 distinct solutions.

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