Answer to Question #231915 in Combinatorics | Number Theory for Alexi

Question

What is the largest integer n which makes n3 + 100 is divisible by n + 10

Answer 1

$$n^3+100=n^3+1000-900$$

$$=(n+10)(n^2-n+100)-900$$

The term $(n+10)(n^2-n+100)$ is divisible by $n+10.$

The Divisors of 900 are as follows:

1, 2, 3, 4, 5, 6, 9, 10, 12, 15, 18, 20, 25, 30, 36, 45, 50, 60, 75, 90, 100, 150, 180, 225, 300, 450, and 900.

We take $n+10=900.$ Then $n=890.$

The largest $n$ which makes $n^3+100$ is divisible by $n+10$ is therefore $890.$