# N^3 + 100

1. Oct 12, 2008

### Trail_Builder

1. The problem statement, all variables and given/known data

Find the largest positive integer n such that n^3 + 100 is divisible by n + 10.

2. Relevant equations

3. The attempt at a solution

The hint ive been given is to use (mod n + 10) to get rid of the n.

but i dont quite see how it would work :S

all my attempts have gotten nowhere, lol.

a little prod in the right direction would be nice :)

cheers

2. Oct 12, 2008

### gabbagabbahey

My method would be to just use any polynomial dividing technique to find the remainder of (n^3+100)/(n+10). If n^3+100 is divisible by n+10, then the remainder will have to be an integer.

3. Oct 13, 2008

### Trail_Builder

right, i think i may have a solution to this problem. can someone please check it for me :) thanks.

say that n^3 + 100 is divisible by n + 10.

then n^3 + 100 = 0 (mod n + 10)

n = -10 (mod n + 10)

so (-10)^3 + 100 = 0 (mod n + 10)

... 0 = 900 (mod n + 10)

we want to maximise n, and because the above line essentially means that 900 is an integer multiple of (n + 10), the maximum n would be when 900 = n +10

so n = 890...

is this reasoning correct?

thanks:)

4. Oct 13, 2008

### gabbagabbahey

Looks good to me I got the same thing using polynomial division.