What is the largest positive integer n for which n^3+100 is divisible by n+1?
By division we find that 𝑛3+100=(𝑛+10)(𝑛2−10𝑛+100)−900n3+100=(n+10)(n2−10n+100)−900.
Therefore, if 𝑛+10n+10 divides 𝑛3+100n3+100, then it must also divide 900. Since we are looking for largest 𝑛n, 𝑛n is maximized whenever 𝑛+10n+10 is, and since the largest divisor of 900 is 900, we must have 𝑛+10=900⇒𝑛=890n+10=900⇒n=890
The largest 𝑛 is therefore 890
ok so the n3 and n2 is just \(n^3 or N^2\)