Question

greatest positive integer m which make m3 +100 divisible by m+10​

Answer 1

By division we find that m3+100=(m+10)(m2−10m+100)−900.

Therefore, if m+10 divides m3+100, then it must also divide 900. Since we are looking for largest m, m is maximized whenever m+10 is, and since the largest divisor of 900 is 900, we must have m+10=900⇒m=890

The largest m is therefore 890