Find the largest positive integer n such that n3 + 100, is divisible by n + 10.

Find the largest positive integer n such that n³ + 100, is divisible by n + 10.

A 890

B 920

C 990

D 940

E None of these

Correct Answer: Option A

We know that, a³ + b³ = (a + b) (a² –ab + b²)

∴ n³ + 10³ = n³ + 1000 = (n + 10) (n² – 10n + 100)

Thus, (n + 10) divides (n³ + 1000).

Given equation,

n³ + 100 = (n³ + 1000) – 900

Since, (n + 10) divides (n³ + 100) and (n³ + 1000), it must divide 900 also.

So, largest value of (n + 10) = 900

∴ n = 890