N = 823 − 623 − 203 is divisible by

Concept: if three numbers a, b, c satisfy a + b + c = 0, then a3+ b3+ c3 = 3abc. This identity collapses a sum/difference of cubes into a simple product, which is far easier to factor for a divisibility check than expanding each cube directly.

Application:

1. Let a = 82, b = −62, c = −20, so that a3+ b3+ c3 = 823+ (−62)3+ (−20)3 = 823 − 623 − 203 = N.

2. Check the sum of the bases: a + b + c = 82 − 62 − 20 = 0, so the identity applies.

3. Substitute into a3+ b3+ c3 = 3abc: N = 3 × 82 × (−62) × (−20) = 3 × 82 × 62 × 20.

4. Factor the bases: 82 = 2 × 41 and 62 = 2 × 31, so N = 3 × (2 × 41) × (2 × 31) × 20.

5. N therefore carries both 41 and 31 as factors (along with 2, 3, 5, and others), so N is divisible by both 31 and 41.

Cross-check: direct computation gives N = 305,040. Dividing this out confirms the result — 305,040 ÷ 31 = 9,840 and 305,040 ÷ 41 = 7,440 exactly, while it leaves a remainder when divided by 13, 67, 17, or 7. So 31 and 41 is the only pair among the given options that actually divides N.