Question

The expression $N = 55^3 + 17^3 - 72^3$ is exactly divisible by:

• A. 7 \& 13
• B. 3 \& 13
• C. 17 \& 7
• D. 3 \& 17

Hint:

Use modular arithmetic to check divisibility quickly without full expansion.

Answer 1

The Correct Option is D

Recognize: $55 + 17 = 72$, so $55^3 + 17^3 - 72^3$ = $(55^3 + 17^3) - (72^3)$. From sum of cubes: $p^3+q^3 = (p+q)(p^2 - pq + q^2)$. Here $p+q=72$: $55^3+17^3 = 72 \times (55^2 - 55\times 17 + 17^2)$. Thus: $N = 72 \times K - 72^3 = 72(K - 72^2)$. Clearly divisible by $72 = 3 \times 24$, so divisible by $3$. Check mod $17$: $55 \equiv 4, 17 \equiv 0, 72 \equiv 4 \ (\text{mod }17)$. $55^3 \equiv 4^3 = 64 \equiv 13$, $17^3 \equiv 0$, $72^3 \equiv 13$, so $N \equiv 13+0-13 \equiv 0$ mod 17. Hence divisible by $3$ and $17$. \[ \boxed{\text{3 and 17}} \]