Question

Let N = \( 55^{3} + 17^{3} - 72^{3} \). N is divisible by:

• A. both 7 and 13
• B. both 3 and 13
• C. both 17 and 7
• D. both 3 and 17

Hint:

Notice that \(55 + 17 = 72\). Use the identity for \(a^3 + b^3 - (a+b)^3\).

Answer 1

The Correct Option is D

Step 1: Spot the pattern:
With \(a = 55\) and \(b = 17\) we have \(a + b = 72\). So \(N = a^3 + b^3 - (a+b)^3\).

Step 2: Expand:
\((a+b)^3 = a^3 + 3a^2 b + 3ab^2 + b^3\), so \(N = -3a^2 b - 3ab^2 = -3ab(a+b)\).

Step 3: Substitute:
\(N = -3 \times 55 \times 17 \times 72 = -201960\). Its prime factorisation is \(2^3 \times 3^3 \times 5 \times 11 \times 17\) (with a minus sign).

Step 4: Check the options:
The factors 3 and 17 both appear, so N is divisible by both 3 and 17. The primes 7 and 13 do not appear (for example \(201960 = 7 \times 28851 + 3\), not a multiple of 7). So options (1), (2), (3) are wrong.

Final Answer:
The correct option is (4): both 3 and 17