Question

It is given that $(2^{32} + 1)$ is completely divisible by a whole number $w$. Which of the following numbers is completely divisible by this number $w$?

• A. $(2^{16} + 1)$
• B. $(2^{16} - 1)$
• C. $(7 \times 2^{23})$
• D. $(2^{96} + 1)$

Hint:

Use $a^3+1=(a+1)(a^2-a+1)$ when powers are multiples of 3.

Answer 1

The Correct Option is D

Concept: Use identity: \[ a^3 + 1 = (a+1)(a^2 - a + 1) \]
Step 1: Rewrite expression.
\[ 2^{96} + 1 = (2^{32})^3 + 1 \]
Step 2: Apply identity.
\[ (2^{32})^3 + 1 = (2^{32} + 1)(2^{64} - 2^{32} + 1) \]
Step 3: Conclusion.
\[ (2^{96} + 1) { is divisible by } (2^{32} + 1) \]
Hence, it is divisible by $w$.