Question

Which one of the following numbers is exactly divisible by $\left(11^{13}+1\right)$?

• A. $11^{26}+1$
• B. $11^{33}+1$
• C. $11^{39}-1$
• D. $11^{52}-1$

Hint:

Use exponent divisibility: $a^m - 1$ is divisible by both $(a^k-1)$ and $(a^k+1)$ if $m = 2k$.

Answer 1

The Correct Option is D

We need to determine which expression is divisible by: \[ 11^{13} + 1 \] This is a classic number theory pattern based on factorization identities: \[ a^m - 1 \text{ is divisible by } a^n - 1 \text{ if } n \mid m \] and \[ a^{2k} - 1 = (a^k - 1)(a^k + 1) \] Since the expression to divide is $11^{13}+1$, notice: \[ 11^{26} - 1 = (11^{13}-1)(11^{13}+1) \] Thus, anything of the form $11^{26k} - 1$ is divisible by $(11^{13}+1)$. Check option (D): \[ 11^{52}-1 = 11^{4\times 13} - 1 \] Since 13 divides 52, \[ 11^{52} - 1 \text{ is divisible by } 11^{13} - 1 \text{ and also by } 11^{13} + 1 \] Thus option (D) is exactly divisible by $11^{13} + 1$. Other options fail because: - (A) is $11^{26}+1$ → not divisible
- (B) $11^{33}+1$ wrong sign
- (C) $11^{39}-1$, exponent 39 not divisible by 26 or 13
Hence, only Option (D) is correct.