Solving Divisibility: $11^{13} + 1$

The question asks us to identify which number from the given options is exactly divisible by $11^{13} + 1$. We can solve this using algebraic properties and modular arithmetic.

Identify the Divisor and Substitute

1. Let the divisor be $D = 11^{13} + 1$.
2. To simplify, let $x = 11^{13}$. Then the divisor $D$ can be written as $x+1$.
3. We will now check each option for divisibility by $x+1$. We use the property that $x \equiv -1 \pmod{x+1}$.

Analyze Each Option

1. Option 1: $11^{26}+1$

   Rewrite this in terms of $x$: $11^{26}+1 = (11^{13})^2 + 1 = x^2+1$. Now, check the remainder when divided by $x+1$: $x^2+1 \pmod{x+1}$ Since $x \equiv -1 \pmod{x+1}$, substitute $x=-1$: $(-1)^2+1 \equiv 1+1 \equiv 2 \pmod{x+1}$. The remainder is 2, so $11^{26}+1$ is not divisible by $11^{13}+1$.

2. Option 2: $11^{33}+1$

   Rewrite this in terms of $x=11^{13}$: $11^{33}+1 = 11^{2 \times 13 + 7} + 1 = (11^{13})^2 \times 11^7 + 1 = x^2 \times 11^7 + 1$. Check the remainder when divided by $x+1$: $x^2 \times 11^7 + 1 \pmod{x+1}$ Substitute $x \equiv -1 \pmod{x+1}$: $(-1)^2 \times 11^7 + 1 \equiv 1 \times 11^7 + 1 \equiv 11^7+1 \pmod{x+1}$. Since $11^7+1$ is not 0, this option is not divisible by $11^{13}+1$.

3. Option 3: $11^{39}-1$

   Rewrite this in terms of $x$: $11^{39}-1 = (11^{13})^3 - 1 = x^3-1$. Check the remainder when divided by $x+1$: $x^3-1 \pmod{x+1}$ Substitute $x \equiv -1 \pmod{x+1}$: $(-1)^3-1 \equiv -1-1 \equiv -2 \pmod{x+1}$. The remainder is -2, so $11^{39}-1$ is not divisible by $11^{13}+1$.

4. Option 4: $11^{52}-1$

   Rewrite this in terms of $x$: $11^{52}-1 = (11^{13})^4 - 1 = x^4-1$. Check the remainder when divided by $x+1$: $x^4-1 \pmod{x+1}$ Substitute $x \equiv -1 \pmod{x+1}$: $(-1)^4-1 \equiv 1-1 \equiv 0 \pmod{x+1}$. The remainder is 0. Therefore, $11^{52}-1$ is exactly divisible by $11^{13}+1$.

Conclusion

Based on the analysis, the number $11^{52}-1$ is exactly divisible by $11^{13}+1$.