Which one of the following numbers is exactly divisible by (11¹³ + 1)?

The correct option is 11⁵² - 1.

To understand why 11⁵² - 1 is exactly divisible by (11¹³ + 1), we can analyze this problem using algebraic factorization rules.

Let us define a variable:
$$x={11}^{13}$$
Using this substitution, the divisor in the question is:
$${11}^{13}+1=x+1$$
We need to check which of the given options, when expressed in terms of $x$, has $x+1$ as a factor.

Let's write the correct option, 11⁵² - 1, in terms of $x$:
$${11}^{52}-1={\left({11}^{13}\right)}^4-1=x^4-1$$
Using the algebraic identity for the difference of squares, $a^2-b^2=(a-b)(a+b)$, we can factorize $x^4-1$ step-by-step:
$$x^4-1={\left(x^2\right)}^2-1^2$$
$$x^4-1=(x^2-1)(x^2+1)$$
Now, we factorize the term $x^2-1$ further:
$$x^2-1=(x-1)(x+1)$$
Substituting this back, we get the complete factorization:
$$x^4-1=(x-1)(x+1)(x^2+1)$$
Since $(x+1)$ is clearly one of the factors of $x^4-1$, the expression $x^4-1$ is exactly divisible by $x+1$.

Re-substituting $x={11}^{13}$, we have:
$${11}^{52}-1=({11}^{13}-1)({11}^{13}+1)({11}^{26}+1)$$
Thus, 11⁵² - 1 is exactly divisible by (11¹³ + 1).