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

The correct answer is Option 4: 11⁵² - 1.

Step-by-step Explanation:

We are asked to find which of the given expressions is exactly divisible by ${11}^{13}+1$.

Let us recall the algebraic expansion identity for the difference of two powers:

$x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+...+y^{n-1})$

More specifically, when $n$ is an even integer, say $n=2k$, we can write:

$a^{2k}-b^2=(a^k{)}^2-(b^k{)}^2=(a^k+b^k)(a^k-b^k)$

This shows that for any even power $n$, the expression $x^n-1$ is always divisible by $x^{n/2}+1$.

Let $x={11}^{13}$. We want to check which option contains $x+1={11}^{13}+1$ as a factor.

Consider the fourth option: ${11}^{52}-1$.

Notice that $52=13\times 4$. We can rewrite ${11}^{52}-1$ using exponent rules as:

${11}^{52}-1=({11}^{26}{)}^2-1^2$

Applying the difference of squares formula, $A^2-B^2=(A+B)(A-B)$:

${11}^{52}-1=({11}^{26}+1)({11}^{26}-1)$

Now factor the term ${11}^{26}-1$ further as a difference of squares:

${11}^{26}-1=({11}^{13}{)}^2-1^2=({11}^{13}+1\left)\right({11}^{13}-1)$

Substituting this back into the expression for ${11}^{52}-1$, we get:

${11}^{52}-1=({11}^{26}+1)({11}^{13}+1)({11}^{13}-1)$

Since $({11}^{13}+1)$ is explicitly one of the factors of ${11}^{52}-1$, it follows that ${11}^{52}-1$ is completely and exactly divisible by $({11}^{13}+1)$.