Understanding the Divisor: $11^3 + 1$

The problem asks us to identify which number among the options is exactly divisible by the expression $11^3 + 1$. First, let's calculate the value of the divisor.

We calculate $11^3$: $$11^3 = 11 \times 11 \times 11 = 121 \times 11 = 1331$$

Therefore, the divisor is: $$11^3 + 1 = 1331 + 1 = 1332$$

Our task is to find which of the given options is exactly divisible by $1332$.

Number Divisibility Rules with Exponents

We can determine divisibility using properties of exponents and modular arithmetic. A useful technique is to analyze the expression modulo the divisor. The property $a^m \equiv -1 \pmod{a^m + 1}$ is particularly helpful.

Let $a = 11$ and $m = 3$. Our divisor is $a^3 + 1$. We will check each option by finding its remainder when divided by $11^3 + 1$. This is equivalent to evaluating the expression modulo $11^3 + 1$, using the fact that $11^3 \equiv -1 \pmod{11^3 + 1}$.

Analyzing Each Option

Option 1: Check Divisibility of $11^{39} - 1$

Let the expression be $E = 11^{39} - 1$. We want to find the remainder of $E$ when divided by $11^3 + 1$. Let $x = 11$. The expression becomes $x^{39} - 1$. We can rewrite the exponent $39$ as $3 \times 13$. So, $x^{39} = (x^3)^{13}$. Now, we use the modular property $x^3 \equiv -1 \pmod{x^3 + 1}$: $$x^{39} = (x^3)^{13} \equiv (-1)^{13} \pmod{x^3 + 1}$$ $$x^{39} \equiv -1 \pmod{x^3 + 1}$$ Now substitute this back into the expression $E$: $$E = x^{39} - 1 \equiv (-1) - 1 \pmod{x^3 + 1}$$ $$E \equiv -2 \pmod{x^3 + 1}$$ Since the remainder is $-2$ (not $0$), $11^{39} - 1$ is not exactly divisible by $11^3 + 1$.

Option 2: Check Divisibility of $11^{33} + 1$

Let the expression be $E = 11^{33} + 1$. We want to find the remainder of $E$ when divided by $11^3 + 1$. Let $x = 11$. The expression becomes $x^{33} + 1$. We can rewrite the exponent $33$ as $3 \times 11$. So, $x^{33} = (x^3)^{11}$. Using the property $x^3 \equiv -1 \pmod{x^3 + 1}$: $$x^{33} = (x^3)^{11} \equiv (-1)^{11} \pmod{x^3 + 1}$$ $$x^{33} \equiv -1 \pmod{x^3 + 1}$$ Now substitute this back into the expression $E$: $$E = x^{33} + 1 \equiv (-1) + 1 \pmod{x^3 + 1}$$ $$E \equiv 0 \pmod{x^3 + 1}$$ Since the remainder is $0$, $11^{33} + 1$ is exactly divisible by $11^3 + 1$.

Option 3: Check Divisibility of $11^{26} + 1$

Let the expression be $E = 11^{26} + 1$. We want to find the remainder of $E$ when divided by $11^3 + 1$. Let $x = 11$. The expression becomes $x^{26} + 1$. We can rewrite the exponent $26$ in terms of $3$: $26 = 3 \times 8 + 2$. So, $x^{26} = x^{3 \times 8 + 2} = (x^3)^8 \cdot x^2$. Using the property $x^3 \equiv -1 \pmod{x^3 + 1}$: $$x^{26} = (x^3)^8 \cdot x^2 \equiv (-1)^8 \cdot x^2 \pmod{x^3 + 1}$$ $$x^{26} \equiv 1 \cdot x^2 = x^2 \pmod{x^3 + 1}$$ Now substitute this back into the expression $E$: $$E = x^{26} + 1 \equiv x^2 + 1 \pmod{x^3 + 1}$$ Substitute $x = 11$: $$E \equiv 11^2 + 1 \pmod{11^3 + 1}$$ $$E \equiv 121 + 1 = 122 \pmod{11^3 + 1}$$ Since the remainder is $122$ (not $0$), $11^{26} + 1$ is not exactly divisible by $11^3 + 1$.

Option 4: Check Divisibility of $11^{52} - 1$

Let the expression be $E = 11^{52} - 1$. We want to find the remainder of $E$ when divided by $11^3 + 1$. Let $x = 11$. The expression becomes $x^{52} - 1$. We can rewrite the exponent $52$ in terms of $3$: $52 = 3 \times 17 + 1$. So, $x^{52} = x^{3 \times 17 + 1} = (x^3)^{17} \cdot x^1$. Using the property $x^3 \equiv -1 \pmod{x^3 + 1}$: $$x^{52} = (x^3)^{17} \cdot x^1 \equiv (-1)^{17} \cdot x^1 \pmod{x^3 + 1}$$ $$x^{52} \equiv (-1) \cdot x = -x \pmod{x^3 + 1}$$ Now substitute this back into the expression $E$: $$E = x^{52} - 1 \equiv (-x) - 1 \pmod{x^3 + 1}$$ Substitute $x = 11$: $$E \equiv -11 - 1 \pmod{11^3 + 1}$$ $$E \equiv -12 \pmod{11^3 + 1}$$ Since the remainder is $-12$ (not $0$), $11^{52} - 1$ is not exactly divisible by $11^3 + 1$.

Conclusion Summary

By applying modular arithmetic and exponent properties, we found that only $11^{33} + 1$ yields a remainder of $0$ when divided by $11^3 + 1$. Therefore, $11^{33} + 1$ is the number that is exactly divisible by $11^3 + 1$.