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

**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 $a^m\equiv -1(\mathrm{mod}{a}^m+1)$ 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 $11^3\equiv -1(\mathrm{mod}11^3+1)$.

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 $x^3\equiv -1(\mathrm{mod}{x}^3+1)$:
$$x^{39}=(x^3{)}^{13}\equiv (-1{)}^{13}(\mathrm{mod}{x}^3+1)$$

$$x^{39}\equiv -1(\mathrm{mod}{x}^3+1)$$
Now substitute this back into the expression $E$:
$$E=x^{39}-1\equiv (-1)-1(\mathrm{mod}{x}^3+1)$$

$$E\equiv -2(\mathrm{mod}{x}^3+1)$$
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 $x^3\equiv -1(\mathrm{mod}{x}^3+1)$:
$$x^{33}=(x^3{)}^{11}\equiv (-1{)}^{11}(\mathrm{mod}{x}^3+1)$$

$$x^{33}\equiv -1(\mathrm{mod}{x}^3+1)$$
Now substitute this back into the expression $E$:
$$E=x^{33}+1\equiv (-1)+1(\mathrm{mod}{x}^3+1)$$

$$E\equiv 0(\mathrm{mod}{x}^3+1)$$
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 $x^3\equiv -1(\mathrm{mod}{x}^3+1)$:
$$x^{26}=(x^3{)}^8\cdot x^2\equiv (-1{)}^8\cdot x^2(\mathrm{mod}{x}^3+1)$$

$$x^{26}\equiv 1\cdot x^2=x^2(\mathrm{mod}{x}^3+1)$$
Now substitute this back into the expression $E$:
$$E=x^{26}+1\equiv x^2+1(\mathrm{mod}{x}^3+1)$$
Substitute $x=11$:
$$E\equiv 11^2+1(\mathrm{mod}11^3+1)$$

$$E\equiv 121+1=122(\mathrm{mod}11^3+1)$$
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 $x^3\equiv -1(\mathrm{mod}{x}^3+1)$:
$$x^{52}=(x^3{)}^{17}\cdot x^1\equiv (-1{)}^{17}\cdot x^1(\mathrm{mod}{x}^3+1)$$

$$x^{52}\equiv (-1)\cdot x=-x(\mathrm{mod}{x}^3+1)$$
Now substitute this back into the expression $E$:
$$E=x^{52}-1\equiv (-x)-1(\mathrm{mod}{x}^3+1)$$
Substitute $x=11$:
$$E\equiv -11-1(\mathrm{mod}11^3+1)$$

$$E\equiv -12(\mathrm{mod}11^3+1)$$
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 $11^3+1$.