Find the remainder when `x=11^(13^(15…….^(91)))` divided by 120.

Options

A. `1`
B. `0`
C. `119`
D. `11`

Explanation

To find the remainder when \( x = 11^{13^{15^{\cdots^{91}}}} \) is divided by 120, we can follow these steps:

### Step 1: Simplify the exponent
We start with \( x = 11^{13^{15^{\cdots^{91}}}} \). The exponent is a tower of powers, which we will denote as \( n = 13^{15^{\cdots^{91}}} \).

### Step 2: Use properties of modular arithmetic
We will use the fact that \( 11^{\phi(120)} \equiv 1 \mod 120 \), where \( \phi \) is the Euler's totient function. First, we need to calculate \( \phi(120) \).

### Step 3: Calculate \( \phi(120) \)
The prime factorization of 120 is:
\[ 120 = 2^3 \times 3^1 \times 5^1 \]
Using the formula for the totient function:
\[ \phi(120) = 120 \left(1 - \frac{1}{2}\right)\left(1 - \frac{1}{3}\right)\left(1 - \frac{1}{5}\right) \]
Calculating this:
\[ \phi(120) = 120 \times \frac{1}{2} \times \frac{2}{3} \times \frac{4}{5} = 120 \times \frac{8}{30} = 32 \]

### Step 4: Reduce the exponent modulo \( \phi(120) \)
Now we need to find \( n \mod 32 \). We will simplify \( n = 13^{15^{\cdots^{91}}} \mod 32 \).

### Step 5: Calculate \( 13 \mod 32 \)
First, we find \( 13 \mod 32 \), which is simply 13.

### Step 6: Calculate \( 15^{\cdots^{91}} \mod 16 \)
Next, we need to reduce the exponent \( 15^{17^{\cdots^{91}}} \mod 16 \) because \( \phi(32) = 16 \).

### Step 7: Calculate \( 15 \mod 16 \)
Since \( 15 \equiv -1 \mod 16 \), we need to evaluate the exponent \( 17^{19^{\cdots^{91}}} \mod 2 \).

### Step 8: Calculate \( 17^{19^{\cdots^{91}}} \mod 2 \)
Since any odd number raised to any power remains odd, we have \( 17^{19^{\cdots^{91}}} \equiv 1 \mod 2 \).

### Step 9: Determine the result of \( 15^{\text{odd}} \)
Thus, \( 15^{17^{\cdots^{91}}} \equiv 15 \mod 16 \).

### Step 10: Calculate \( n \mod 32 \)
Now we have \( n \equiv 13^{15} \mod 32 \). We can calculate \( 13^{15} \mod 32 \) using successive squaring:
- \( 13^1 \equiv 13 \)
- \( 13^2 \equiv 169 \equiv 9 \mod 32 \)
- \( 13^4 \equiv 9^2 \equiv 81 \equiv 17 \mod 32 \)
- \( 13^8 \equiv 17^2 \equiv 289 \equiv 1 \mod 32 \)

Thus, \( 13^{15} = 13^8 \cdot 13^4 \cdot 13^2 \cdot 13^1 \equiv 1 \cdot 17 \cdot 9 \cdot 13 \mod 32 \).

Calculating this:
\[ 17 \cdot 9 = 153 \equiv 25 \mod 32 \]
\[ 25 \cdot 13 = 325 \equiv 5 \mod 32 \]

### Step 11: Calculate \( 11^n \mod 120 \)
Now we have \( n \equiv 5 \mod 32 \), so we need to calculate \( 11^5 \mod 120 \).

Calculating \( 11^5 \):
\[ 11^2 = 121 \equiv 1 \mod 120 \]
Thus,
\[ 11^4 \equiv 1^2 \equiv 1 \mod 120 \]
\[ 11^5 \equiv 11^4 \cdot 11 \equiv 1 \cdot 11 \equiv 11 \mod 120 \]

### Final Result
The remainder when \( x = 11^{13^{15^{\cdots^{91}}}} \) is divided by 120 is \( \boxed{11} \).