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 use properties of modular arithmetic and the Chinese Remainder Theorem.

### Step-by-step solution:

**Step 1: Factor 120**
120 can be factored into its prime factors:
\[ 120 = 2^3 \times 3 \times 5 \]

**Step 2: Calculate \( x \mod 8 \)**
First, we find \( x \mod 8 \):
\[ 11 \equiv 3 \mod 8 \]
Thus,
\[ x \equiv 3^{13^{15^{\cdots^{91}}}} \mod 8 \]
Now, we need to find \( 13^{15^{\cdots^{91}}} \mod 2 \) (since \( \phi(8) = 4 \)):
\[ 13 \equiv 1 \mod 2 \]
So,
\[ 13^{15^{\cdots^{91}}} \equiv 1 \mod 2 \]
This means \( 13^{15^{\cdots^{91}}} \) is odd. Therefore, we need to compute:
\[ 3^1 \equiv 3 \mod 8 \]

**Step 3: Calculate \( x \mod 3 \)**
Next, we find \( x \mod 3 \):
\[ 11 \equiv 2 \mod 3 \]
Thus,
\[ x \equiv 2^{13^{15^{\cdots^{91}}}} \mod 3 \]
Now, we need to find \( 13^{15^{\cdots^{91}}} \mod 2 \):
\[ 13 \equiv 1 \mod 2 \]
So,
\[ 13^{15^{\cdots^{91}}} \equiv 1 \mod 2 \]
This means:
\[ 2^1 \equiv 2 \mod 3 \]

**Step 4: Calculate \( x \mod 5 \)**
Now, we find \( x \mod 5 \):
\[ 11 \equiv 1 \mod 5 \]
Thus,
\[ x \equiv 1^{13^{15^{\cdots^{91}}}} \equiv 1 \mod 5 \]

**Step 5: Combine results using the Chinese Remainder Theorem**
Now we have the following system of congruences:
1. \( x \equiv 3 \mod 8 \)
2. \( x \equiv 2 \mod 3 \)
3. \( x \equiv 1 \mod 5 \)

We can solve this step by step:

- From \( x \equiv 1 \mod 5 \), we can write:
\[ x = 5k + 1 \]

- Substituting into \( x \equiv 2 \mod 3 \):
\[ 5k + 1 \equiv 2 \mod 3 \implies 2k + 1 \equiv 2 \mod 3 \implies 2k \equiv 1 \mod 3 \]
The multiplicative inverse of 2 modulo 3 is 2, so:
\[ k \equiv 2 \mod 3 \implies k = 3m + 2 \]
Thus,
\[ x = 5(3m + 2) + 1 = 15m + 10 + 1 = 15m + 11 \]

- Now substituting into \( x \equiv 3 \mod 8 \):
\[ 15m + 11 \equiv 3 \mod 8 \implies 7m + 3 \equiv 3 \mod 8 \implies 7m \equiv 0 \mod 8 \]
Since 7 and 8 are coprime, \( m \equiv 0 \mod 8 \):
\[ m = 8n \]
Thus,
\[ x = 15(8n) + 11 = 120n + 11 \]

**Final Step: Conclusion**
This means:
\[ x \equiv 11 \mod 120 \]
Thus, the remainder when \( x \) is divided by 120 is:
\[ \boxed{11} \]