The tens digit of the `2^(248)` is :

Options

A. 1
B. 7
C. 5
D. none of these

Answer

To find the tens digit of \(2^{248}\), we need to determine the last two digits of \(2^{248}\). This can be done by calculating \(2^{248} \mod 100\).

### Step-by-Step Solution:

1. **Divide the problem**: We will find \(2^{248} \mod 100\). Since \(100 = 4 \times 25\), we can use the Chinese Remainder Theorem (CRT) after finding \(2^{248} \mod 4\) and \(2^{248} \mod 25\).

2. **Calculate \(2^{248} \mod 4\)**:
\[ 2^{248} \mod 4 = 0 \]
(since any power of 2 greater than or equal to \(2^2\) is divisible by 4).

3. **Calculate \(2^{248} \mod 25\)**:
- First, we need to find \(\phi(25)\) (Euler's Totient Function):
\[ \phi(25) = 25 \left(1 - \frac{1}{5}\right) = 25 \times \frac{4}{5} = 20 \]
- By Euler's theorem, since \(2\) and \(25\) are coprime:
\[ 2^{20} \equiv 1 \mod 25 \]
- Now we reduce \(248\) modulo \(20\):
\[ 248 \mod 20 = 8 \]
- Therefore, we need to calculate \(2^8 \mod 25\):
\[ 2^8 = 256 \]
\[ 256 \mod 25 = 6 \]

4. **Combine results using CRT**:
- We have:
\[ 2^{248} \equiv 0 \mod 4 \]
\[ 2^{248} \equiv 6 \mod 25 \]
- We need to find a number \(x\) such that:
\[ x \equiv 0 \mod 4 \]
\[ x \equiv 6 \mod 25 \]
- Let \(x = 25k + 6\). To satisfy \(x \equiv 0 \mod 4\):
\[ 25k + 6 \equiv 0 \mod 4 \]
\[ 25 \equiv 1 \mod 4 \implies k + 2 \equiv 0 \mod 4 \implies k \equiv 2 \mod 4 \]
- Thus, \(k = 4m + 2\) for some integer \(m\). Substituting back:
\[ x = 25(4m + 2) + 6 = 100m + 56 \]
- Therefore:
\[ x \equiv 56 \mod 100 \]

5. **Final Result**: The last two digits of \(2^{248}\) are \(56\). Hence, the tens digit is:
\[ \text{Tens digit} = 5 \]