When `252^126 + 244^152` is divided by 10, the remainder will be?

Options

A. 8
B. 6
C. 0
D. 3

Answer

To find the remainder of \( 252^{126} + 244^{152} \) when divided by 10, we can focus on the last digits of each term. The last digit of a number determines its remainder when divided by 10.

### Step 1: Identify the last digits
- The last digit of \( 252 \) is \( 2 \).
- The last digit of \( 244 \) is \( 4 \).

### Step 2: Calculate the last digit of \( 252^{126} \)
We need to find the last digit of \( 2^{126} \). The last digits of powers of \( 2 \) follow a pattern:
- \( 2^1 = 2 \) (last digit is 2)
- \( 2^2 = 4 \) (last digit is 4)
- \( 2^3 = 8 \) (last digit is 8)
- \( 2^4 = 16 \) (last digit is 6)
- \( 2^5 = 32 \) (last digit is 2)

The last digits repeat every 4 terms: \( 2, 4, 8, 6 \).

To find \( 2^{126} \), we calculate \( 126 \mod 4 \):
\[ 126 \div 4 = 31 \quad \text{remainder } 2 \]
Thus, \( 126 \mod 4 = 2 \).

From the pattern, the last digit of \( 2^{126} \) corresponds to the last digit of \( 2^2 \), which is \( 4 \).

### Step 3: Calculate the last digit of \( 244^{152} \)
Now we find the last digit of \( 4^{152} \). The last digits of powers of \( 4 \) follow this pattern:
- \( 4^1 = 4 \) (last digit is 4)
- \( 4^2 = 16 \) (last digit is 6)
- \( 4^3 = 64 \) (last digit is 4)
- \( 4^4 = 256 \) (last digit is 6)

The last digits repeat every 2 terms: \( 4, 6 \).

To find \( 4^{152} \), we calculate \( 152 \mod 2 \):
\[ 152 \div 2 = 76 \quad \text{remainder } 0 \]
Thus, \( 152 \mod 2 = 0 \).

From the pattern, the last digit of \( 4^{152} \) corresponds to the last digit of \( 4^2 \), which is \( 6 \).

### Step 4: Add the last digits
Now we can add the last digits of \( 252^{126} \) and \( 244^{152} \):
\[ 4 + 6 = 10 \]

### Step 5: Find the remainder when divided by 10
The last digit of \( 10 \) is \( 0 \). Therefore, the remainder when \( 252^{126} + 244^{152} \) is divided by \( 10 \) is \( 0 \).

### Final Answer
The remainder is \( \boxed{0} \).