The unit digit in the sum of `(124)^(372) + (124)^(373)` is

Options

A. 5
B. 4
C. 2
D. 0

Answer

To find the unit digit in the sum of \( (124)^{372} + (124)^{373} \), we can follow these steps:

### Step 1: Identify the unit digit of the base
The unit digit of \( 124 \) is \( 4 \). Therefore, we only need to consider the unit digit of \( 4^{372} + 4^{373} \).

### Step 2: Find the pattern of unit digits of powers of 4
Next, we will determine the unit digits of the powers of \( 4 \):
- \( 4^1 = 4 \) (unit digit is 4)
- \( 4^2 = 16 \) (unit digit is 6)
- \( 4^3 = 64 \) (unit digit is 4)
- \( 4^4 = 256 \) (unit digit is 6)

From this, we can see that the unit digits of powers of \( 4 \) alternate between \( 4 \) and \( 6 \):
- Odd powers of \( 4 \) have a unit digit of \( 4 \).
- Even powers of \( 4 \) have a unit digit of \( 6 \).

### Step 3: Determine the unit digits of \( 4^{372} \) and \( 4^{373} \)
Since \( 372 \) is an even number, the unit digit of \( 4^{372} \) is \( 6 \).
Since \( 373 \) is an odd number, the unit digit of \( 4^{373} \) is \( 4 \).

### Step 4: Add the unit digits
Now, we can add the unit digits:
\[ 6 + 4 = 10 \]

### Step 5: Find the unit digit of the sum
The unit digit of \( 10 \) is \( 0 \).

### Conclusion
Therefore, the unit digit in the sum of \( (124)^{372} + (124)^{373} \) is \( 0 \).

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