When 'n' is divisible by 5 the remainder is 2. What is the remainder when `n^2` is divided by 5 ?

Options

A. 2
B. 3
C. 1
D. 4

Answer

To solve the problem, we need to find the remainder when \( n^2 \) is divided by 5, given that \( n \) leaves a remainder of 2 when divided by 5.

### Step-by-Step Solution:

1. **Understanding the Given Condition**:
Since \( n \) is divisible by 5 and leaves a remainder of 2, we can express \( n \) in terms of 5:
\[ n = 5k + 2 \]
where \( k \) is some integer.

2. **Finding \( n^2 \)**:
Next, we need to calculate \( n^2 \):
\[ n^2 = (5k + 2)^2 \]
Expanding this using the binomial theorem:
\[ n^2 = 25k^2 + 20k + 4 \]

3. **Dividing \( n^2 \) by 5**:
Now, we need to find the remainder when \( n^2 \) is divided by 5. We can simplify \( n^2 \) modulo 5:
\[ n^2 \mod 5 = (25k^2 + 20k + 4) \mod 5 \]
Since \( 25k^2 \) and \( 20k \) are both multiples of 5, they will leave a remainder of 0 when divided by 5. Therefore:
\[ n^2 \mod 5 = 0 + 0 + 4 = 4 \]

4. **Conclusion**:
Thus, the remainder when \( n^2 \) is divided by 5 is:
\[ \boxed{4} \]