Find the smallest number such that when it is divided by 5, 6 and it leaves a remainder 3 in each case.

Options

A. 243
B. 123
C. 117
D. 792

Answer

To find the smallest number that leaves a remainder of 3 when divided by 5 and 6, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**: We need a number \( N \) such that:
- \( N \mod 5 = 3 \)
- \( N \mod 6 = 3 \)

2. **Reformulating the Conditions**: If \( N \) leaves a remainder of 3 when divided by 5 and 6, we can express \( N \) as:
- \( N = 5k + 3 \) for some integer \( k \)
- \( N = 6m + 3 \) for some integer \( m \)

3. **Subtracting the Remainder**: To simplify the problem, we can subtract 3 from \( N \):
- Let \( M = N - 3 \)
- Then, \( M \) must be divisible by both 5 and 6:
- \( M \mod 5 = 0 \)
- \( M \mod 6 = 0 \)

4. **Finding the Least Common Multiple (LCM)**: The smallest number \( M \) that is divisible by both 5 and 6 is the least common multiple (LCM) of 5 and 6.
- The LCM of 5 and 6 is \( 30 \).

5. **Calculating \( N \)**: Since \( M = 30 \), we can find \( N \):
- \( N = M + 3 = 30 + 3 = 33 \)

6. **Verifying the Result**:
- Check \( 33 \mod 5 \):
- \( 33 \div 5 = 6 \) remainder \( 3 \) (Correct)
- Check \( 33 \mod 6 \):
- \( 33 \div 6 = 5 \) remainder \( 3 \) (Correct)

Thus, the smallest number \( N \) that leaves a remainder of 3 when divided by both 5 and 6 is **33**.