Institution: Bangladesh Public Service Commission (BPSC) Exam: 17th BCS Preliminary Exam 1996
Topic: Arithmetic (পাটিগণিত), Subject: Mathematics (সাধারণ গণিত)

Best Answer Explained in Below-:

To solve the problem, we need to find the smallest number among the given options (A: 71, B: 41, C: 31, D: 39) that, when divided by 3, 5, and 6, leaves a remainder of 1.

When a number \( x \) leaves a remainder of 1 when divided by another number \( n \), it can be represented as:
\[ x = kn + 1 \]
where \( k \) is some integer.

Firstly, we will check each option:

1. **For 71**:
- \( 71 \mod 3 = 2 \)
- \( 71 \mod 5 = 1 \)
- \( 71 \mod 6 = 5 \)
- Not a valid choice.

2. **For 41**:
- \( 41 \mod 3 = 2 \)
- \( 41 \mod 5 = 1 \)
- \( 41 \mod 6 = 5 \)
- Not a valid choice.

3. **For 31**:
- \( 31 \mod 3 = 1 \) (valid)
- \( 31 \mod 5 = 1 \) (valid)
- \( 31 \mod 6 = 1 \) (valid)
- All conditions met, thus valid.

4. **For 39**:
- \( 39 \mod 3 = 0 \)
- \( 39 \mod 5 = 4 \)
- \( 39 \mod 6 = 3 \)
- Not a valid choice.

Since only 31 meets all criteria of giving a remainder of 1 when divided by 3, 5, and 6, it is the correct and smallest choice. Thus, the answer is C: 31.