What is the least possible number which when divided by 13 leaves the remainder 3 and when it is divided by 5 it leaves the remainder 2.

Answer

To find the least possible number which, when divided by 13, leaves a remainder of 3, and when divided by 5, leaves a remainder of 2, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem:**
We need to find a number \( x \) such that:
- \( x \mod 13 = 3 \)
- \( x \mod 5 = 2 \)

2. **Expressing the Conditions:**
From the first condition, we can express \( x \) in terms of a quotient \( q \):
\[ x = 13q + 3 \]
where \( q \) is a non-negative integer.

3. **Substituting into the Second Condition:**
We substitute \( x \) into the second condition:
\[ 13q + 3 \mod 5 = 2 \]
To simplify this, we first calculate \( 13 \mod 5 \):
\[ 13 \mod 5 = 3 \]
Therefore, we can rewrite the equation:
\[ (3q + 3) \mod 5 = 2 \]

4. **Simplifying the Equation:**
We can simplify \( 3q + 3 \) as follows:
\[ 3q + 3 \equiv 2 \mod 5 \]
Subtracting 3 from both sides gives:
\[ 3q \equiv -1 \mod 5 \]
Since \(-1\) is equivalent to \(4\) in modulo \(5\), we have:
\[ 3q \equiv 4 \mod 5 \]

5. **Finding the Value of \( q \):**
Now we need to find \( q \) such that \( 3q \equiv 4 \mod 5 \). We can test values of \( q \):
- For \( q = 0: \quad 3(0) \equiv 0 \)
- For \( q = 1: \quad 3(1) \equiv 3 \)
- For \( q = 2: \quad 3(2) \equiv 1 \)
- For \( q = 3: \quad 3(3) \equiv 4 \) (this works)

Thus, \( q = 3 \) satisfies the equation.

6. **Calculating \( x \):**
Now we substitute \( q = 3 \) back into the equation for \( x \):
\[ x = 13(3) + 3 = 39 + 3 = 42 \]

7. **Final Verification:**
- Check \( 42 \mod 13 \):
\[ 42 \div 13 = 3 \quad \text{(remainder 3)} \]
- Check \( 42 \mod 5 \):
\[ 42 \div 5 = 8 \quad \text{(remainder 2)} \]

Both conditions are satisfied.

### Conclusion:
The least possible number is \( \boxed{42} \).