To find the least number which, when divided by 5, 6, 8, 9, and 12 leaves a remainder of 1, and when divided by 13 leaves no remainder, we can follow these steps:
### Step-by-Step Solution:
1. **Identify the Problem**: We need to find a number \( x \) such that:
- \( x \equiv 1 \mod 5 \)
- \( x \equiv 1 \mod 6 \)
- \( x \equiv 1 \mod 8 \)
- \( x \equiv 1 \mod 9 \)
- \( x \equiv 1 \mod 12 \)
- \( x \equiv 0 \mod 13 \)
2. **Find the LCM of the Divisors**: First, we need to find the least common multiple (LCM) of the numbers 5, 6, 8, 9, and 12.
- The prime factorization of each number is:
- \( 5 = 5^1 \)
- \( 6 = 2^1 \times 3^1 \)
- \( 8 = 2^3 \)
- \( 9 = 3^2 \)
- \( 12 = 2^2 \times 3^1 \)
- The LCM is calculated by taking the highest power of each prime:
- \( LCM = 2^3 \times 3^2 \times 5^1 = 8 \times 9 \times 5 = 360 \)
3. **Set Up the Equation**: Since \( x \) leaves a remainder of 1 when divided by 5, 6, 8, 9, and 12, we can express \( x \) in terms of the LCM:
\[
x = 360k + 1 \quad \text{for some integer } k
\]
4. **Condition for Divisibility by 13**: Now, we need \( x \) to be divisible by 13:
\[
360k + 1 \equiv 0 \mod 13
\]
This simplifies to:
\[
360k \equiv -1 \mod 13
\]
5. **Calculate \( 360 \mod 13 \)**: We need to find \( 360 \mod 13 \):
\[
360 \div 13 = 27 \quad \text{(integer part)}
\]
\[
27 \times 13 = 351
\]
\[
360 - 351 = 9 \quad \Rightarrow \quad 360 \equiv 9 \mod 13
\]
Thus, the equation becomes:
\[
9k \equiv -1 \mod 13
\]
or equivalently:
\[
9k \equiv 12 \mod 13
\]
6. **Find the Inverse of 9 modulo 13**: We need to find the multiplicative inverse of 9 modulo 13. We can test values:
- \( 9 \times 3 = 27 \equiv 1 \mod 13 \)
So, the inverse is 3.
7. **Multiply Both Sides by the Inverse**:
\[
k \equiv 12 \times 3 \mod 13
\]
\[
k \equiv 36 \mod 13 \quad \Rightarrow \quad k \equiv 10 \mod 13
\]
Thus, \( k = 10 \) is the smallest non-negative solution.
8. **Calculate \( x \)**:
\[
x = 360 \times 10 + 1 = 3600 + 1 = 3601
\]
9. **Final Verification**:
- Check if \( 3601 \) leaves a remainder of 1 when divided by 5, 6, 8, 9, and 12:
- \( 3601 \mod 5 = 1 \)
- \( 3601 \mod 6 = 1 \)
- \( 3601 \mod 8 = 1 \)
- \( 3601 \mod 9 = 1 \)
- \( 3601 \mod 12 = 1 \)
- Check if \( 3601 \) is divisible by 13:
- \( 3601 \div 13 = 277 \quad \text{(exact division)} \)
Thus, the least number that satisfies all conditions is \( \boxed{3601} \).
