To solve the problem of finding the least multiple of 23 that, when divided by 18, 21, and 24, leaves remainders of 7, 10, and 13 respectively, we can follow these steps:
### Step-by-Step Solution:
1. **Understanding the Remainders**:
We need to find a number \( x \) such that:
- \( x \equiv 7 \mod 18 \)
- \( x \equiv 10 \mod 21 \)
- \( x \equiv 13 \mod 24 \)
2. **Finding the Differences**:
To simplify the problem, we can express these congruences in terms of their differences:
- From \( x \equiv 7 \mod 18 \), we can write \( x = 18k + 7 \) for some integer \( k \).
- From \( x \equiv 10 \mod 21 \), we can write \( x = 21m + 10 \) for some integer \( m \).
- From \( x \equiv 13 \mod 24 \), we can write \( x = 24n + 13 \) for some integer \( n \).
Now, we can find the differences:
- \( 18 - 7 = 11 \)
- \( 21 - 10 = 11 \)
- \( 24 - 13 = 11 \)
This shows that all the differences are the same (11).
3. **Finding the LCM**:
Next, we need to find the least common multiple (LCM) of 18, 21, and 24:
- The prime factorization of the numbers is:
- \( 18 = 2 \times 3^2 \)
- \( 21 = 3 \times 7 \)
- \( 24 = 2^3 \times 3 \)
The LCM is found by taking the highest power of each prime:
- LCM = \( 2^3 \times 3^2 \times 7 = 504 \)
4. **Setting Up the Equation**:
We want \( x \) to be of the form:
\[
x = 504k - 11
\]
where \( k \) is a positive integer, and \( x \) must also be a multiple of 23.
5. **Finding the Least Value of \( k \)**:
We need to find the smallest \( k \) such that \( 504k - 11 \) is a multiple of 23:
\[
504k - 11 \equiv 0 \mod 23
\]
First, we find \( 504 \mod 23 \):
\[
504 \div 23 = 21 \quad \text{(remainder 21)}
\]
Thus, \( 504 \equiv 21 \mod 23 \). Now we rewrite the equation:
\[
21k - 11 \equiv 0 \mod 23
\]
Rearranging gives:
\[
21k \equiv 11 \mod 23
\]
6. **Finding the Inverse**:
To solve for \( k \), we need the multiplicative inverse of 21 modulo 23. The inverse can be found using the Extended Euclidean Algorithm, which gives us:
\[
21 \times 11 \equiv 1 \mod 23
\]
Thus, we can multiply both sides by the inverse of 21 to find \( k \).
7. **Calculating \( k \)**:
After calculating, we find that the least \( k \) that satisfies the equation is \( k = 6 \).
8. **Calculating \( x \)**:
Now substituting \( k \) back into the equation for \( x \):
\[
x = 504 \times 6 - 11 = 3024 - 11 = 3013
\]
9. **Final Check**:
We check if \( 3013 \) is indeed a multiple of 23:
\[
3013 \div 23 = 131 \quad \text{(exact)}
\]
Thus, \( 3013 \) is a multiple of 23.
### Final Answer:
The least multiple of 23 that satisfies the given conditions is **3013**.
