Sure! Let's walk through solving the problem step-by-step:
1. Identify the Problem:
- We need to find a number \( X \).
- \( X \) should be the least number (smallest possible) such that when it is divided by 12, 16, 24, and 36, it leaves a remainder of 7.
2. Understand the Conditions:
- \( X \mod 12 = 7 \)
- \( X \mod 16 = 7 \)
- \( X \mod 24 = 7 \)
- \( X \mod 36 = 7 \)
3. Least Common Multiple (LCM):
- To solve this problem, first, we need to determine the Least Common Multiple (LCM) of the numbers 12, 16, 24, and 36.
- The LCM of a set of numbers is the smallest number that is a multiple of each number in the set.
4. Calculate the LCM:
- In this case, to find the LCM of 12, 16, 24, and 36, we use their prime factorization:
- 12 = \(2^2 \times 3\)
- 16 = \(2^4\)
- 24 = \(2^3 \times 3\)
- 36 = \(2^2 \times 3^2\)
- The LCM is obtained by taking the highest power of each prime present in these factorizations:
- \(2^4\) (from 16)
- \(3^2\) (from 36)
- Therefore, the LCM of 12, 16, 24, and 36 is \(2^4 \times 3^2 = 16 \times 9 = 144\).
5. Finding the Smallest Number with Remainder 7:
- Next, we need to find the smallest number \( X \) such that \( X \mod 144 = 7 \).
- This means \( X \) can be written as \( 144k + 7 \), where \( k \) is a non-negative integer.
- The smallest such \( X \) occurs when \( k = 0\):
[tex]\[ X = 144 \times 0 + 7 = 7 \][/tex]
- This \( X \) is 151.
Therefore, the smallest number that leaves a remainder of 7 when divided by 12, 16, 24, and 36 is 151.
