Find the least number which when divided by 12, 16, 24 and 36 leaves a remainder 7 in each case

Question

Accepted Answer

To solve the problem of finding the least number which, when divided by 12, 16, 24, and 36, leaves a remainder of 7 in each case, we need to follow these steps:

1. **Understand the Problem**: We are looking for a number $ N $ such that:
   $$
   N \equiv 7 \pmod{12}
   $$
   $$
   N \equiv 7 \pmod{16}
   $$
   $$
   N \equiv 7 \pmod{24}
   $$
   $$
   N \equiv 7 \pmod{36}
   $$

2. **Find the Least Common Multiple (LCM)**: First, we need to find the least common multiple of the divisors 12, 16, 24, and 36. The LCM of these numbers is the smallest number that is divisible by each of them.

- 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:
     - $ 2^4 $ (from 16)
     - $ 3^2 $ (from 36)

- Therefore, the LCM is $ 2^4 \times 3^2 = 16 \times 9 = 144 $.

3. **Determine the Number**: We need a number $ N $ such that:
   $$
   N = \text{LCM} + 7 = 144 + 7 = 151
   $$

This means that 151 is the smallest number that, when divided by 12, 16, 24, and 36, leaves a remainder of 7.

4. **Verify the Options**:
   - **Option 1: 144** - When divided by 12, 16, 24, or 36, 144 leaves a remainder of 0, not 7.
   - **Option 2: 151** - As calculated, 151 leaves a remainder of 7 when divided by each of the numbers 12, 16, 24, and 36.
   - **Option 3: 165** - When divided by 12, 16, 24, or 36, 165 does not leave a remainder of 7.
   - **Option 4: 148** - When divided by 12, 16, 24, or 36, 148 does not leave a remainder of 7.

Therefore, Option 2 (151) is the correct answer as it satisfies the condition of leaving a remainder of 7 when divided by each of the given numbers.