What is the smallest number which when divided by 72 and 96, respectively, leaves a remainder 5 in each case?

Options

A. 283
B. 571
C. 581
D. 293

Answer

To find the smallest number which, when divided by 72 and 96, leaves a remainder of 5 in each case, we can follow these steps:

### Step 1: Understand the Problem
We need to find a number \( N \) such that:
- \( N \mod 72 = 5 \)
- \( N \mod 96 = 5 \)

This means that when \( N \) is divided by 72 and 96, it leaves a remainder of 5.

### Step 2: Adjust the Conditions
Since \( N \) leaves a remainder of 5 when divided by both numbers, we can express \( N \) in terms of multiples of 72 and 96:
- \( N = 72k + 5 \) for some integer \( k \)
- \( N = 96m + 5 \) for some integer \( m \)

### Step 3: Eliminate the Remainder
To find a common \( N \), we can rewrite the equations without the remainder:
- \( N - 5 = 72k \)
- \( N - 5 = 96m \)

This implies that \( N - 5 \) must be a common multiple of 72 and 96.

### Step 4: Find the LCM of 72 and 96
To find the smallest \( N - 5 \), we need to calculate the least common multiple (LCM) of 72 and 96.

**Finding the LCM:**
1. Prime factorization of 72: \( 72 = 2^3 \times 3^2 \)
2. Prime factorization of 96: \( 96 = 2^5 \times 3^1 \)

The LCM is found by taking the highest power of each prime:
- For \( 2 \): \( \max(3, 5) = 5 \) → \( 2^5 \)
- For \( 3 \): \( \max(2, 1) = 2 \) → \( 3^2 \)

Thus, the LCM is:
\[ \text{LCM} = 2^5 \times 3^2 = 32 \times 9 = 288 \]

### Step 5: Calculate \( N \)
Since \( N - 5 = 288 \), we can find \( N \):
\[ N = 288 + 5 = 293 \]

### Conclusion
The smallest number which, when divided by 72 and 96, respectively, leaves a remainder of 5 in each case is **293**.

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