A man has some bananas. If he distributes 6 or 8 bananas to each person then 4 bananas are left behind. If he gives 10 or 12 bananas to each person, 4 bananas are left behind this time also. Find the minimum number of bananas he has?

Options

A. A)484
B. B)304
C. C)124
D. D)260

Answer

To find the minimum number of bananas that the man has, we can set up the problem using the information provided. Let's break it down step by step.

### Step 1: Understand the conditions
The man has some bananas, and we know two conditions based on how he distributes them:
1. When he distributes 6 or 8 bananas to each person, 4 bananas are left.
2. When he distributes 10 or 12 bananas to each person, 4 bananas are also left.

### Step 2: Set up the equations
Let \( N \) be the total number of bananas. From the first condition, we can write:
- \( N - 4 \) is divisible by 6.
- \( N - 4 \) is divisible by 8.

From the second condition, we have:
- \( N - 4 \) is divisible by 10.
- \( N - 4 \) is divisible by 12.

### Step 3: Find the least common multiple (LCM)
To find a number that satisfies all these conditions, we need to find the least common multiple (LCM) of the divisors:
- The LCM of 6 and 8.
- The LCM of 10 and 12.

#### Finding LCM of 6 and 8:
- Prime factorization of 6: \( 2^1 \times 3^1 \)
- Prime factorization of 8: \( 2^3 \)

The LCM is determined by taking the highest power of each prime:
- LCM(6, 8) = \( 2^3 \times 3^1 = 24 \)

#### Finding LCM of 10 and 12:
- Prime factorization of 10: \( 2^1 \times 5^1 \)
- Prime factorization of 12: \( 2^2 \times 3^1 \)

The LCM is:
- LCM(10, 12) = \( 2^2 \times 3^1 \times 5^1 = 60 \)

### Step 4: Find the common multiple
Now, we need to find a number \( N - 4 \) that is a common multiple of both LCMs (24 and 60). To do this, we find the LCM of 24 and 60.

#### Finding LCM of 24 and 60:
- Prime factorization of 24: \( 2^3 \times 3^1 \)
- Prime factorization of 60: \( 2^2 \times 3^1 \times 5^1 \)

The LCM is:
- LCM(24, 60) = \( 2^3 \times 3^1 \times 5^1 = 120 \)

### Step 5: Calculate \( N \)
Since \( N - 4 \) must be a multiple of 120, we can express it as:
- \( N - 4 = 120k \) for some integer \( k \).

Thus, we have:
- \( N = 120k + 4 \)

To find the minimum number of bananas, we set \( k = 1 \):
- \( N = 120 \times 1 + 4 = 124 \)

### Conclusion
The minimum number of bananas the man has is **124**.