What is the least number which when increased by 8 is exactly divisible by 4, 5, 6 and 7?

Options

A. 322
B. 312
C. 412
D. 422

Answer

To find the least number which, when increased by 8, is exactly divisible by 4, 5, 6, and 7, we can follow these steps:

### Step 1: Find the LCM of 4, 5, 6, and 7
To solve the problem, we first need to find the least common multiple (LCM) of the numbers 4, 5, 6, and 7.

- The prime factorization of each number is:
- 4 = 2^2
- 5 = 5^1
- 6 = 2^1 × 3^1
- 7 = 7^1

- To find the LCM, we take the highest power of each prime that appears in the factorizations:
- The highest power of 2 is 2^2 (from 4)
- The highest power of 3 is 3^1 (from 6)
- The highest power of 5 is 5^1 (from 5)
- The highest power of 7 is 7^1 (from 7)

Thus, the LCM is:
\[ \text{LCM} = 2^2 \times 3^1 \times 5^1 \times 7^1 = 4 \times 3 \times 5 \times 7 \]

Calculating this step-by-step:
- \( 4 \times 3 = 12 \)
- \( 12 \times 5 = 60 \)
- \( 60 \times 7 = 420 \)

So, the LCM of 4, 5, 6, and 7 is 420.

### Step 2: Set up the equation
Let \( x \) be the least number we are looking for. According to the problem, when \( x \) is increased by 8, it should be divisible by the LCM we found.

This can be expressed as:
\[ x + 8 = 420k \]
for some integer \( k \).

### Step 3: Solve for \( x \)
Rearranging the equation gives:
\[ x = 420k - 8 \]

To find the least number, we can start with \( k = 1 \):
\[ x = 420(1) - 8 = 420 - 8 = 412 \]

### Step 4: Conclusion
Thus, the least number which when increased by 8 is exactly divisible by 4, 5, 6, and 7 is:
\[ \boxed{412} \]