Find the least number, which when increased by 7 is divisible by 3, 7, 28 and 15.

Options

A. 420
B. 427
C. 413
D. 127

Answer

To solve the problem of finding the least number which when increased by 7 is divisible by 3, 7, 28, and 15, we can follow these steps:

### Step 1: Find the Least Common Multiple (LCM)
We first need to find the LCM of the numbers 3, 7, 28, and 15.

- **Prime Factorization**:
- \(3 = 3^1\)
- \(7 = 7^1\)
- \(28 = 2^2 \times 7^1\)
- \(15 = 3^1 \times 5^1\)

- **Identify the highest power of each prime**:
- For \(2\): The highest power is \(2^2\) from 28.
- For \(3\): The highest power is \(3^1\) from both 3 and 15.
- For \(5\): The highest power is \(5^1\) from 15.
- For \(7\): The highest power is \(7^1\) from both 7 and 28.

- **Calculate the LCM**:
\[ \text{LCM} = 2^2 \times 3^1 \times 5^1 \times 7^1 = 4 \times 3 \times 5 \times 7 \]

Calculating this step-by-step:
1. \(4 \times 3 = 12\)
2. \(12 \times 5 = 60\)
3. \(60 \times 7 = 420\)

Thus, the LCM of 3, 7, 28, and 15 is **420**.

### Step 2: Find the Required Number
The problem states that we need a number which when increased by 7 is divisible by 3, 7, 28, and 15. This means:
\[ x + 7 = 420k \quad \text{(where \(k\) is a positive integer)} \]
To find \(x\), we rearrange the equation:
\[ x = 420k - 7 \]

### Step 3: Find the Least Value of \(x\)
To find the least number \(x\), we can take \(k = 1\):
\[ x = 420 \times 1 - 7 = 420 - 7 = 413 \]

Thus, the least number which when increased by 7 is divisible by 3, 7, 28, and 15 is **413**.

### Final Answer
The least number is **413**.

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