Find the least number, which when divided by 27, 42, 63 and 84 leaves the remainder 21 in each case.

Options

A. 760
B. 745
C. 777
D. 767

Answer

To find the least number which, when divided by 27, 42, 63, and 84, leaves a remainder of 21 in each case, we can follow these steps:

### Step 1: Subtract the remainder from each divisor
Since we want the number to leave a remainder of 21, we first subtract 21 from each of the divisors:
- 27 - 21 = 6
- 42 - 21 = 21
- 63 - 21 = 42
- 84 - 21 = 63

### Step 2: Find the LCM of the adjusted divisors
Now, we need to find the least common multiple (LCM) of the numbers obtained in Step 1: 6, 21, 42, and 63.

#### Finding the LCM:
1. **Prime factorization**:
- 6 = 2 × 3
- 21 = 3 × 7
- 42 = 2 × 3 × 7
- 63 = 3^2 × 7

2. **Take the highest power of each prime factor**:
- For 2: highest power is \(2^1\) (from 6 and 42)
- For 3: highest power is \(3^2\) (from 63)
- For 7: highest power is \(7^1\) (from 21, 42, and 63)

3. **Calculate the LCM**:
\[ \text{LCM} = 2^1 × 3^2 × 7^1 = 2 × 9 × 7 = 126 \]

### Step 3: Add the remainder back to the LCM
Now that we have the LCM, we need to add the remainder (21) back to it:
\[ \text{Least number} = \text{LCM} + 21 = 126 + 21 = 147 \]

### Final Answer
The least number which when divided by 27, 42, 63, and 84 leaves a remainder of 21 is **147**.

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