Find the greatest number by which the numbers 29, 43 and 71 when divide, leaves remainder 5, 7 and 11 respectively

Options

A. 6
B. 12
C. 16
D. 18

Answer

To find the greatest number by which the numbers 29, 43, and 71 leave remainders of 5, 7, and 11 respectively, we can follow these steps:

### Step 1: Subtract the remainders from the numbers
- For 29, the remainder is 5. So, we subtract:
\[ 29 - 5 = 24 \]
- For 43, the remainder is 7. So, we subtract:
\[ 43 - 7 = 36 \]
- For 71, the remainder is 11. So, we subtract:
\[ 71 - 11 = 60 \]

### Step 2: Find the highest common factor (HCF) of the results
Now we need to find the HCF of the numbers 24, 36, and 60.

### Step 3: Prime factorization of each number
- **For 24**:
\[ 24 = 2^3 \times 3^1 \]
- **For 36**:
\[ 36 = 2^2 \times 3^2 \]
- **For 60**:
\[ 60 = 2^2 \times 3^1 \times 5^1 \]

### Step 4: Identify the common factors
Now we identify the common prime factors:
- The common factor for 2 is \(2^2\) (the minimum power of 2 in all three factorizations).
- The common factor for 3 is \(3^1\) (the minimum power of 3 in all three factorizations).

### Step 5: Calculate the HCF
Now we multiply the common factors:
\[ HCF = 2^2 \times 3^1 = 4 \times 3 = 12 \]

### Conclusion
The greatest number by which 29, 43, and 71 leave remainders of 5, 7, and 11 respectively is **12**.

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