Find the greatest number that divides 79, 88 and 160 leaving remainder 7 in each case

Answer

To find the greatest number that divides 79, 88, and 160 leaving a remainder of 7 in each case, we can follow these steps:

### Step-by-Step Solution:

1. **Subtract the Remainder from Each Number**:
Since we want to find a number that leaves a remainder of 7, we first subtract 7 from each of the numbers:
- \( 79 - 7 = 72 \)
- \( 88 - 7 = 81 \)
- \( 160 - 7 = 153 \)

2. **List the New Numbers**:
The new numbers we have after subtraction are:
- 72
- 81
- 153

3. **Find the Highest Common Factor (HCF)**:
Now we need to find the HCF of the numbers 72, 81, and 153.

- **Factorization of 72**:
- \( 72 = 2^3 \times 3^2 \) (Factors: 2, 2, 2, 3, 3)

- **Factorization of 81**:
- \( 81 = 3^4 \) (Factors: 3, 3, 3, 3)

- **Factorization of 153**:
- \( 153 = 3 \times 51 = 3 \times 3 \times 17 = 3^2 \times 17 \) (Factors: 3, 3, 17)

4. **Identify Common Factors**:
The common factor among the factorizations is \( 3 \). The lowest power of 3 present in all three numbers is \( 3^1 \).

5. **Conclusion**:
Therefore, the HCF of 72, 81, and 153 is \( 3 \).

6. **Final Answer**:
The greatest number that divides 79, 88, and 160 leaving a remainder of 7 in each case is:
\[ \text{Greatest Number} = 3 \]