Find the greatest number which divides 29,60 and 103 leaving remainders 5,12 and 7, respectively.

Answer

To find the greatest number that divides 29, 60, and 103 leaving remainders of 5, 12, and 7 respectively, we can follow these steps:

### Step 1: Adjust the Numbers by Subtracting the Remainders
We need to subtract the given remainders from each of the numbers to find the new numbers that the greatest number will divide.

- For 29 with a remainder of 5:
\[ 29 - 5 = 24 \]

- For 60 with a remainder of 12:
\[ 60 - 12 = 48 \]

- For 103 with a remainder of 7:
\[ 103 - 7 = 96 \]

### Step 2: Find the HCF of the Adjusted Numbers
Now, we need to find the highest common factor (HCF) of the numbers 24, 48, and 96.

#### Step 2.1: Factorize Each Number
- **Factorization of 24**:
\[ 24 = 2^3 \times 3^1 \]

- **Factorization of 48**:
\[ 48 = 2^4 \times 3^1 \]

- **Factorization of 96**:
\[ 96 = 2^5 \times 3^1 \]

#### Step 2.2: Identify Common Factors
To find the HCF, we take the lowest power of each common prime factor:
- For the prime factor 2, the lowest power is \(2^3\).
- For the prime factor 3, the lowest power is \(3^1\).

### Step 3: Calculate the HCF
Now we multiply the common factors:
\[ \text{HCF} = 2^3 \times 3^1 = 8 \times 3 = 24 \]

### Conclusion
The greatest number that divides 29, 60, and 103 leaving remainders of 5, 12, and 7 respectively is **24**.

---