Find the largest number that divides 1989 and 967 and leaves a remainder of 5 and 7 respectively.

Answer

To find the largest number that divides 1989 and 967 and leaves a remainder of 5 and 7 respectively, we can follow these steps:

### Step 1: Adjust the numbers by subtracting the remainders
We need to subtract the remainders from the original numbers to find the numbers that the largest divisor must divide.

- For 1989, subtract the remainder 5:
\[ 1989 - 5 = 1984 \]

- For 967, subtract the remainder 7:
\[ 967 - 7 = 960 \]

### Step 2: Find the Highest Common Factor (HCF) of the adjusted numbers
Now, we need to find the HCF of 1984 and 960.

#### Step 2.1: Prime factorization of 1984
To find the HCF, we can start with the prime factorization of 1984:
- Divide by 2:
\[ 1984 \div 2 = 992 \]
- Divide by 2 again:
\[ 992 \div 2 = 496 \]
- Divide by 2 again:
\[ 496 \div 2 = 248 \]
- Divide by 2 again:
\[ 248 \div 2 = 124 \]
- Divide by 2 again:
\[ 124 \div 2 = 62 \]
- Divide by 2 again:
\[ 62 \div 2 = 31 \]
- 31 is a prime number.

Thus, the prime factorization of 1984 is:
\[ 1984 = 2^6 \times 31 \]

#### Step 2.2: Prime factorization of 960
Now, we can find the prime factorization of 960:
- Divide by 2:
\[ 960 \div 2 = 480 \]
- Divide by 2 again:
\[ 480 \div 2 = 240 \]
- Divide by 2 again:
\[ 240 \div 2 = 120 \]
- Divide by 2 again:
\[ 120 \div 2 = 60 \]
- Divide by 2 again:
\[ 60 \div 2 = 30 \]
- Divide by 2 again:
\[ 30 \div 2 = 15 \]
- Divide by 3:
\[ 15 \div 3 = 5 \]
- 5 is a prime number.

Thus, the prime factorization of 960 is:
\[ 960 = 2^6 \times 3 \times 5 \]

### Step 3: Identify the common factors
Now, we can find the HCF by identifying the common prime factors:
- The common factor is \(2^6\).

### Step 4: Conclusion
Thus, the HCF of 1984 and 960 is:
\[ HCF = 2^6 = 64 \]

Therefore, the largest number that divides 1989 and 967 and leaves a remainder of 5 and 7 respectively is **64**.

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