Question

The greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively is

• A. 127
• B. 235
• C. 123
• D. 305

Hint:

If a number \( N \) divides \( A \) and \( B \) leaving remainders \( r_1 \) and \( r_2 \) respectively, then \( N \) must be a common divisor of \( A-r_1 \) and \( B-r_2 \). To find the greatest such number, you need to calculate the HCF (or GCD) of \( A-r_1 \) and \( B-r_2 \).

Answer 1

The Correct Option is A

Let the required greatest number be \( N \).
The problem states that when 1657 is divided by \( N \), the remainder is 6. This means that the number \( 1657 - 6 = 1651 \) is exactly divisible by \( N \).
Similarly, when 2037 is divided by \( N \), the remainder is 5. This means that the number \( 2037 - 5 = 2032 \) is exactly divisible by \( N \).
So, \( N \) is a common divisor of 1651 and 2032. Since we need the greatest such number, we need to find the Highest Common Factor (HCF) or Greatest Common Divisor (GCD) of 1651 and 2032.
We can use the Euclidean algorithm to find the HCF.
\[ 2032 = 1651 \times 1 + 381 \] \[ 1651 = 381 \times 4 + 127 \] \[ 381 = 127 \times 3 + 0 \] The last non-zero remainder is the HCF.
Therefore, HCF(1651, 2032) = 127.
The greatest number is 127.