The greatest number which on dividing 1657 and 2037 leaves a remainder 6 and 5 respectively is A. 271B. 172C. 721D. 127

To find the greatest number that satisfies the given conditions, we need to apply the concept of the remainder when dividing two numbers.

Let's assume the greatest number is 'x.'

According to the problem, when 1657 is divided by 'x,' the remainder is 6. This can be expressed as:

1657 = a*x + 6, where 'a' is some positive integer (quotient)

Similarly, when 2037 is divided by 'x,' the remainder is 5:

2037 = b*x + 5, where 'b' is some positive integer (quotient)

Now, we need to find the greatest common divisor (GCD) of the differences between these two equations.

Subtracting the second equation from the first:

1657 - 2037 = ax + 6 - bx - 5
-380 = (a - b)*x + 1

Since 'x' is a positive integer and we want to find the greatest value for 'x,' we need to find the greatest divisor of 380 that is one less than a multiple of 'x.'

Prime factorizing 380, we get: 380 = 2^2 * 5 * 19

Now, we look for a factor of 380 that is one less than a multiple of 'x.'

380 = 19 * 20, where 20 is one more than a multiple of 'x' (20 = 2 * 10).

So, 'x' = 19.

Therefore, the correct option is:

A. 271