Question

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

Verified
128.4k+ views
Hint: In this question, first of all, find the number for which we have to find the H.C.F by subtracting the remainders 6 and 5 from 1657 and 2037 respectively. Then find the H.C.F of the two obtained numbers which are our required greatest number. So, use this concept to reach the problem of the solution.

Complete step-by-step solution:
Here we have to find the greatest number which on dividing 1657 and 2037 leaves a remainder 6 and 5 respectively.
So, required number = Highest common factor of $\left( {1657 - 6} \right)$ and $\left( {2037 - 5} \right)$
= H.C.F of 1651 and 2032
Now, consider the prime factors of 1651 $= 13 \times 127$
And the prime factors of 2032 $= 2 \times 2 \times 2 \times 127$
In 1651 there were one 13 and one 127 while in 2032 there were three 2`s and one 127 prime factors. So, there is only one common factor for both 1651 and 2032 i.e., 127.
Therefore, the H.C.F. of 1651 and 2032 is 127.
Thus, our required number is 127.

Hence, the correct option is D. 127

Note: In mathematics, the greatest common divisor (G.C.D) or highest common factor (H.C.F) of two or more integers, which are not all zero, is the largest positive integer that divides each of the two integers.