Question

FInd the largest no. which devides 62,132 and 237 to leave the same remainder in each case.

Answer 1

If a number 'a' and a number 'b' are divisible by a number 'n' then, a+b and a-b is also divisible by n

=> let the number be n which divides 62, 132 and 237 leaving a reminder r

=> the required number then becomes H.C.F of 62-r, 132-r and 237-r

=> it could also be the H.C.F of
(132-r)-(62-r) and (237-r)-(132-r)

i.e. 70 and 105

=> H.C.F of 70 and 105 = 35

Answer 2

Thank you for asking this question:

We will assume the largest number as x which will give as the same remainder which will be divided by 62,132 and 237.

Take 62=ax+r......(1)

       132=bx+r.. .  (2)

       237=cx+r......(3)

(2)-(1),(3)-(2) and (3)-(1)

=>70=px,105=qx and 175=rx

Where p=b-a,q=c-b and r=c-a

Now x=H.C.F of 70,105,175

= 35

So 35 is the answer to this question.

If there is any confusion please leave a comment below.