Question

4. find the largest number which divides 62, 132 and 237 to leave the same remainder in each case.

Answer 1

Dear Student,

Answer:35

Solution:

As the given numbers are 62, 132 and 237.

As we have to find largest number which divides 62,132 and 237 leaving same remainder in each case.

let us assume that remainder left is m

So, numbers which are completely divisible are

62-m , 132-m, 237-m

Now subtract the pairs

132-m -62+m = 70

237-m-132+m = 105

Now find the HCF of 70 and 105

$105 = 70 \times 1 + 35 \\ 70 = 35 \times 2 + 0$
So, HCF is 35

So, the largest number which divides 62,132 and 237 and leaves same remainder is 35

Proof:

$62 = 35 \times 1 + 27 \\ \\ 132 = 35 \times 3 + 27 \\ \\ 237 = 35 \times 6 + 27$

Hope it helps you.

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.