Question

Find the greatest number which when divides 63 121 and 208 leaves same reminder .​

Answer 1

Answer:

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

So, HCF is 35

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

Hope it helps you.

Step-by-step explanation: