Question

Find the argest umber which divides 62, 132 and 237 to leave the same remainder in each case.

Answer 1

Let us say that the number 'd' divides 62, 132, and 237 to leave the remainder 'r'

=> 62 - r, 132 - r, and 237 - r are divisible by 'd'

=> (132 - r) - (62 - r) and (237 - r) - (132 - r) are divisible by 'd'

=> 70 and 105 are divisible by 'd'

=> Highest Common Factor of 70 and 105 = d = 35

Another method

62 mod n = 132 mod n = 237 mod n

If two numbers have the same remainder, then the difference of their remainders is 0. 
so (132-62 ) mod n = 0. Or 70 mod n = 0. So n divides 70 evenly. 
How large can n be? Well maybe 70. 
237 mod 70 = 27 but 132 mod 70 is 62. Nope. 
Ok try 35.
62 mod 35 = 27; 132 mod 35 = 27 and 237 mod 37 = 27 
So 35 is the answer.

i hope u understand.