There are three numbers 116, 135 and 211. What is the largest possible number which divides them leaving the same remainder in each case? 

There are three numbers 116, 135 and 211. What is the largest possible number which divides them leaving the same remainder in each case?  Correct Answer 19

Given:

Three numbers = 116, 135 and 211

Concept Used:

The largest number which when divides the numbers a, b and c, leaving the same remainders is HCF of (a - b), (b - c) and (c - a).

Calculation:

According to the question,

The largest possible number which divides 116, 135 and 211 leaving the same remainder in each case = HCF of (135 - 116), (211 - 135) and (211 - 116)

⇒ HCF of 19, 76 and 95

Now,

19 = 19 × 1,

76 = 19 × 2 × 2,

95 = 19 × 5

So, the HCF of 19, 76 and 95 is 19.

And so, the HCF of 116, 135 and 211 is 19.

∴ The largest possible number which divides 116, 135 and 211 leaving the same remainder in each case is 19.