A four-digit number 4ob5 is divisible by 55.

Question:

A four-digit number 4ob5 is divisible by 55. Then, the value of b-a is

(a) 0

(b) 1 

(c) 4

(d) 5

Solution:

(b) Given, a four-digit number 4ab5 is divisible by 55. Then, it is also divisible by 11.

The difference of sum of its digits in odd places and sum of its digits in even places is either 0 or multiple of 11.

i.e. (4 + b) – (a + 5) is 0 or a multiple of 11, if 4 + b — a — 5 = 0 => b-a = 1

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