The digits of a three-digit number A are written in the reverse order to form another three-digit number B. If B > A and B-A is perfectly divisible by 7, then which of the following is necessarily true?

a. 100< A<299

b. 106<A <305

c. 112<A <311

d. 118< A <317

Let A = 100x + 10y + z and B = 100z + 10y + x .

According to given condition B - A = 99(z - x)

As (B - A) is divisible by 7 .

So clearly (z - x) should be divisible by 7.

z and x can have values 8,1 or 9,2 , such that 8-1=9-2=7 and y can have value from 0 to 9.

So Lowest possible value of A lowest x,y and z which is is 108 and the highest possible value of A is 299**Write Here**