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?
1. 100 < A < 299 2. 106 < A < 305
3. 112 < A < 311 4. 118 < A < 317
If the digits of A are abc, Then B is cba.
100c+10b+a - 100a+10b+c = 7k
99c-99a = 7k
99(c-a) = 7k
so, c-a is a multiple of 7. In fact, since c and a are both single digits, c-a=7.
so, A must begin with 1 or 2, making c either 8 or 9
So, of the given choices, 100<A<299
correct answer is 2. 106 < A < 305posted by amit sharma
Since A=101,102,103,104,105,300,301,302,303,304 all don't work,
Both answers are true.
Since actually 107 works and 299 works,
107 <= A <= 299
You are right. I should not have excluded 299 from the solution set.