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? Choose the correct choice:
A) 100<A<299 B) 106<A<305 C) 112<A<311 D) 118<A<317 E) None
Explanation:   Let A = abc the B = cba   Given, B – A is exactly divisible by 7.   Therefore, 100c + 10b + a – (100a + 10b + c) = 99c – 99a = 99(c – a)   So 99(c – a) should be divisible by 7.   We know that 99 is not divisible by 7. So (c – a) should be divisible by 7 which implies c – a is 0, 7, 14...   c – a = 0 is not possible as c = a implies A = B but given that B &gt; A   c – a = 14 is not possible as the maximum difference between c and a = 9 – 1 = 8 only.   So c – a = 7.   If c = 9, a = 2   c = 8, a = 1   b can take any value from 0 to 9   Therefore, minimum value of abc = 109, maximum value = 299   From the given options, option B satisfies this.   Correct option: 106&lt;A&lt;305
<result>B</result>
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?