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

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  1. 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

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  2. correct answer is 2. 106 < A < 305

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  3. Hmmm.
    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.

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