Question

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. \(112 < A < 311\)
• C. \(106 < A < 305\)
• D. \(118 < A < 317\)

Hint:

Write \(A = 100a + 10b + c\); then \(B - A = 99(c - a)\). Work out which pairs \((a, c)\) make this divisible by 7.

Answer 1

The Correct Option is C

Step 1: Express the difference:
Let \(A = 100a + 10b + c\) with \(a \ge 1\). Then \(B = 100c + 10b + a\) and \(B - A = 99(c - a)\).

Step 2: Use divisibility by 7:
99 is not divisible by 7, so 7 must divide \(c - a\). Since \(B > A\), \(c > a\); with digits, \(c - a\) lies between 1 and 8, so \(c - a = 7\).

Step 3: List the possibilities:
\(a = 1, c = 8\) or \(a = 2, c = 9\) (a cannot be 0, and \(a = 3\) would need \(c = 10\)). So A has the form \(1b8\) or \(2b9\), where \(b\) is any digit 0 to 9.
The smallest A is 108, the largest is 299. All 20 such numbers lie between 108 and 299.

Step 4: Check the options:
(1) \(100 < A < 299\) fails for \(A = 299\).
(3) \(106 < A < 305\) holds for all A from 108 to 299, so it is necessarily true.
(2) \(112 < A < 311\) fails for \(A = 108\).
(4) \(118 < A < 317\) fails for \(A = 108\).

Final Answer:
The correct option is (3): \(106 < A < 305\)