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?
Explanation:
Let A = 100x + 10y + z (x ≠ 0, x, y, z are single-digit numbers)
∴ B = 100z + 10y + x
∴ B – A = 99(z – x)
As (B – A) is divisible by 7 and 99 is not, (z – x) is divisible by 7.
∴ z and x can have values (8, 1) or (9, 2). [Since B > A, z > x]
y can have any value from 0 to 9.
A = 1y8 or 2y9
∴ Lowest possible value of A is 108 and the highest possible value of A is 299.
Hence, option (b).
Help us build a Free and Comprehensive Preparation portal for various competitive exams by providing us your valuable feedback about Apti4All and how it can be improved.