Question

The digit of a positive no of three digit are on a.p and their sum is 15the no obtained by reversing digit is 594 less than the original no . Find the no
Please answer fast

Answer 1

Let the number be (a-d), a, (a+d)

(a+d)×100+a×10+(a−d)=111a+99d

on reversing

(a−d)×100+a×10+(a+d)a+d+a+a−d⟹=111a−99d=15a=5 and d = 3

Hence the no is 111×5+99×3=852