Question

The digit of a positive number of three digits are in AP and their sum is 15 . the numbers obtained by reversing the digits in 594 less then the original number. find the number.

Answer 1

HEY MATE!!!

THIS IS UR ANSWER...:-)

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

HOPE IT HELPS!!!=)