Question

the digits of a positive integer, having three digits are in A.P. and their sum is 15.the number obtained by reversing the digits is 594 less than the original number. Find the number.

Answer 1

Let digits of the number be (a – d), a and (a + d) respectively.
∴ The required number is 100 (a – d) + 10a + (a + d) .
Given : The sum of the digits = 15
➪ (a-d) + a + (a+d) = 15
3a=15➪  a = 5
Now, the number on reversing the digits is 100 (a + d) + 10a + (a – d) .
According to the question
100(a – d) + 10a + a + d = 100 (a + d) + 10a + (a – d) + 594
on solving we get, d = -3
The digits of the number are (5 – (–3)), 5, (5 + (–3) = 8, 5, 2
And the required number is 8 × 100 + 5 × 10 + 2 = 852