Question

The digits of a positive integer ,having three digits are in AP and their sum is 15 .The number obtaining by reversing the digit is 594 less than the original number .find the number

Answer 1

Let the digit at the hundredth place of the number be (a + d).
Digit at the tens place be a.
Digit at the ones place be (a – d).
Sum of the digits = 15
(a + d) + a + (a – d) = 15
3a = 15
⇒ a = 5

The number formed by the digits
= 100 (5 + d) + 10 (5) + (5 – d)
= 555 + 99d

The number formed by reversing the digits
= 100 (5 - d) + 10 (5) + (5 + d)
= 555 – 99d

Given that the number formed by reversing the digits is 594 less
the original number.

⇒ (555 + 99d) – (555 - 99d) = 594
⇒ 198 d = 594
⇒ d = 3
(a + d) = (5 + 3) = 8,
a = 5,
(a – d) = (5 – 3) = 2.

Hence, the number formed by the digits = 100(8) + 10(5) + 1(2) = 852