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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

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Solution

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

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