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

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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 – d) = (5 – 3) = 2.

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