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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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