Question

The digits of a three digit positive number are in AP and sum is 15 on subtracting 594 from number the digits get reversed find the number
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Answer 1

HEY BUDDY HERE IS UR ANSWER !!!

Sol: 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.

Hope u like the process !!

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