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 ...​

Answer 1

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Answer 2

Answer:

Let The Digits at ones, tens and hundreds place be

(a - d) a and (a + d) respectively.

Then the number is

(a + d) × 100 + a × 10 + (a - d) = 111a - 99d

The number obtained by reversing the digits is

(a - d) × 100 + a × 10 + (a + d) = 111a - 99d

It is given that the sum of the digits is 15.

(a - d) + a + ( a + d) = 15. (i)

Also it is given that the number obtained by reversing the digits is 594 less than the original number.

Therefore,

111a - 99d = 111a + 99d - 594. (ii)

⇒ 3a = 15 and 198d = 594

⇒ a = 5 and d = 3

So, the number is 111 × 5 + 99 × 3 = 852

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