1.

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»

Let digits of the number be (a – d), a and (a + d) respectively.
∴ The required number is 100 (a – d) + 10a + (a + d) .
Given : The sum of the digits = 15
➪ (a-d) + a + (a+d) = 15
3a=15➪  a = 5
Now, the number on reversing the digits is 100 (a + d) + 10a + (a – d) .
According to the question
100(a – d) + 10a + a + d = 100 (a + d) + 10a + (a – d) + 594
on solving we get, d = -3
The digits of the number are (5 – (–3)), 5, (5 + (–3) = 8, 5, 2
And the required number is 8 × 100 + 5 × 10 + 2 = 852


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