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.

Explanation

Let the digits at ones, tens, and hundreds place be (a-d), a, and (a+d), respectively. Then the number is
`(a+d)xx100+axx10+(a-d)=111a+99d`
The number obtained by reversing the digits is
`(a-d)xx100+axx10+(a+d)=111a-99d`
It is given that
(a-d)+a+(a+d)=15
and 111a-99d=111a+99d-594
`therefore`3a=15 and 198d=594
`rArr` a=5 and d=3
So, the number is `111xx5+99xx3=852`