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 3 digits in A.P. at units, tens and hundredth places are a-d, a and a+d.
According to the first condition,
`(a-d)+a+(a+d)=15 rArr 3a=15`
`a=5 " " ...(1)`
The number is (a-d)+10a+100(a+d)
i.e., `111a+99d " " ...(2)`
The number, on reversing the digits is
(a+d)+10a+100(a-d) i.e., 111a-99d
`:. ` According to the 2nd condition,
`111a-99d=(111a+99d)-594`
`rArr 198d=594 rArr d=3 " " ...(3)`
`:.` Required number = 111a + 99d [from (2)]
` =111(5) + 99(3) " "` [from (1) and (3) ]
` =555 + 297`
`=852`