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 digit in the unit's plce be `a-d,` digit in the ten's place be a and the digit in the hundred's place be `a+d.`
Sum of digits `=a-d+a+a+d=15" " [" Given "]`
`implies 3a= 15`
`therefore a=5 " " "....(i)"` `therefore` Original number `=(a-d)+10a+100(a+d)`
`=111a-99d=555+99d`
and number formed by reversing the digits
`(a+d)+10a+100(a-d)`
`=111a-99d=555+99d`
Given, `(555+99d)-(555-99d)=594 implies 198d=594`
`therefore " "d=3`
Hence, original number `=555+99xx3=852`