The digits of a three digit number N are in A.P.If sum of the digits is 15 and the number obtained by re-versing the digits of the number is 594 less than the original number, then 1000N−252 is equal to

Let three digits of N be: Hundred’s digit = a – dten’s digit = a,and unit’s digit = a+d(a–d)+a+(a+d)=15 fi a=5andN=100(5−d)+50+(5+d)                                         (1)Number obtained by reversing digits of N isN1=100(5+d)+50+(5−d)                                       (2)Now,          594=N−N1=100(−2d)+2d⇒    2d=−6⇒d=−3Thus, N = 852 and1000N−252=1000600=53