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The digits of a positive integer having three digits are in arithmetic progression and their sum is 15. The number obtained by reversing the digits is 594

less than the original number.Find the number.


First equation a + a+d + a+2d = 15 or a+d=5 

second equation 
(100a + 10(a+d) + a+2d - (100(a+2d) + 10(a+d) + a = 594 

solve , 
I got d=-3 and a=8 , which works 
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