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.
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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
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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