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.

Let a - d, a, a + d be the three digits of a three digit number respectively at unit place, ten's place, hundreth place.

∴       The number = 100 (a + d) + 10 + a + a - d

Since the sum of digits is 15

∴  a -d + a + a + d = 15           or      3a = 15            or          a = 5

The number obtained by reversing the digits = 100 (a - d) + 10a + (a + d)             ...(ii)

According to the given condition

100 (a - d) + 10 a + a + d = 100 (a + d) + 10a + a - d - 594

or    100a - 100d + 10a + a + d = 100a + 100d + 10a + a - d - 594

or           198d = 594      or           d = 3

∴  Original number = 100 (5 + 3) + 10 (5) + 5 - 3 = 800 + 50 + 2 = 852

Verification: On reversing the digits, the number is 258

                  852 - 258 = 594