Permutations and Combinations

Permutations and Combinations

Question

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.

Answer

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

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