∴       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

" />∴       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

" />∴       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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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.

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