In a 3-digit number unit's digit is one more than the hundred's digit and ten's digit is one less than the hundred's digit. If the sum of the original 3 - digit number and numbers obtained by changing the order of digits cyclically is 2664. Find the number.

Let the hundred’s digit be x

Unit’s = x + 1 and ten’s digit = x – 1

The number formed = 100 x + 10 (x – 2) + x + 1

= 100x + 10x – 20 + x + 1

= 111x – 19

After changing the numbers cyclically the first case would be

Unit’s digit = x – 2, ten’s digit = x and hundred’s digit = x + 1

The number formed = 100 (x + 1) + 10 x + x – 2

= 100x + 100 + 10x + x – 2

= 111x + 98

On changing the digits cyclically for the second time

Unit digit = x, ten’s digit =x + 1 and hundred’s digit = x – 2 and the number formed is

100 (x – 2) + 10 (x + 1) + x

= 100 x – 200 + 10x + 10 + x

= 111x – 190

As per the conditions in the question

111x – 19 + 111x + 198 + 111x – 190 = 2664

⇒ 333x – 11 = 2664

⇒ 333x = 2664 + 11

Hence the original number = 111 x – 19 = 111 x 8 – 119 = 888 – 119

Original Number = 769