Question

In a 3-digit number, the 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.

Answer 1

Hint: In this question, we would assume the hundred’s place to be x and then start applying the conditions given in the question over the ones and tens place.Also, while adding the numbers after changing the order we will take all possible numbers that can be made by changing the digits cyclically. This will eventually lead us to the answer.

Complete step-by-step answer:

We have been given that, 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.
So, let the hundred’s digit be x.
Then,
Ten’s digit = x-1
And
One’s digit = x+1
So, the number is = 100x + 10(x-1) + (x+1)
= 111x – 9….............Equation (1)
Now, the numbers obtained by changing the order of digits cyclically will be,
100(x-1) + 10(x+1) + x = 111x - 90 ….............Equation (2)
And
100(x+1) + 10x + (x-1) = 111x + 99….............Equation (3)
We have been also given that the sum of the original 3-digit number and numbers obtained by changing the order of digits cyclically is 2664.
So, Equation (1) + Equation (2) + Equation (3) = 2664
Therefore, (111x - 9) + (111x – 90) + (111x + 99) = 2664
333x = 2673
x = 8
Therefore, the required number is 111(8) -9 = 879.

Note: Whenever we face such types of problems the value point to remember is that we need to have a good grasp over linear equations in one variable and number theory. We must also firstly assume a digit and then proceed by applying the conditions as given in the question. This helps in getting us the required expressions and gets us on the right track to reach the answer.