Question

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

Answer 1

Step 1: Establish the equations:

Let hundred's digit be $x$

$\therefore$ ten's digit $=x-1$

and, unit's digit $=x+1$

Now on changing the order of digits cyclically, the first form of Number is

$100x+100\left(x-\right)+\left(x+1\right)=111x-9$

The second form of Number is

$\left(x+1\right)100+10x+\left(x-1\right)=111x+99$

The third form of number is

$\left(x-1\right)100+\left(x+1\right)10+x=111x-90$

Step 2: Solve for $x$:

Upon adding the three equations we get,

$111x-9+111x+99+111x-90=2664$

$\Rightarrow$ $333x=2664$

$\Rightarrow$ $x=8$

Step 3: Estimate the number:

So, the ten's digit is

$x-1=8-1=7$

And, the unit's digit is

$x+1=8+1=9$

Therefore, the number is

$8\times 100+7\times 10+9=879$

Hence, the required number is $879$.