Question

the sum of the revered number and the original number is 154. Find the original number, if the ones digit in it is 2 less than the tens digit.

Answer 1

1) We write the number that we want to find as:

$10x+y$

Where x is the tens digit, and y is the unit digit.

2) Its reversed number is:

$10y+x$

3) The sum of the number and its reverse is equal to 154:

$\begin{gathered} (10x+y)+(10y+x)=154, \\ 11x+11y=154, \\ 11\cdot(x+y)=154, \\ x+y=\frac{154}{11}, \\ x+y=14 \end{gathered}$

4) One of the digits in the number is 2 less than the tens digit, so we have:

$y=x-2$

Replacing this in the previous equation that we found and solving for x:

$\begin{gathered} x+(x-2)=14, \\ 2x-2=14, \\ 2x=14+2, \\ 2x=16, \\ x=\frac{16}{2}, \\ x=8 \end{gathered}$

So:

$y=x-2=8-2=6$

The number that we were looking for is:

$10x+y=10\cdot8+6=86$

Verification:

$86+68=154$

Answer: 86