Question

The sum of the digits of a 2-digit number is 10. If 18 is added to the number, the digits of the new number are those of the original number but in reverse order. Find the original number.

Answer 1

Answer:

46

Step-by-step explanation:

Let $(ab)$ represent a number with $a$ in the ten's position and $b$ is the one's position.

This means $(ab)$ actually has value of $10a+b$.

We are given the sum of those digits of $(ab)$ is 10; this means $a+b=10$.

It says if 18 is added to the number $(ab)$, then the result is $(ba)$.

So $(ab)$ has value $10a+b$ and

$(ba)$ has value $10b+a$.

We are given then:

$(ab)+18=(ba)$

$10a+b+18=10b+a$

Subtract $10a$ on both sides:

$b+18=10b+a-10a$

Simplify:

$b+18=10b-9a$

Subtract $b$ on both sides:

$18=10b-b-9a$

$18=9b-9a$

Divide both sides by 9:

$2=b-a$

Rearrange by commutative property:

$2=-a+b$

So the system of equations we want to solve is:

$a+b=10$

$-a+b=2$

-------------------------Add equations together (this will eliminate the variable $a$ and allow you to go ahead and solve for $b$:

$0+2b=12$

$2b=12$

Divide both sides by 2:

$b=\frac{12}{2}$

Simplify:

$b=6$

If $b=6$ and $a+b=10$, then $a=4$.  $a=4$ since 4+6=10.

So the original number is (46).

18 more than 46 is 18+46=(64) which is what we wanted.

We also have the sum of 4 and 6 is 10 as well.