Thread: The sum of the digits of a two digit number is thirteen...

1. The sum of the digits of a two digit number is thirteen...

The sum of the digits of a two digit number is thirteen. If the digits are reversed and the two numbers are added, the sum is 27 less than twice the original number. What is the unit's digit of the original number?

If the two digits are reversed and added wouldn't it still equal 13?
If it is 13, I could just add 27 and divide by 2, couldn't I?
$\displaystyle 13+27=40$
$\displaystyle 40/2=20$

The answer that the book says is 5, but I don't have a clue how it's 5.

2. Originally Posted by iamanoobatmath

If the two digits are reversed and added wouldn't it still equal 13?
If it is 13, I could just add 27 and divide by 2, couldn't I?
$\displaystyle 13+27=40$
$\displaystyle 40/2=20$

The answer that the book says is 5, but I don't have a clue how it's 5.

Let the digits be $\displaystyle x$ and $\displaystyle y$ so

$\displaystyle (1) \;\;x + y = 13$.

If the digits are reversed and added then

$\displaystyle (2) \;\;(10 x + y) + (10y + x) = 2(10x + y) - 27$.

(1) an (2) give two equations for the two unkowns, when solved, gives

$\displaystyle x = 5, y = 8$.

3. The sum of the digits of a two digit number is thirteen. If the digits are reversed and the two numbers are added, the sum is 27 less than twice the original number. What is the unit's digit of the original number?
Originally Posted by iamanoobatmath

If the two digits are reversed and added wouldn't it still equal 13?
The digits are reversed and the two numbers are added, not the digits. danny arrigo's solution is excellent.

If it is 13, I could just add 27 and divide by 2, couldn't I?
$\displaystyle 13+27=40$
$\displaystyle 40/2=20$

The answer that the book says is 5, but I don't have a clue how it's 5.