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

• Dec 25th 2008, 07:41 PM
iamanoobatmath
The sum of the digits of a two digit number is thirteen...
Quote:

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?
$13+27=40$
$40/2=20$

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

(Angry)
• Dec 25th 2008, 08:37 PM
Jester
Quote:

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?
$13+27=40$
$40/2=20$

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

(Angry)

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

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

If the digits are reversed and added then

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

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

$x = 5, y = 8$.
• Dec 26th 2008, 10:13 AM
HallsofIvy
Quote:

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?
Quote:

Originally Posted by iamanoobatmath
$13+27=40$
$40/2=20$