Thread: Sum of the digits of a two-digit number is ten..

Hey,

My problem is:

The sum of the digits of a two-digit number is ten. If the digits are reversed, the new number is one less than twice the original number. Find the original number.

The answer is 37, but how exactly would someone go about finding that? Thanks.

Originally Posted by Averee

Hey,

My problem is:

The sum of the digits of a two-digit number is ten. If the digits are reversed, the new number is one less than twice the original number. Find the original number.

The answer is 37, but how exactly would someone go about finding that? Thanks.

Let the number be "xy", that is to say 10*x + y.

Then we know that
x + y = 10

and
"yx = 2*xy - 1" or 10*y + x = 2*(10*x + y) - 1

From the first condition
y = 10 - x

Inserting this into the second condition:
10*[10 - x] + x = 2*(10*x + [10 - x]) - 1

100 - 10x + x = 2*(10x + 10 - x) - 1

100 - 9x = 2*(9x + 10) - 1

100 - 9x = 18x + 20 - 1

100 - 9x = 18x + 19

-27x = -81

x = -81/(-27) = 3

Thus
y = 10 - (3) = 7

and the number is 37, as advertised.

-Dan

Thank you, that makes sense.