|
Question 994057: "The sum of the digits of a certain two digit number is 13. When you reverse its digits you increase the number by 45. Find the number."
I solved this and wanted to put it up for everyone else who would have trouble with this problem. You can solve it yourselves if you want, but this is just what I got.
I got 49. The explanation: http://screencast.com/t/fGexMR44y0iR
and the proof/check work: http://screencast.com/t/jpZV9m0vt
Answer by addingup(3677)   (Show Source):
You can put this solution on YOUR website! Lets call the two digits in the number x and y, where x is any number times 10 (like in 15, 25, 35, etc.) and the y is a single digit from 1 to 9.
----------------------
Then, the value of the original number is 10x + y. If you reverse the digits, the new value will be 10y + x. So now we have the equations that we need. The problem says that:
(10y + x) = (10x + y) + 45 Now subtract one y and one x from both sides:
9y = 9x+45 Now divide both sides by 9:
y = x+5 Now we know that in our number one digit is 5 more than the other. And the problem tells us that the sum of the two digits is 13. Let's put this information into an equation.
We just said that y= x+5. Therefore, lets substitute in x+y= 13:
x+x+5= 13 Add x on one side and subtract 5 from both sides:
2x= 8 Divide both sides by 2:
x= 4 This is our "tens" digit, and the addition of our two digits sums 13. Thus:
13-4= 9 This is our second digit. Our number is 49.
Proof: The problem says that if we reverse the digits the new number is 45:
94-49= 45 We have the correct solution.
|
|
|
| |
