Question 994057
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.
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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.