SOLUTION: The value of a certain two digit number is 9 times the sum of it's digits. If the digits are reversed, the resulting number is 63 less than the original number. What is the origina

Question 550872: The value of a certain two digit number is 9 times the sum of it's digits. If the digits are reversed, the resulting number is 63 less than the original number. What is the original number? Please, please help me answer this question!
Found 2 solutions by fcabanski, stanbon:
Answer by fcabanski(1391) (Show Source): You can put this solution on YOUR website!
Look at the second part first. xy - 63 = yx. Look at the possibilities for y-3=x:

y=1 so 1-3=x=-2 or, if a ten had been carried from the tens digit it's 11-3=8.

Could x=8? 8, carry one ten to the ones digit so that becomes 7-6 = 1.

Yes, if x=8 and y=1 it's 81-63=18.

Check it in the first part.

xy = 81 = 9 (8+1)=9*9. Yes!

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Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
The value of a certain two digit number is 9 times the sum of it's digits. If the digits are reversed, the resulting number is 63 less than the original number.
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Let the number be 10t+u.
----
Equations:
10t+u = 9(t+u)
10u+t = 10t+u-63
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Simplify each equation:
t = 8u
9t = 9u + 63
-----
Modify:
t = 8u
t = u + 7
---
Solve for "u":
8u = u+7
u = 1
----
Solve for "t":
t = 8u
t = 8
---
Original Number: 10t+u = 81
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Cheers,
Stan H.
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