SOLUTION: The sum of the digits of a two-digit number is 13. If the digits are reversed, the new number is 4 less than twice the original number. Find the original number. I an very confu

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Question 133852: The sum of the digits of a two-digit number is 13. If the digits are reversed, the new number is 4 less than twice the original number. Find the original number.
I an very confused with this problem.
It's be greatly appreciated if I could get help!
Answer by solver91311(24713) About Me  (Show Source):
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10s digit: x
1s digit: y

The original number: n%5Bo%5D=10x%2By, because a number like 24 could be represented as 2%2A10%2B4

Reversing the digits makes a new number: n%5Bn%5D=10y%2Bx

Twice the original number is 2%2810x%2By%29 and 4 less than that is 2%2810x%2By%29-4 which we know to be equal to the new number, so:

10y%2Bx=2%2810x%2By%29-4

Simplify:
10y%2Bx=20x%2B2y-4
10y-2y%2Bx-20x=-4
8y-19x=-4


Since we know that the sum of the digits is 13, we can say x%2By=13, or in other words, y=13-x

Substitute:
8%2813-x%29-19x=-4
104-8x-19x=-4
-27x=-108
x=4

So the 10s digit of the original number (and the ones digit of the new number) is 4. Therefore the ones digit of the original number must be 13-4=9, and the original number must be 49.

Check:
Number is: 49
4%2B9=13
2%2A49-4=98-4=94 which is the original number with the digits reversed.