SOLUTION: Four times the ones digit of a positive, two digit integer is 25 greater than the sum of the digits. Reversing the digits increases the number by 63. What is the number?

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Question 1153592: Four times the ones digit of a positive, two digit integer is 25 greater than the sum of the digits. Reversing the digits increases the number by 63. What is the number?
Found 2 solutions by josgarithmetic, greenestamps:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
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Four times the ones digit of a positive, two digit integer is 25 greater than the sum of the digits. Reversing the digits increases the number by 63. What is the number?
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u, the ones digit
t, the tens digit
u+10t, the two-digit number

The description:
system%284u=25%2Bu%2Bt%2C10u%2Bt=63%2B%28u%2B10t%29%29
Simplify.


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system%283u-t=25%2C9u-9t=63%29
-
system%283u-t=25%2Cu-t=7%29

Finish to solve. E1-E2
%283u-t%29-%28u-t%29=25-7
3u-t-u%2Bt=18
2u=18
highlight%28u=9%29

u-t=7
-t=7-u
t=u-7
highlight%28t=2%29

The number is 29.

Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


Here is a quick path to the answer, if a formal algebraic solution is not required.

The difference between a 2-digit number and the 2-digit number with the digits reversed is 9 times the difference of the two digits.

The given number is increased by 63 = 9*7 when the digits are reversed, so the difference between the digits is 7. That means the 2-digit number can be only 18 or 29.

18 doesn't satisfy the condition that the ones digit is 25 more than the sum of the digits; 29 does.

ANSWER: 29