Question 1150396: The sum of the digits of a two digit counting number is 15 , when the digits are reversed the number is 27 more than the original number . what was the original number?
Found 2 solutions by ikleyn, Alan3354:
Answer by ikleyn(52781)   (Show Source):
You can put this solution on YOUR website! .
Let "a" be the tens digit of the 2-digit number, and
let "b" be the ones digit of the number.
Then the number is
n = 10a + b, (1)
while the reversed digit number is
m = 10b + a. (2)
The sum of the digits "a" and "b" is 15
a + b = 15. (3)
According to the condition, the reversed number is 27 more than the original number. It means
(10b+a) - (10a+b) = 27, or
9b - 9a = 27,
9*(b-a) = 27,
b-a = 27/9 = 3. (4)
Thus we have the system of two equations
b + a = 15 (3')
b - a = 3 (4')
To solve it, from equation (3') express a = 15-b and express it into equation (4'). You will get then
b - (15-b) = 3
2b = 3+15 = 18,
b = 18/2 = 9.
Then from equation (1'), a = 15-9 = 6.
Thus the number is 69.
CHECK. The reversed number MINUS the original number is 96 - 69 = 27. ! Correct !
Solved.
Answer by Alan3354(69443)  (Show Source):
You can put this solution on YOUR website! The sum of the digits of a two digit counting number is 15, when the digits are reversed the number is 27 more than the original number . what was the original number?
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Reversing the digits of a 2 digit integer changes the value by 9 times the difference between the digits.
================
T + U = 15
T - U = 3
---------------- Add
2T = 18
T = 9
U = 6
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Since reversing increases the value, the Tens is the smaller number.
---> 69
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