SOLUTION: A two-digit number is eight times the sim of its digits. When the number is added to the number obtained by reversing the digits, the sum is 99. Find the original number.
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Question 557651: A two-digit number is eight times the sim of its digits. When the number is added to the number obtained by reversing the digits, the sum is 99. Find the original number.
(I tried 10x+y=8(x+y), 10x+y+10y+x=99) It didn't work though...
Thanks!
Found 3 solutions by josmiceli, solver91311, KMST:
Answer by josmiceli(19441) (Show Source): You can put this solution on YOUR website!
You had the right equations
Let = the tens digit
Let = units digit
given:
(1)
(2)
----------------------
(1)
(1)
(1)
(1)
and
(2)
(2)
Substitute (1) into (2)
(2)
(2)
(2)
(2)
and, since
(1)
(1)
(1)
The number is 72
check:
(1)
(1)
(1)
and
(2)
(2)
(2)
(2)
OK
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
Of course it does.
Eq. 1: )
Eq. 2: 
Put Equation 1 into standard form:
Eq. 1: 
Put Equation 2 into standard form and reduce:
Eq. 2: 
Eq. 2: 
Multiply Equation 2 by -2:
Eq. 2: 
Add Equation 1 to Equation 2:
Sum Equation: 
And solve:
Then
7 plus 2 is 9 and 72 is 8 times 9. 72 plus 27 is 99. Checks.
John
My calculator said it, I believe it, that settles it
Answer by KMST(5328) (Show Source): You can put this solution on YOUR website!
Let tens digit be x and units digit be y.
The value of the number is .
The sum of its digits is
"A two-digit number is eight times the sum of its digits" translates into
--> --> -->
Since x and y are digits, the only solution is with .
(tells you that is a multiple of 7 or zero, so x must be a multiple of 7, and can only be 7. So , which leads to ).
So the number is 72.
When you reverse the digits, you get 27, which added to 72 is 99, but why that was added to the problem is a mystery to me.
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