SOLUTION: There is a two-digit number such that the sum of its digits is 6, while the product of the digits is 1/3 the original number. Find this number.

Question 209635: There is a two-digit number such that the sum of its digits is 6, while the product of the digits is 1/3 the original number. Find this number.
Found 2 solutions by checkley77, Alan3354:
Answer by checkley77(12844) (Show Source): You can put this solution on YOUR website!
x+y=6
x=6-y
xy=(10x+y)/3
(6-y)y=(10[6-y]+y)/3
6y-y^2=(60-10y+y)/3
3(6y-y^2)=60-9y
18y-3y^2=60-9y
-3y^2+18y+9y-60=0
-3y^2+27y-60=0
-3(y^2-9y+20)=0
-3(y-4)(y-5)=0
y-4=0
y=4 ans.
x=6-4
x=2 ans (24)
y-5=0
y=5 ans.
x=6-5
x=1 ans (15)
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
There is a two-digit number such that the sum of its digits is 6, while the product of the digits is 1/3 the original number. Find this number.
--------
t + u = 6 (Tens digit + Units digit = 6)
t*u = (10t + u)/3 (the original number is 10t+u)
3tu = 10t + u
u = 6-t
3t(6-t) = 10t + 6-t
18t-3t^2 = 9t+6
3t^2 - 9t + 6 = 0
t^2 - 3t + 2 = 0
(t-1)*(t-2) = 0
t = 1, u = 5
t = 2, u = 4
--------
24 or 15, they both work

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