<PRE><B>Question 209635</B>
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.
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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
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24 or 15, they both work

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