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\begin{document}


\bigskip

\begin{problem}
\bigskip Two mathematicians are having a conversation. Mathematician A asks
B about his kids. B answers: "I have three children, the product of their
ages is 36." \ A says: "I still don't know how old your children are." \
Then B tells A the sum of his three kids' ages. A answers: "I still don't
know how old they are. Then B adds: "The youngest one has red hair." Now A
knows how old the kids are. Do you?
\end{problem}

\bigskip

\begin{solution}
The first information we have is that the product of the three kids' ages is 
$36$. \ This gives us all possibilities. We need to find all possible ways
we can express $36$ as a product of $3$ integers. Let us fix the age of the
youngest kid.

If the youngest kid's age is $1$, we have to express $\dfrac{36}{1}=36$ as a
product of two integers.

\begin{tabular}{lll}
& $36$ &  \\ 
$1$ &  & $36$ \\ 
$2$ &  & $18$ \\ 
$3$ &  & $12$ \\ 
$4$ &  & $9$ \\ 
$6$ &  & $6$%
\end{tabular}

Thus we have so far:

\begin{tabular}{|l|l|l|}
\hline
youngest & middle & oldest \\ \hline
$1$ & $1$ & $36$ \\ \hline
$1$ & $2$ & $18$ \\ \hline
$1$ & $3$ & $12$ \\ \hline
$1$ & $4$ & $9$ \\ \hline
$1$ & $6$ & $6$ \\ \hline
\end{tabular}

If the youngest kid's age is $2$, we have to express $\dfrac{36}{2}%
=\allowbreak 18$ as a product of two integers.

\begin{tabular}{lll}
& $18$ &  \\ 
$1$ &  & $18$ \\ 
$2$ &  & $9$ \\ 
$3$ &  & $6$%
\end{tabular}

Notice that the first row is repetition of a previously found case. \ The
youngest kid being $2$ years old mean the middle one cannot be $1$ year.

Thus we have:

\begin{tabular}{|l|l|l|}
\hline
youngest & middle & oldest \\ \hline
$2$ & $2$ & $9$ \\ \hline
$2$ & $3$ & $6$ \\ \hline
\end{tabular}

If the youngest kid's age is $1$, we have to express $\dfrac{36}{3}=12$ as a
product of two integers, where the smaller one is at least $3$.

\begin{tabular}{lll}
& $12$ &  \\ 
$3$ &  & $4$%
\end{tabular}

Thus we have:

\begin{tabular}{|l|l|l|}
\hline
youngest & middle & oldest \\ \hline
$3$ & $3$ & $4$ \\ \hline
\end{tabular}

Higher numbers will give us only repetition of previous cases. How do we
know?....Thus we have so far:

\begin{tabular}{|l|l|l|}
\hline
youngest & middle & oldest \\ \hline
$1$ & $1$ & $36$ \\ \hline
$1$ & $2$ & $18$ \\ \hline
$1$ & $3$ & $12$ \\ \hline
$1$ & $4$ & $9$ \\ \hline
$1$ & $6$ & $6$ \\ \hline
$2$ & $2$ & $9$ \\ \hline
$2$ & $3$ & $6$ \\ \hline
$3$ & $3$ & $4$ \\ \hline
\end{tabular}

B tells A the sum of the kid's ages. We can add all these tripples.

\begin{tabular}{|l|l|l|l|}
\hline
youngest & middle & oldest & sum of ages \\ \hline
$1$ & $1$ & $36$ & $38$ \\ \hline
$1$ & $2$ & $18$ & $21$ \\ \hline
$1$ & $3$ & $12$ & $16$ \\ \hline
$1$ & $4$ & $9$ & $14$ \\ \hline
$1$ & $6$ & $6$ & $13$ \\ \hline
$2$ & $2$ & $9$ & $13$ \\ \hline
$2$ & $3$ & $6$ & $11$ \\ \hline
$3$ & $3$ & $4$ & $10$ \\ \hline
\end{tabular}

After B told A the sum of the ages, and A still didn't know the ages. This
means that the sum was NOT ENOUGH information to distinguish two
possibilities. Since the only sum that appears twice is $13$, with $1,6,6$
and $2,2,9$, these are the only possibilities. (Any other sum was mentioned,
A would have known the answer.) Since B said the YOUNGEST one has red hair,
there is a youngest kid. This rules out $2,2,9$ and so the kids are $1$ and $%
6$ and $6$ years old.
\end{solution}

\end{document}