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%TCIDATA{<META NAME="Title" CONTENT="Problem 22 - Answers - Math 118">}
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\begin{document}


\begin{enumerate}
\item Let $N$ be the smallest positive integer with the following condition:
its last digit is $4,$ and if we remove this last digit and place it as its
first digit, the result is $4N$ (four times the number). \ \ Find the sum of
the digits of this number.

A) \ $15$ \ \ \ \ \ \ \ B) \ $18$ \ \ \ \ \ \ C) \ $21$ \ \ \ \ \ \ \ D) \ $%
22$ \ \ \ \ \ \ E) \ $24$

Solution: B) \ We start multiplying this number by $4$, starting at the last
digit.

\ 
\begin{tabular}{lllllll}
$4$ & $\cdot $ & \_ & \_ & \_ & \_ & $4$ \\ \cline{3-7}
&  &  &  &  &  & $6$%
\end{tabular}
\ \ \ \ so the digit before the $4$ is a $6.$ \ Insert this digit and
continue multiplying.

\ 
\begin{tabular}{lllllll}
$4$ & $\cdot $ & \_ & \_ & \_ & $6$ & $4$ \\ \cline{3-7}
&  &  &  &  & $5$ & $6$%
\end{tabular}
\ \ \ \ so the next digit is $5.$ \ \ Insert this digit and continue
multiplying.

\ 
\begin{tabular}{lllllll}
$4$ & $\cdot $ & \_ & \_ & $5$ & $6$ & $4$ \\ \cline{3-7}
&  &  &  & $2$ & $5$ & $6$%
\end{tabular}
\ \ \ \ so the next digit is $2.$ \ \ Insert this digit and continue
multiplying.

\ 
\begin{tabular}{lllllll}
$4$ & $\cdot $ & \_ & $2$ & $5$ & $6$ & $4$ \\ \cline{3-7}
&  &  & $0$ & $2$ & $5$ & $6$%
\end{tabular}
\ \ \ \ so the next digit is $0.$ \ \ Insert this digit and continue
multiplying.

\ 
\begin{tabular}{llllllll}
$4$ & $\cdot $ & \_ & $0$ & $2$ & $5$ & $6$ & $4$ \\ \cline{3-8}
&  &  & $1$ & $0$ & $2$ & $5$ & $6$%
\end{tabular}
\ \ \ \ so the next digit is $1.$ \ \ Insert this digit and continue
multiplying.

\ 
\begin{tabular}{llllllll}
$4$ & $\cdot $ & $1$ & $0$ & $2$ & $5$ & $6$ & $4$ \\ \cline{3-8}
&  & $4$ & $1$ & $0$ & $2$ & $5$ & $6$%
\end{tabular}
\ \ \ \ I think this is our number $N.$

$\left( 4\right) \left( 102564\right) =\allowbreak 410\,256$

Source: \ I think I saw this problem years and years ago in a KOMAL. \
(KOMAL is a Hungarian mathematical journal.)
\end{enumerate}

\end{document}