Division of a three-place number into a five-place number, each digit have been replaced by a code letter, reconstruct and show solution is unique.

Question

I'm currently working in the following excercise:

In the following division of a three-place number into a five-place number each digit have been replaced by a code letter. Assuming only that the remainder, Y, is not zero, reconstruct the problem and show that the solution is unique.

I've started, for the first operation with:

$$U \cdot LMN = RTYX$$

I've been trying to check if the potential numbers may be divisible by ten in order to find the first component but I've been unable to, and I'm not sure about it to be the correct way. Another way I tried to follow is to check the term $RSTUN$ divisibility by many numbers but I haven't reached any good result yet.

I would really appreciate any hint or help about how to proceed in this problem, thanks in advance.

Answer 1

\begin{array}{r} &&&&&&&U&X\\ &&&&-&-&-&-&- \\ L&M&N&)&R&S&T&U&N \\ &&&&R&T&Y&X \\ &&&&-&-&-&- \\ &&&&&T&Y&Y&N \\ &&&&&T&Y&Y&J \\ &&&&&-&-&-&- \\ &&&&&&&&Y \end{array}

Out of the ten letters, $\{L,M,N,R,S,T,U,X,Y\}$

• L, R, T, U can't be $0$ because they are the first digit of a number.
• X, Y can't be zero or because of the column $\quad \begin{array}{c}U\\X\\-\\Y\end{array}\quad$
• Similarly J (and Y) can't be $0$.
• N can't be $0$ because UN mod $10$ equals X and X $\ne 0$.
• S can't be $0$ because of $\quad \begin{array}{cC}R&S\\ R&T\\ -&-\\ &T \end{array}$

Thus $M = 0$.

This means that

• $UN = 10Y + X$
• $XN = 10Y + J$
• $UL = 10R + T$
• $XL = 10T + Y$

As @RossMillikan said, $T \in \{1,2,3,4\}$. Note that, if $Y \ge 5$, then $S$ has to be odd. And if $Y \le 4$, then $S$ has to ber even. That implies these possibilities.

\begin{array}{c} T & S & Y \\ \hline 1 & 3 & 5 \\ 2 & 4 & 1 \\ 2 & 5 & 6 & *\\ 3 & 6 & 1 & *\\ 3 & 7 & 6 \\ 4 & 8 & 2 \\ 4 & 9 & 7 & * \end{array}

Because $M=0$, then $XL = 10T+Y$ needs to be a product of two digits. So we can remove the starred items.

\begin{array}{c} T & S & Y \\ \hline 1 & 3 & 5 \\ 2 & 4 & 1 \\ 3 & 7 & 6 \\ 4 & 8 & 2 \end{array}