Henry Ernest Dudeney/Modern Puzzles/72 - Alphabetical Sums/Solution

Modern Puzzles by Henry Ernest Dudeney: $72$

      RSR
   ------
 PR)MTVVR
    MVR
    ---
     KKV
     KMD
     ---
      MVR
      MVR
      ---

We have immediately that $R \times R = R \pmod {10}$.

However, because $R \times PR = MTV$ it is clear $R \ne 1$ and (obviously) $R \ne 0$.

So $R = 5$ or $R = 6$.

To get $V$ in the fifth line it is clear that $D = 0$.

Thus we have that $K = 2 M$.

In order for $D$ to be $0$:

if $R = 5$ then $S$ must be even.

if $R = 6$ then $S = 5$.

Note that $KMD > MVR$, which implies that $S > R$.

Hence it cannot be the case that $R = 6$ and $S = 5$.

And thus we conclude $R = 5$ and $S = 6$ or $8$.

The divisor $PR$ is a factor of $KMD$, which can only be $210, 420, 630$ or $840$.

Therefore $PR$ is a factor of the lowest common multiple of these numbers, $2520$.

Of the factors of $2520$, only $15$, $35$ and $45$ are $2$-digit numbers ending in $5$.

However $15 \times RSR < 15 \times 600 = 9000 < 10 \, 000$, so $PR \ne 15$.

Finally we check the remaining $4$ possibilities:

$35 \times 565 = 19 \, 775$

$45 \times 565 = 25 \, 425$

$35 \times 585 = 20 \, 475$

$45 \times 585 = 26 \, 325$

and only $19 \, 775$ matches the pattern $MTVVR$, so it must be our dividend.

The rest of the puzzle can be filled in accordingly.

$\blacksquare$