# Solution

(related to Problem: The Ten Counters)

As I pointed out, it is quite easy so to arrange the counters that they shall form a pair of simple multiplication sums, each of which will give the same product — in fact, this can be done by anybody in five minutes with a little patience. But it is quite another matter to find that pair which gives the largest product and that which gives the smallest product.

Now, in order to get the smallest product, it is necessary to select as multipliers the two smallest possible numbers. If therefore, we place $1$ and $2$ as multipliers, all we have to do is to arrange the remaining eight counters in such a way that they shall form two numbers, one of which is just double the other; and in doing this we must, of course, try to make the smaller number as low as possible. Of course, the lowest number we could get would be $3,045;$ but this will not work, neither will $3,405,$ $3,450,$ etc., and it may be ascertained that $3,485$ is the lowest possible. One of the required answers is $3,485 \times 2 = 6,970,$ and $6,970 \times 1 = 6,970.$

The other part of the puzzle (finding the pair with the highest product) is, however, the real knotty point, for it is not at all easy to discover whether we should let the multiplier consist of one or of two figures, though it is clear that we must keep, so far as we can, the largest figures to the left in both multiplier and multiplicand. It will be seen that by the following arrangement so high a number as $58,560$ may be obtained. Thus, $915 \times 64 = 58,560,$ and $732 \times 80 = 58,560.$

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### References

#### Project Gutenberg

1. Dudeney, H. E.: "Amusements in Mathematics", The Authors' Club, 1917

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