In this case we use the nought in addition to the $1,$ $2,$ $3,$ $4,$ $5,$ $6,$ $7,$ $8,$ $9.$ The puzzle is, as in the last case, so to arrange the ten counters that the products of the two multiplications shall be the same, and you may here have one or more figures in the multiplier, as you choose. The above is a very easy feat, but it is also required to find the two arrangements giving pairs of the highest and lowest products possible. Of course, every counter must be used, and the cipher may not be placed to the left of a row of figures where it would have no effect. Vulgar fractions or decimals are not allowed.
Solutions: 1
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