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.

THE TEN APPLES. THE TEN PRISONERS. facebooktwittergoogle_plusredditpinterestlinkedinmail