80.—­THE THREE GROUPS.

There appeared in “Nouvelles Annales de Mathematiques” the following puzzle as a modification of one of my “Canterbury Puzzles.”  Arrange the nine digits in three groups of two, three, and four digits, so that the first two numbers when multiplied together make the third.  Thus, 12 x 483 = 5,796.  I now also propose to include the cases where there are one, four, and four digits, such as 4 x 1,738 = 6,952.  Can you find all the possible solutions in both cases?

81.—­THE NINE COUNTERS.

[Illustration: 

(1)(5)(8)    (7)(9)
(2)(3)    (4)(6)

]

I have nine counters, each bearing one of the nine digits, 1, 2, 3, 4, 5, 6, 7, 8 and 9.  I arranged them on the table in two groups, as shown in the illustration, so as to form two multiplication sums, and found that both sums gave the same product.  You will find that 158 multiplied by 23 is 3,634, and that 79 multiplied by 46 is also 3,634.  Now, the puzzle I propose is to rearrange the counters so as to get as large a product as possible.  What is the best way of placing them?  Remember both groups must multiply to the same amount, and there must be three counters multiplied by two in one case, and two multiplied by two counters in the other, just as at present.

82.—­THE TEN COUNTERS.

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.

83.—­DIGITAL MULTIPLICATION.

Here is another entertaining problem with the nine digits, the nought being excluded.  Using each figure once, and only once, we can form two multiplication sums that have the same product, and this may be done in many ways.  For example, 7 x 658 and 14 x 329 contain all the digits once, and the product in each case is the same—­4,606.  Now, it will be seen that the sum of the digits in the product is 16, which is neither the highest nor the lowest sum so obtainable.  Can you find the solution of the problem that gives the lowest possible sum of digits in the common product?  Also that which gives the highest possible sum?

84.—­THE PIERROT’S PUZZLE.

[Illustration]