(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.

THE NINE ALMONDS. THE NINE SCHOOLBOYS. facebooktwittergoogle_plusredditpinterestlinkedinmail