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.

Amusements in Mathematics · 第 101 / 970 题