THE NINE COUNTERS.
(
Money Puzzles)
[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.
Answer:
In this case a certain amount of mere "trial" is unavoidable. But there
are two kinds of "trials"--those that are purely haphazard, and those
that are methodical. The true puzzle lover is never satisfied with mere
haphazard trials. The reader will find that by just reversing the
figures in 23 and 46 (making the multipliers 32 and 64) both products
will be 5,056. This is an improvement, but it is not the correct answer.
We can get as large a product as 5,568 if we multiply 174 by 32 and 96
by 58, but this solution is not to be found without the exercise of some
judgment and patience.
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