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.
Solutions: 1
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