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Sixteen cards numbered 1 to 16 lay on a table. Anne and Clare each take four cards. Anne's cards add up to 23, while Clare's add up to 48. Bernard and David also take four cards each. The product of Bernard's cards is 4480, while the product of David's cards is 1050.

Which four cards did each of the four friends take?

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We start with

$4480 = 2^7 \times 5 \times 7$
$1050 = 2 \times 3 \times 5^2 \times 7$

which means that

the cards of B and D contain $5, 10, 15, 7, 14$

and thus

the product of the remaining three cards of B and D is $2^6$.

On the other hand, we also know that

the sum of the remaining three cards of B and D is $14$, by calculation,

which leads to

the remaining three cards of B and D are $2, 4, 8$.

Now D has

three cards that are multiples of either $5$ or $7$, hence the remaining card is a power of $2$,

which can only be

$2$.

This uniquely determines the cards of D:

$2, 5, 7, 15$,

and hence the cards of B:

$4, 8, 10, 14$.

For A and C, we note that

their cards are $1, 3, 6, 9, 11, 12, 13, 16$

which means that

$1, 3, 6$ should all belong to A, otherwise the smallest sum would be $1 + 3 + 9 + 11 = 24$.

Therefore the cards of A are

$1, 3, 6, 13$

and the cards of C are

$9, 11, 12, 16$.

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