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Happy Venndredi all,

A relatively straightforward one! The three overlapping ellipses form seven curved regions and there are seven tiles. Your task is to place exactly one tile per region so that the tiles in each ellipse satisfy the corresponding rule. Can you find a configuration that works and if so is it unique?

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1 Answer 1

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The only foursomes that multiply to one less than a square are: $$1\times2\times3\times4=5^2-1\\ 1\times2\times3\times8=7^2-1\\ 1\times2\times4\times6=7^2-1\\ 1\times4\times8\times9=17^2-1\\ 2\times3\times6\times8=17^2-1$$

Clearly $x\ne5,7$: else we can't find four of these numbers adding to $x-1$. So $x=17$.

Suppose the four numbers used in the bottom oval are $1, 4, 8, 9$. That leaves $2, 3, 6$. For the upper-left oval to have $2$ less than than the upper-right, we need the upper-left to have $2, 8$ or $4, 6$ to the other's $3, 9$, or to have $3, 9$ to the other's $6, 8$. None of those possibilities has numbers available for the middle-top two segments that will allow the sums to be $16$ and $18$.

So the four numbers used in the bottom oval are $2, 3, 6, 8$. That leaves $1, 4, 9$. For the upper-left oval to have $2$ less than than the upper-right, we need the upper-left to have $1, 8$ to the other's $2, 9$, or to have $3, 4$ to the other's $1, 8$, or to have $4, 6$ to the other's $3, 9$. Only the first of those possibilities has numbers available for the middle-top two segments that will allow the sums to be $16$ and $18$: $3$ and $4$. So we obtain (in reading order, right to left and top to bottom,) 1, 4, 9, 8, 3, 2, 6

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