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Can you write a number from 1 to 16 in each of the triangles, using each number exactly once, such that the sum of the two numbers in the two cells that share an edge is always odd?

triangle as such:

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Took a look at the picture. It appears the answer is no. If the sum of the numbers in adjacent triangles is odd, then one triangle contains an odd number while the other contains an even. Light green triangles are adjacent only to dark green triangles and vice-versa. So if light green triangles contain odd numbers, the dark green triangles contain even numbers. There are equal numbers of odd and even numbers in the range 1 through 16, but there are different numbers of dark and light green triangles.

By the pigeonhole principle, one of the dark green triangles must contain a number of the wrong parity. Therefore, there will be an even sum somewhere.

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No, if a corner tringle takes odd value, it forces 9 of the triangles to take odd values and 6 tirangles to take even values.But there are only 8 odd values from 1-16. Hence not possible to fill the triangles with distinct values between 1-16.

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Start with E at the position 7, this determines a pattern of E and O for the remaining cells so that all cells with a common edge are different types. This E must be surrounded by all O cells, etc. The resulting pattern has 10 Es and 6 Os. If we place an O at position 7 then we get 10 Os and 6 Es. In any case the placement of all distinct 16 numbers is impossible.

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9 and 7 or 10 and 6? – Mike Jun 15 '12 at 19:14
Sorry--10 and 6 – i. m. soloveichik Jun 15 '12 at 19:37

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