You can use math to help you solve puzzles like this!
The technique is illustrated with an example.
Suppose you are given the digits $\,1\,$ through $\,9\,,$ and you want each side of the triangle to sum to $\,23\,.$
If we can figure out the numbers that go in the corners, then it's easy from there.
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Let the numbers in the corners be $\,a\,,$ $\,b\,,$ and $\,c\,.$ Let $\,x\,$ represent the sum of the numbers between $\,a\,$ and $\,b\,.$ Let $\,y\,$ represent the sum of the numbers between $\,a\,$ and $\,c\,.$ Let $\,z\,$ represent the sum of the numbers between $\,b\,$ and $\,c\,.$ Since the sum along each side must equal $\,23\,,$ we have $$ (a + x + b) + (a + y + c) + (b + z + c) = 3(23) = 69 $$ Solving this equation for $\,x+y+z\,$ gives $$ x + y + z = 69 - 2(a + b + c) $$ Since the digits $\,1\,$ through $\,9\,$ must be used to fill all the circles, we also have $$ a + x + b + z + c + y = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45 $$ Solving this equation for $\,x+y+z\,$ gives $$ x + y + z = 45 - (a + b + c) $$ Let $\,S := a + b + c\,.$ Equating the two expressions for $\,x+y+z\,$ gives $$ 45 - S = 69 - 2S $$ Solving for $\,S\,$ yields $\,S = 24\,.$ The only three available numbers that sum to $\,24\,$ are $\,7\,,$ $\,8\,,$ and $\,9\,,$ so these must go in the corners. Then, the rest is easy! |
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