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The way to solve an arithmagon is to take the 3 box numbers, add them up and divide by 2. That gives your "center" number. For each corner circle, take the opposite box and subtract from the center number.

As long as you are allowed to have fractions and/or negative numbers, I think this method always works. Do you have an example of an impossible arithmagon?

For example:
a -- 7 -- b
.\ ........ /
. 8 ..... 9
... \ .. /
..... c

Add up 7+8+9 = 24
Divide by 2 = 12

a = 12 - 9 = 3
b = 12 - 8 = 4
c = 12 - 7 = 5

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11y ago

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