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I'm having trouble with a math problem. I need to arrange 6 numbers on a certain diagram:

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At every intersection of two circles, I have to put one of these six numbers: 4, 5, 5, 6, 6, or 7. The sum of all of the numbers on each circle must be the same. Is it possible to arrange these numbers this way?

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You mean no number that belongs to all three? –  Jason Chen May 9 at 4:06
    
@JoelReyesNoche: Oh, you misunderstood. Each section does not get its own number. The intersections get numbers, and each number is on one of the circles. –  Jason Chen May 9 at 4:08

2 Answers 2

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The sum of the numbers is $33$. Each occurs twice, total $66$. Three circles, four numbers adding up to $22$ on each. The $7$ must be matched with $6,5,4$. The $7$ is on the intersection of two circles; each of these circles must include a $4$; but there is only one $4$, so it must be on the other intersection of the same two circles. The rest is easy, and there are various solutions.

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Edit: I misunderstood the question. Anyway, I'm keeping my answer here in case it provides additional ideas...

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The sum of the numbers in each circle is 17.

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Technically, all of the numbers are in intersections, and the diagram just makes it look like it's "alone." –  Jason Chen May 9 at 4:10
    
This wasn't actually what the question looked like. You need to put each number on the circumference of the circles, where the circles meet. –  Jason Chen May 9 at 4:15

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