A Question that I need help for

A cube has its vertex named A,B,C,D,E,F,G,H. All of them are real positive numbers. A near sum is the sum of all numbers of the vertexes that connects to the original vertex. The value of the near sums are (3,6,9,12,15,18,21,24). If the value of the near sum of A is 1, then what is the value of the near sum of F?

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TopNewestJust to clarify, is \(1\) the

near sumof \(A\) or thevalueof \(A?\) Because it seems that there are exactly 8 near sums you listed out, none of which are \(1.\) – Steven Yuan · 2 years, 3 months agoLog in to reply

@Alfian Edgar Notice that the near sums have complements which sum to 27, ie the near sum of a vertex is 27 minus the near sum of the opposite vertex; shouldn't be too hard to prove. Then, \(nearsum(F) = 27-nearsum(A)\).

But if \(A = 1\), then I don't know. My brain isn't working today. – Jake Lai · 2 years, 3 months ago

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– Math Man · 2 years, 3 months ago

lol, then nearsum(F)= 27-3 = 24Log in to reply

I'm sorry, what I meant was near sum of a=3, the original question was using the average of the near sums lol. It's my mistake. – Alfian Edgar · 2 years, 3 months ago

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– Jake Lai · 2 years, 3 months ago

Then hopefully my reply to Steven will help you out a bit. Good luck and happy solving!Log in to reply