the other three problems were answered, this one has not been answered yet. Do not read the comments if you want to try.https://twitter.com/BarbarianCap/status/1236411619174948865 …
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a+c+f+i can be expressed as 2n+1 b+c+d+e can be expressed as 2m+1 a+d+g+j can be expressed as 2o+1 b+f+h+j can be expressed as 2p+1 i+h+g+e can be expressed as 2q+1 where n...q are integers > 0
Sum all of the above sums 2a+2b+2c + ... + 2i + 2j = 2n+1+2m+1+2o+1+2p+1+2q+1 = 2(n+m+o+p+q) + 5 = 2(n+m+o+p+q+2) + 1 The sum of sums can be express as 2r+1, therefore it's odd, but this is impossible because it equals 2a+2b+2c + ... + 2i + 2j Contradiction. It's impossible.
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Did I pass Russian math, grade 2?
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Same. If all sums are odd, then the total is odd. But then you also have that the total is 2x(a+b+c...) which is even. So, not possible. (not grade 2 btw, no grade assigned)
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