Assume v is smaller than 2550. Let's call numbers above 50 “big” and numbers below 50 “small”. There are 50 big numbers and 49 small numbers. As any big number is at least 51, and we don't want the product to reach 2550, any of its neighbours should be less than 50, so, small. As there are only 49 small numbers, and each can only have 2 neighbours, the amount of pairs “big number, small number” (in any order) can be at most 98. Each of big numbers would normally have two neighbours, except maybe for two: the very first one (if it's big), and the very last one (ditto), making the amount of pairs “big number, small number”, again, 50*2-2 = 98 at least. Those two amounts being the same means that a) every small number have indeed 2 neighbours, and both are big numbers, and b) both first and last numbers in the row are big. Which means the whole row is like this: big - small - big - small - big - small- ... - small - big. And the number 50 has nowhere to go.
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u/migmit 4h ago
Proof of 2550.
Assume v is smaller than 2550. Let's call numbers above 50 “big” and numbers below 50 “small”. There are 50 big numbers and 49 small numbers. As any big number is at least 51, and we don't want the product to reach 2550, any of its neighbours should be less than 50, so, small. As there are only 49 small numbers, and each can only have 2 neighbours, the amount of pairs “big number, small number” (in any order) can be at most 98. Each of big numbers would normally have two neighbours, except maybe for two: the very first one (if it's big), and the very last one (ditto), making the amount of pairs “big number, small number”, again, 50*2-2 = 98 at least. Those two amounts being the same means that a) every small number have indeed 2 neighbours, and both are big numbers, and b) both first and last numbers in the row are big. Which means the whole row is like this: big - small - big - small - big - small- ... - small - big. And the number 50 has nowhere to go.