Find 4 consecutive numbers that divide 𝑤, 𝑥, 𝑦, 𝑧 respectively, where 𝑤, 𝑥, 𝑦, and 𝑧 are also 4 consecutive positive numbers greater than 1, or prove it's impossible.
Bonus: What if w > the first consecutive number / 2?
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1$\begingroup$ What about $1, 2, 3, 4$ dividing $1, 2, 3, 4$ respectively? $\endgroup$– WhatsUpMay 2, 2021 at 20:47
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1$\begingroup$ Even if you want them to be different, it's pretty easy to see other examples such as $1, 2, 3, 4$ dividing $13, 14, 15, 16$. I actually don't see the puzzling point here. $\endgroup$– WhatsUpMay 2, 2021 at 20:49
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$\begingroup$ @WhatsUp Ha, I see you made the exact same point as me pretty much simultaneously here! :) $\endgroup$– StivMay 2, 2021 at 20:51
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$\begingroup$ @Stiv Yes, it seems to be the most natural reaction (: $\endgroup$– WhatsUpMay 2, 2021 at 20:52
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$\begingroup$ I was expecting it to exclude 1. $\endgroup$– BoesfMay 3, 2021 at 1:35
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1 Answer
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Yes, and moreover, it's possible (in infinitely many ways) for any 4 consecutive numbers. By the Chinese remainder theorem, if a solution for some 4-tuple of numbers does exist, then there is a whole (infinite) family of solutions differing by a multiple of the LCM (lowest common measure) of the 4 given numbers. But we know that a solution exists, namely a trivial one consisting of the 4 numbers themselves. For example, consider the 4 consecutive numbers 10, 11, 12 and 13. The LCM of them is $2^2\cdot3\cdot5\cdot11\cdot13=8580$, so we can get a solution by adding any multiple of 8580 to (10, 11, 12, 13). For example, using $85800=10\cdot8580$, we get: $85810=8581\cdot10$, $85811=7801\cdot11$, $85812=7151\cdot12$ and $85813=6601\cdot13$.
answered May 2, 2021 at 21:19