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I derive the answer G by randomly choosing four integers multiple times, which is a bit time-consuming for ACT Math. Could anyone offer me a quicker method to cope with it?

HypnoticBuggyWraithVirileBevy
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2 Answers2

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I is false since $24, 25, 26, 27$ are all not prime. To see this, I simply went through the primes in my head $2,3,5,7,11,13,17,19,23,29,\ldots$ until I found two which were more than four apart.

III is false since none of $4,5,6,7$ is a factor of the others. To see this, I simply noted that once $n$ gets "large enough", $n-3$, $n-2$, $n-1$, and $n$ are all bigger than $\tfrac{n}{2}$, which means none of them are small enough to be a factor of the others.

II is true since two of the four consecutive numbers must be even, and therefore have $2$ as a common factor.

Hence, the answer is G.

PM 2Ring
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JimmyK4542
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    Another fun way to see that I is false is that you can make arbitrarily large prime gaps with the set $(n!+2,n!+3,\dots ,n!+n)$, since all of these numbers are composite since you can factor out the addend. – Ty Jensen Dec 09 '20 at 04:06
  • Thanks for rolling back the answer to version 1. Also, why exactly does II imply that I is false? Two of the numbers having a common prime factor doesn't stop either of the other two from being prime. – JimmyK4542 Dec 09 '20 at 04:19
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    @ChaitanyaSharma Your edits were incorrect here. "I" is referring to item I in the question, so "I is false", not "I am false". "is all bigger than" is also incorrect since there are multiple subjects. (Also on another suggested edit you turned multiple text blocks into a heading, making the text huge. Don't do that.) – epimorphic Dec 09 '20 at 04:32
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I) should be rejected outright. It should be well known that we can have a gap in primes of more than $4$. There is the excercise that $n! + 2$ through $n! + n$ will be $n-1$ consecutive digits with no primes. But even if you don't think of that the statement should just seem absurd. After all no even number over $2$ is prime so that would mean every other odd number is prime and surely that can't be true! $25$ and $27$ are not prime so $24,25,26,27$ are a counter example.

II) is clear when you realize $2$ is prime and two of the numbers must be even. (Otherwise $3$ being a common factor of the first and last is the only other possible option but its not guarenteed.)

III) is clearly false. If two numbers have a common factor, $k$ the must be a multiple of $k$ apart. As four consecutive numbers are at most $3$ apart $1,2,3$ are the only possible factors they can have in common and if these numbers are larger than $3$ this statement is false.

So II is the only correct one.

fleablood
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