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?
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$\begingroup$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.
$\endgroup$ 3 $\begingroup$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.
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