\begin{align} & 1\cdot 72 \\ & 2\cdot 36 \\ & 3\cdot 24 \\ & 4\cdot 18 \\ & 6\cdot 12 \\ & 8\cdot 9 \end{align} When the divisors of a number are listed in this way, let us call the numbers on the left the "small" divisors and those on the right the "large" divisors.
When one multiplies four consecutive integers, e.g. $17\times18\times19\times20 = 116280$, then the largest small divisor and the smallest large divisor must differ by only $2$ because they are $17\times20$ and $18\times 19$, the outer two and the middle two.
Thus among the small divisors one must find two that differ by only $1$, namely half of each of the above two numbers. In our example these are $170$ and $171$. If I'm not mistaken, no two small divisors bigger than those can be so close together.[This remark was mistaken; see the postscript down below.]
However, going through several instances I find what seems like an unusually large number of small divisors differing by only $1$.
In this present example, I find that $152$ and $153$ are both divisors of $116280$.
With $18\times19\times20\times21$ the two that "must" be there are $189$ and $190$, and we also find $56$ and $57$.
With $19\times20\times21\times22$ the two that "must" be there are $209$ and $210$, and we also find $132$ and $133$, and $76$ and $77$.
I tried this with a bunch of other examples. But I don't have any good guesses as to how frequently consecutive numbers occur as small divisors when one uses a more-or-less random composite number.
Is there anything to my guess that an unusually large number of consecutive numbers are small divisors when one multiplies four consecutive integers, not including the pair that "must" be there?
http://www.wolframalpha.com/input/?i=divisors+116280
PS: Earlier I wrote above "If I'm not mistaken, no two small divisors bigger than those can be so close together." That was an error. Here is a counterexample: $12\times13\times14\times15=32760$ and the pair that "must" be there is $90$ and $91$. But another such pair actually consists of larger numbers: $104$ and $105$. There are also the pairs $20$ and $21$, $35$ and $36$, $39$ and $40$.