Here’s an interesting little problem that came across my desk this afternoon: how much time is \(10!\) seconds? Is it a duration that is best measured in seconds? days? centuries? And, perhaps more importantly, what is the best way of figuring it out? Think about it for a minute, then check below the fold for the answer, which is a surprisingly round number!
One possible way to answer the question is to multiply out \(10!\) using a calculator or other device, then multiply by the appropriate conversion factors to first convert to minutes, then hours, then days, and so on, until something tractable pops out.
On the other hand, \(10!\) is a large number, and multiplying it out is impractical without a calculator. Moreover, there is something inelegant about all of that computation. My approach, which uses unit analysis and prime factorization, is as follows:
\[\begin{aligned}
10!\text{ sec}
&= \frac{10!\text{ sec}}{1} \cdot \frac{1\text{ min}}{60\text{ sec}} \cdot \frac{1\text{ hr}}{60\text{ sec}} \cdot \frac{1\text{ day}}{24\text{ hr}} \cdot \frac{1\text{ week}}{7\text{ day}} && \text{(1)}\\
&= \frac{10!}{1\cdot (2\cdot 2\cdot 3\cdot 5)\cdot (2\cdot 2\cdot 3\cdot 5) \cdot (2\cdot 2\cdot 2\cdot 3) \cdot 7}\text{ weeks} && \text{(2)}\\
&= \frac{10!}{(5\cdot 2)\cdot (3\cdot 3)\cdot (2\cdot 2\cdot 2)\cdot (7)\cdot (5)\cdot (2\cdot 2)\cdot (3)\cdot (2)\cdot 1}\text{ weeks} && \text{(3)}\\
&= \frac{10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}{10\cdot 9\cdot 8\cdot 7\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}\text{ weeks} && \text{(4)}\\
&= 6\text{ weeks}.
\end{aligned}\]
In line (1), we write out the conversion factors that we are going to need. Initially, I wasn’t sure how much would be necessary, but I figured that having a 7 in the denominator might be useful (since \(10!\) has a factor of 7), so I went out to weeks. It turns out that this was the Right Thing to Do™.
Next, we cancel all of the units (seconds in the numerator cancel seconds in the denominator, and so on), leaving only weeks in the numerator and no units in the denominator. We also factor the terms in the denominator, which will prove to be useful in the next step. The result is given in line (2), where each set of grouped terms represents one of the conversion factors.
The number \(10!\) is the product of all of the natural numbers up to 10, i.e. 1,2,…,10. Since all of those numbers must be factors of the numerator, we try to group the terms in the denominator so that we get some cancelation. This rearrangement of denominator is given in line (3), then simplified in line (4). Finally, we simplify obtaining 6 weeks!