Factorial Fragments
Problem
Think about this expression
$$\log n!$$
What can you say about it?
Try some values of $n$.
Choose a base or work generally. What happens if you change the base?
What do you notice?
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Visit the site at undergroundmathematics.org to find more resources, which also offer suggestions, solutions and teacher notes to help with their use in the classroom.
Getting Started
$$\begin{aligned}
\log_2 3! \, &= \log_2 (1\times 2\times 3) \\
&= \log_2 1 + \log_2 2 + \log_2 3 \\
&= 1+\log_2 3
\end{aligned}$$
and
$$\begin{aligned}
\log_3 6!\, &= \log_3 (1\times2\times3\times4\times5\times6) \\
&= \log_3 1 + \log_3 2 + \log_3 3 + \log_3 4 +\log_3 5 + \log_3 6 \\
&= 4\log_3 2 + 2 + \log_3 5
\end{aligned}$$
- What do you notice about the cofficients?
- Do you always get an integer term, whatever base you use?
- If you knew $\log_2 12!$ could you write down what $\log_4 12!$ is?
- What could $n$ be if $\log_5 n!$ has integer term $3$ when expanded as far as possible?
- What can you say about $n$ if $\log_6 n!$ has integer term $4$?
- If you know that $p$ and $q$ are primes larger than $7$ and $a$ is not prime, can you fill in the boxes in the equation below?
$$\log \square! = a\log 2 + \square \log 3 + 3 \log 5 +\square \log 7 + \log p +\log q$$
Note that the numbers in the boxes may or may not be the same. Can you make up a similar question?
Underground Mathematics is funded by a grant from the UK Department for Education and provides free web-based resources that support the teaching and learning of post-16 mathematics. It started in 2012 as the Cambridge Mathematics Education Project (CMEP).
Visit the site at undergroundmathematics.org to find more resources, which also offer suggestions, solutions and teacher notes to help with their use in the classroom.
Student Solutions
Maxwell from Samuel Gilbert Public School in Australia has got us started on this problem by noticing two things:
$\log{n!}=\log{n}+\log{(n-1)}+\log{(n-2)}+....+\log{2}+\log{1}$
and
$\log_a{n!}=m+\log_a{\frac{n!}{a^m}}$
The latter observation allows us to see that the integer term will always be the highest power of $a$ that goes into $n!$, so in particular there may not always be an integer term if $a$ is not a factor of $n!$.
Thomas from BHASVIC Sixth Form College in Brighton has furthered this arguement:
If $a$ does not divide $n!$, then there cannot be any integer term.
He also managed to solve the equation
$$\log n!= a \log 2 + \square \log 3 + 3 \log 5 + \square \log 7 + \log p + \log q$$
for $p < q$ both primes greater than $7$ and $a$ not prime.
First use fact that $n!$ has every integer less than and equal to $n$ as a factor. This means that $p$ and $q$ must be the next two primes after $7$, since otherwise these would be missing from the sum, even though they are factors of $n!$ (obviously need $n \geq q$). So $p=11$, $q=13$, and we must have $n \geq 13$.
We must also have $n<17$ since this prime does not appear in the sum, and $n \geq 15$ since we need $5^3$ to divide $n$, which leaves only two possibilities, $n=15$ and $n=16$.
A quick look at the $\log 2$ term tells us that for $n=15$ we get $a=11$ which is prime, but for $n=16$ we have $a=15$, so $n=16$ is the only possibility. Now calculating the empty boxes, get that $\log 16! = 15 \log 2 + 6 \log 3 + 3 \log 5 + 2 \log 7 + \log 11+ \log 13$.
If you found thinking about this problem interesting, you might like to look up an approximation called Stirling's formula:
$$\log n! \sim n\log n - n$$