Which three-digit number has the greatest number of different factors?
This question comes from an Olympiad, where four marks were given for the solution to this problem. The textbook lists an accepted answer as:
Let $N$ be the three-digit number which has the greatest number of different factors. Since $2\times3\times5\times7\times11>1000$, $N$ at most has four different prime factors.
If $N$ only has one prime factor, then $N=2^9$, and so $N$ has 10 different factors.
If $N$ has exactly two prime factors, then $N=2^5\times3^3$, and so $N$ has 24 different factors.
If $N$ has exactly three prime factors, then $N=2^4\times3^2\times5$, and so $N$ has 30 different factors.
If $N$ has exactly four prime factors, then $N=2^3\times3\times5\times7$, and so $N$ has 32 different factors.
Thus, $N=2^3\times5\times7=\bf840$.
What I don't understand is why $N=2^5\times 3^3$ when $N$ has two prime factors, and similarly why $N=2^3\times3\times5\times7$ when N has three factors. I think it has to do with the maximum 'arrangement' of factors such that $N$ has the most factors possible, but I don't understand how we can calculate that. Furthermore, could we solve a similar problem involving a four-digit number instead using the above method?
As the number must be $N<1000$ its prime factors must be $2\times 3\times 5 \times 7$ when it has two factors it makes no sense considering, for instance, $5^3\times 7$ which is $875$ less than $1000$ but has only $(3+1)(1+1)=8$ factors.
$2^3\times 5 $ is smaller and has the same number of divisors. The maximum number of divisors is attained when $N=2^a\times 3^b$. You can try and see that $d(N)=(a+1)(b+1)$ is maximum and $N<1000$ is the largest when $a=5;\;b=3$, that is $N=2^5\times 3^3= 864$ and $d(864)=6\times 4=24$.
In a similar way you can find the other results and solve the problem with $840=2^3\times5\times7$ which has $32$ divisors.
The issue is called Highly Composed Numbers and if you want to know more look here.