How many natural numbers less than 200 will have 12 factors (a.k.a. divisors)?
I think the answer is $11$.
Firstly there can be at most $3$ distinct prime factors.
$12=1\cdot12
=2\cdot6
=3\cdot4
=2\cdot2\cdot3$
$N=a^{11}
=a\cdot b^5
=a^2\cdot b^3
=a\cdot b\cdot c^2$
Then, $1$ prime factor is not possible because the smallest $2^{11}> 200$. So, the options are:
For $2$ prime factors:
$72, 96, 160$
For $3$ prime factors:
$60,90,150,84,140,126,132,156$.
I think this list is exhaustive. But am I right? Is there any other approach?
Best Answer
I know you guys love your maths (and I do too) but, for something as finite as this (i.e., not something that would consume years of compute time), you can test it with an exhaustive computer program, one that takes about three thousandths of a second on my desktop:
The relevant lines of that output are:
The complete output of that is shown below, and I'll ask in advance for forgiveness for butchering the English language with the phrase
"Quantity with 1 factors is 1"
but I don't see the need for making the program grammatically aware for what is, in essence, a one-off activity.Or you can just take that to mean I'm basically lazy :-)