How many positive divisors of 8400 have at least 4 positive divisors? How many positive divisors of 8400 have at least 4 positive divisors?
I'm a little bit fuzzy on number theory, so could someone please help me out?  Thanks!
 A: Seems easier to list out the ones that have three divisors or fewer, then count up the rest.
The prime factorization is $2^4 \cdot 3 \cdot 5^2 \cdot 7$.
$1$, of course, has one divisor.
$2,3,5,7$ are prime, so they have only two divisors.
$4$ and $25$ are squares of primes and have three divisors.
All the rest have four or more.  There are a total of $5 \cdot 2 \cdot 3 \cdot 2 = 60$ divisors, including the ones above.
A: $$8400=2^4\cdot3\cdot5^2\cdot7$$ has $5\cdot2\cdot3\cdot2=60$ divisors and $1,2,3,5,7,2^2$ and $5^2$ have $1,2$ or $3$ divisors. Hence the answer is
$$60-7=\color{red}{53}$$ 
A: So the prime factorization of 8400 is 2^4 * 3^1 * 5^2 * 7^1.
The total number of divisors is 60.
To make it easier to find the values of the divisors that have less than 4 divisors, we know that 6 is the smallest number that has 4 positive divisors, so we can eliminate any multiples of the primes less than 6, that are: 2, 3, 2^2, and 5. We have 4 so far.
We can also eliminate 7 because all primes have only two divisors, 1 and itself. We now have 5.
But the squares of primes always have 3 divisors (example: 9, 4, 49), but the only primes which can be squares are 2 and 5 (In the prime factorization, 3 and 7 appear only once), so we now have 7.
But we can't forget 1, and we have 8 divisors that have less than 4 divisors.
So we have 60 - 8 = 52 divisors.
(Basically John's but he missed 4)
