How many positive divisors of 8400 have at least 4 positive divisors?

Question

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!

Answer 1

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.

Answer 2

$$8400=2^4\cdot3\cdot5^2\cdot7$$ 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 $$60-7=\color{red}{53}$$

Answer 3

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)