a) Describe all positive integers that have exactly 3 positive divisors. b) How many positive divisors of 8400 have at least 4 positive divi

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

$${factor}{\left({\mathtt{8\,400}}\right)} = {{\mathtt{2}}}^{{\mathtt{4}}}{\mathtt{\,\times\,}}{\mathtt{3}}{\mathtt{\,\times\,}}{{\mathtt{5}}}^{{\mathtt{2}}}{\mathtt{\,\times\,}}{\mathtt{7}}$$

there are 8 prime factors here. (including duplicates)

Any one of these by itself will only have 2 factors.

The product of 2 identical ones will have only 2 factors (As Zac has already explained)

The product of any other combination will have 4 or more factors.

So products of 2 different primes:  4C2 = 6

Product of 3 primes:  222,223,225,227,557,553,552,4C3 = 11

Product of 4 primes:  2222,2223,2225,2227,2255,2235,2237,2257,5523,5527,5537,2357 =12

Product of 5 primes:  22223,22225,22227,22235,22237,22255,22257,22355,22357,23557=10

Product of 6 primes:  222235,222237,222255,222257,222355,222357,222557,223557 = 8

Product of 7 primes:  2222355,2222357,2222557,2223557 = 4

Product of 8 primes:  22223557 = 1

Total =

$${\mathtt{6}}{\mathtt{\,\small\textbf+\,}}{\mathtt{11}}{\mathtt{\,\small\textbf+\,}}{\mathtt{12}}{\mathtt{\,\small\textbf+\,}}{\mathtt{10}}{\mathtt{\,\small\textbf+\,}}{\mathtt{8}}{\mathtt{\,\small\textbf+\,}}{\mathtt{4}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}} = {\mathtt{52}}$$            

That is assuming I didn't miss (or double count) any      

And YES there must be a better way of counting these but I don't know what it is.  

a) A square of a prime number,

Reason: primes are the only numbers to have 2 divisors (1 and itself).

To get an odd-number of primes, the value must be a square number (Such as 25 has the divisors 5*5. Since these are both just "5" this is only going to provide 1 extra divisor, giving an odd number of divisors)

So, the only way to remain as square number which is indivisible by anything besides it's square root (and 1 and itself, of course), it's got to be a product of primes.

Is it 52 of 53? Help????????????????????????

b)  We begin by noting the rules for an integer to have more than four divisors.
The smallest number that has 4 divisors is 6. So, all of the divisors of 8400 that have at least 4 divisors must be greater than or equal to 6. 
Also, the divisor cannot be a squared prime, as these numbers have only 3 divisors. 
Finally, the divisor cannot be prime, since all prime numbers have only 2 divisors. 

 Now, we apply our rules to the divisors of 8400.
The prime factorization of 8400 is 2^4*3*5^2*7. 
Since the divisor must be greater than or equal to 6, it cannot be 2, 2^2, 3, or 5. 
Since the divisor cannot be a squared prime, it cannot be 5^2.
Since the divisor cannot be prime, it cannot be 7, the only prime factor we haven't already cancelled out. 
These are six unwanted divisors. 
Finally, the divisor cannot be 1, since 1 is a divisor of every integer and has less than 4 divisors. 
So, these 7 divisors have less than 4 positive divisors. 

Now we must find the total number of divisors of 8400. To do this, we add 1 to each exponent: 2^4 becomes 2^5, 3^1 becomes 3^2, 5^2 becomes 5^3, and 7^1 becomes 7^2. Now, we multiply all these new exponents: 5*2*3*2 = 60. The total number of divisors of 8400 is 60. 

Subtracting our 7 unwanted divisors from the 60 total divisors yields \(\fbox{53}\) divisors that have at least 4 positive divisors.