Question 1194009
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Let's focus on n = p^2*q for now.


The unique primes are p and q
Though we have p show up twice in the form p^2


The divisors are:
1, p, q, p*q, p^2, p^2q
Clearly there are more than 3 divisors here so we move on.


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With n = p*q, the divisors are: 
1, p, q, p*q
We get close to 3 divisors, but we have one too many.


Now onto n = p*q*r
Divisors: 1, p, q, r, p*q, p*r, q*r, p*q*r
There are 8 divisors here, which we rule this answer choice out as well.
Thing to notice: There are 3 atomic pieces of p,q,r so there are 2^3 = 8 different divisors. 
Why does this work? I'll leave it for you to think about. Hint: Think of the power set.


Lastly n = p^2
The divisors are: 1, p, p^2
This is exactly 3 divisors, so you have chosen the correct answer. This is the only answer that fits the 3 divisors pattern.


In summary, you are correct to think choice 4 is the only answer. The quick reasoning is that squaring any prime will have exactly 3 factors.


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Some concrete numeric examples:


For n = p^2*q we could pick p = 5 and q = 7
n = 5^2*7 = 175
Divisors: 1, 5, 7, 25, 35, 175
Number of divisors: 6


Now onto the form n = p*q
Let's go with p = 11 and q = 13
n = 11*13 = 143
Factors: 1, 11, 13, 143
Number of divisors: 4


Form: n = p*q*r
Let p = 2, q = 3, r = 7
n = 2*3*7 = 42
Divisors: 1, 2, 3, 6, 7, 14, 21, 42
Number of divisors: 8


Form: n = p^2
Let p = 17
n = p^2 = 17^2 = 289
Divisors: 1, 17, 289
Number of divisors: 3


A handy tool to use is WolframAlpha. 
Type in something like "divisors of 289" and it will give the correct list in increasing order.
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