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a) Describe all positive integers that have exactly 3 positive divisors.

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

 Apr 14, 2018

I believe this is answer already, but I think I have a better solution:


$\bold{Part (a)}$: A Logical Approach


First of all, a number with an odd number of positive divisors must be a perfect square.


For example, $9=3^2$, and its factors are: 1, 3, and 9. A total of 3 numbers. 


However, a number that is not a perfect square, will have an even number of factors.


For example, $6=2\cdot3$, and its factors are : 1, 2, 3, and 6. A total of 3 numbers. 


Logically, this makes sense, because when you look at 6’s factors, $1\cdot6=6$ and $2\cdot3=6$.


When you look at 9’s factors, $1\cdot9=9$ and $3\cdot3=9$.


In the factors of 6, every number has another number to pair itself with, making pairs of 2, which makes an even number of



On the other hand, in the factors of 9, 1 and 9 can pair with each other to make 9, but 3 must pair with itself to make 9, hence

the even number of factors. 


This problem does not ask of odd number of factors, but asks for 3 factors. This must mean that the number must be a prime

number squared. This is because when you take a composite number, $n$, and square it, you will come across miscellaneous factors other than 1, $\sqrt{n}$. and n. For example, take 16 or $4^2$. The factors are 1, 2, 4, 8, and 16. There are an odd number of factors, just not exactly three. 


This is because when you are taking $4^2$, you are actually take (2^2)^2, and you will see other powers of two, for example

$2^3$, 8.


So the answer for part (a), is a positive integer that have exactly 3 positive divisors, must be the square of a prime number:


$$p^2$$where $p$ is any prime.




The first step in this problem is to prime factorize 8400. 




Instead of counting all of the possible numbers, it is easier to count the numbers that have three divisors or fewer.


1, has one divisor.


All the prime numbers: 2,3,5, and 7 have two divisors. 


All the perfect squares of primes: 4 and 25 have three divisors. 


Now the rest have four or more divisors.


There are a total of $(4+1)(1+1)(2+1)(1+1)=60$ positive divisors of 8400.


But since this list includes those who are four or less, the answer will be:




I hope this helped!

 Apr 14, 2018
edited by GYanggg  Apr 14, 2018

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