How many positive integers from $1-1000$ have $5$ divisors?

Any answers would be greatly appreciated. If you have any questions, I will edit for clarification.


closed as off-topic by T. Bongers, Sahiba Arora, Claude Leibovici, user99914, user223391 Jan 6 '18 at 17:27

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    $\begingroup$ Only perfect squares have an odd number of divisors. Of the perfect squares, what characteristics separates perfect squares with 3 divisors, 5 divisors and more than 5 divisors? $\endgroup$ – Doug M Jan 6 '18 at 0:57
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    $\begingroup$ Exactly five? Or at least five? $\endgroup$ – JakeRobb Jan 6 '18 at 4:56
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    $\begingroup$ Ask Python : [n for n in range(1001) if sum(1 for d in range(1, n+1) if n % d == 0) == 5]. It returns [16, 81, 625]. $\endgroup$ – Eric Duminil Jan 6 '18 at 10:05

(Filling in some details, on the off chance that other answers are unfamiliar to potential readers!)

Consider the prime factorization of $75 = 3^1 \cdot 5^2$. How many factors does $75$ have? Well, in forming a factor of $75$, we have to choose how many $3$s to include - there is one available, but we could also choose none; so, the number of choices is $1+1=2$ - and how many $5$s to include - there are two available, but we could also choose none; so, the number of choices is $2+1=3$. The aforementioned $2$ choices and $3$ choices yield $2 \cdot 3 = 6$ combinations using the prime factors, and these are all the factors of $75$:

$$3^0 \cdot 5^0, 3^0 \cdot 5^1, 3^0 \cdot 5^2, 3^1 \cdot 5^0, 3^1 \cdot 5^1, 3^1 \cdot 5^2$$

More generally, a number with prime factorization $p^a \cdot q^b$ for distinct primes $p$ and $q$ has $(a+1)(b+1)$ factors, since you can form these factors by making one of $a+1$ choices for how many $p$s to include (ranging from $0$ to all $a$) and one of $b+1$ choices for how many $q$s to include (ranging from $0$ to all $b$). Even more generally, you arrive at the formula in Bernard's answer.

So: In one's investigation of $5$, it becomes clear that this can arise as the number of factors if, and only if, it came from a number of the form $p^4$ for a prime $p$. All that remains now is to observe how many primes to the fourth power are present between $1$ and $1000$; this is precisely what is done in Mohammad Riazi-Kermani's answer: just $2^4 = 16$, $3^4 = 81$, and $5^4 = 625$. Already we have $6^4 = 1296 > 1000$, so the three listed numbers are exhaustive.

As a follow up exercise, you may wish to count how many factors $4^4$ and $6^4$ have.

(Which would you guess has more? Etc.)



If an integer $n$ has a decomposition as a product of distinct prime powers: $$n=\prod_{i=1}^r p_i^{\alpha_i},$$ the number of divisors of $n$ equal to $\:\displaystyle\prod_{i=1}^r(\alpha_i+1)$.

If this product is equal to $5$, what can you conclude?


The only way that a number has exactly $5$ positive divisors is for the number to be a fourth power of a prime number. Thus the answer is 3. The numbers are $$ 2^4,3^4,5^4$$


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