# Determine which positive integers have exactly 36 positive divisors.

Determine which positive integers have exactly 36 positive divisors.


Let $n=p_1^a{_2}p_1^a{_2}...p_s^a{_s}$. Then $\tau(n)= \prod_{j=1}^{s} a_j+1$. We need $\tau(n)=36$. So we list the factor pairs of 36 as, 1*36, 2*18, 3*12, 4*9,and 6*6. The prime factorization of 36 is $2^23^2$. So $\tau(36=2^23^2)=\tau(2^2)\tau(3^2)=(2+1)(3+1)=12$.

Disregard above line, we need tau(n)=36, not tau(36).

I'm going to link a related question (24 divisors) and also show a solution to the question: Determine which positive integers have exactly 4 positive divisors from a Chegg answer, neither of which I understand...

4 divisors

Hey everyone thanks so much for the help here. Here is my final answer for others working on this material.

Let $n=p_1^a{_2}p_1^a{_2}...p_s^a{_s}$. Then $\tau(n)= \prod_{j=1}^{s} a_j+1$. We need $\tau(n)=36$, i.e. $(a_1+1)(a_2+1)...(a_s+1)=36$. Let $p,q,r,s$ be distinct primes. Solving $(a_1+1)(a_2+1)...(a_s+1)=36$, we have $p^{35}, p^{17}q, p^{11}q^2, p^8q^3, p^8qr, p^5q^5, p^5q^2r, p^3q^2r^2, p^2q^2rs$. Checking the following we have \begin{eqnarray*} p^{35} &\rightarrow& (35+1) = 36 \\ p^{17}q &\rightarrow& (17+1)(1+1) = 36\\ p^{11}q^2 &\rightarrow& (11+1)(2+1) = 36 \\ p^8q^3 &\rightarrow& (8+1)(3+1) = 36 \\ p^8qr &\rightarrow& (8+1)(1+1)(1+1) = 36\\ p^5q^5 &\rightarrow& (5+1)(5+1) = 36 \\ p^5q^2r &\rightarrow& (5+1)(2+1)(1+1) = 36 \\ p^3q^2r^2 &\rightarrow& (3+1)(2+1)(2+1) = 36 \\ p^2q^2rs &\rightarrow& (2+1)(2+1)(1+1)(1+1) = 36. \end{eqnarray*} We have checked all possible combination of exponents for primes that are factors of 36. Hence positive integers that have exactly $36$ positive divisors are of the form listed above.

• Are we looking for the least positive integer with this property or are we seeking all positive integers with this property? – user328442 Nov 7 '17 at 0:21
• Specifically what I don't understand about the linking question: take the first answer for example. With the case, 2*2*2*3, why is the answer stating "minimum for this factorization is 2^2⋅3⋅5⋅7"? Notably, where the heck did the 5 and 7 come from and also, how does 2^2⋅3⋅5⋅7 relate to the original number of divisors 24?? – user424890 Nov 7 '17 at 0:22
• @user328442 the latter, all positive integers with said property. For example, the back of my textbook gives, "all squares of primes" as an answer to a similar question. Notably 3 divisors. – user424890 Nov 7 '17 at 0:24
• the first 4 primes are 2, 3, 5 and 7 and so an integer will be relatively small. The important part is that we have 3 choices for how many times 2 shows up in a factorization of a divisor of this integer, we have 2 options for 3, 2 for 5 and 2 for 7. This means that there are 3*2*2*2 =24 divisors of that integer. – user328442 Nov 7 '17 at 0:24
• See, there lies my confusion. 7 does not show up in the prime factorization of 24... so why am I looking into it? Similarly with 5. – user424890 Nov 7 '17 at 0:27

Our goal is to find all integers with exactly 36 divisors. To do this, first note that any positive integer will have a unique prime factorization (up to order in which primes appear, of course). This means that in general, a positive integer $n$ will have the form

$$n = p_1^{a_1} \cdots p_k^{a_k}$$ where each prime here is distinct and each $a_i$ is a positive integer.

So, any divisor of $n$ will have the form

$$d = p_1^{b_1} \cdots p_k^{b_k}$$

where $b_i$ can range between $0$ and $a_i$.

Now, how many ways can choose a $b_i$? Well, there are $a_i +1$ options. So, we have $(a_1+1) \cdots (a_k+1)$ divisors of $n$. If we have $(a_1+1) \cdots (a_k+1) = 24$ then $n$ will have exactly $24$ divisors.

If we are looking for positive integers with exactly 36 divisors then any positive integer with the exponents of its primes satisfying $(a_1+1) \cdots (a_k+1) = 36$ will work. Note that $36 = 2^2 3^2$.

With this in mind, we can have $36 = 2*2*3*3$ which would correspond to 4 distinct primes, two being square and the other 2 being first powers. Another possibility is $36 = 2*2*3^2$ which corresponds to 3 distinct primes in the factorization, two being first powers and one being an eighth power. Do you see the idea here?

All such positive integers will look like $p^{35}, p^{17}q, p^{11}q^2, p^8q^3, p^8qr, p^5q^5, p^5q^2r, p^3q^2r^2, p^2q^2rs$ where $p, q, r, s$ are any distinct primes.

We have infinitely many choices of primes and so we will have infinitely many integers with exactly 36 divisors.

• This all makes sense! Thank you. However, what is the next step? Am I otherthinking this and my conclusion should be simply, "There exist infinitely many exponents of primes such that (a_1+1)...(a_k+1)=36"? – user424890 Nov 7 '17 at 0:58
• I’ll edit my post with a fuller answer. I’m a bit busy at the moment and so I apologize for that. – user328442 Nov 7 '17 at 0:59
• @JasonBennett see my edit – user328442 Nov 7 '17 at 1:14
• You should write how to describe them all. They are (for 24 divisors) $p^{23}$ or $p^{11}q$ or $p^{7}q^2$ or $p^5q^3$ or $p^5qr$ of $p^3q^2r$ or $p^2qrs$ where $p,q,r,s$ can be any primes. – fleablood Nov 7 '17 at 1:15
• Oh gosh I don't know why this was throwing me off so much! I would so hung up with the primes themselves and not the exponents! Thank you so much this answers my question! @user328442 – user424890 Nov 7 '17 at 2:49

the first odd one is 17325. The first one with just two prime factors is $6912 = 2^8 3^3 \; . \;$ The first prime power is $2^{35} = 34359738368$

Mon Nov  6 16:42:52 PST 2017
1260 = 2^2 * 3^2 * 5 * 7
1440 = 2^5 * 3^2 * 5
1800 = 2^3 * 3^2 * 5^2
1980 = 2^2 * 3^2 * 5 * 11
2016 = 2^5 * 3^2 * 7
2100 = 2^2 * 3 * 5^2 * 7
2340 = 2^2 * 3^2 * 5 * 13
2400 = 2^5 * 3 * 5^2
2700 = 2^2 * 3^3 * 5^2
2772 = 2^2 * 3^2 * 7 * 11
2940 = 2^2 * 3 * 5 * 7^2
3060 = 2^2 * 3^2 * 5 * 17
3150 = 2 * 3^2 * 5^2 * 7
3168 = 2^5 * 3^2 * 11
3276 = 2^2 * 3^2 * 7 * 13
3300 = 2^2 * 3 * 5^2 * 11
3420 = 2^2 * 3^2 * 5 * 19
3528 = 2^3 * 3^2 * 7^2
3744 = 2^5 * 3^2 * 13
3840 = 2^8 * 3 * 5
3900 = 2^2 * 3 * 5^2 * 13
4140 = 2^2 * 3^2 * 5 * 23
4284 = 2^2 * 3^2 * 7 * 17
4410 = 2 * 3^2 * 5 * 7^2
4500 = 2^2 * 3^2 * 5^3
4704 = 2^5 * 3 * 7^2
4788 = 2^2 * 3^2 * 7 * 19
4860 = 2^2 * 3^5 * 5
4896 = 2^5 * 3^2 * 17
4950 = 2 * 3^2 * 5^2 * 11
5100 = 2^2 * 3 * 5^2 * 17
5148 = 2^2 * 3^2 * 11 * 13
5220 = 2^2 * 3^2 * 5 * 29
5292 = 2^2 * 3^3 * 7^2
5376 = 2^8 * 3 * 7
5472 = 2^5 * 3^2 * 19
5580 = 2^2 * 3^2 * 5 * 31
5600 = 2^5 * 5^2 * 7
5700 = 2^2 * 3 * 5^2 * 19
5796 = 2^2 * 3^2 * 7 * 23
5850 = 2 * 3^2 * 5^2 * 13
6468 = 2^2 * 3 * 7^2 * 11
6624 = 2^5 * 3^2 * 23
6660 = 2^2 * 3^2 * 5 * 37
6732 = 2^2 * 3^2 * 11 * 17
6804 = 2^2 * 3^5 * 7
6900 = 2^2 * 3 * 5^2 * 23
6912 = 2^8 * 3^3
7260 = 2^2 * 3 * 5 * 11^2
7308 = 2^2 * 3^2 * 7 * 29
7350 = 2 * 3 * 5^2 * 7^2
7380 = 2^2 * 3^2 * 5 * 41
7524 = 2^2 * 3^2 * 11 * 19
7644 = 2^2 * 3 * 7^2 * 13
7650 = 2 * 3^2 * 5^2 * 17
7700 = 2^2 * 5^2 * 7 * 11
7740 = 2^2 * 3^2 * 5 * 43
7776 = 2^5 * 3^5
7812 = 2^2 * 3^2 * 7 * 31
7840 = 2^5 * 5 * 7^2
7956 = 2^2 * 3^2 * 13 * 17
8352 = 2^5 * 3^2 * 29
8448 = 2^8 * 3 * 11
8460 = 2^2 * 3^2 * 5 * 47
8550 = 2 * 3^2 * 5^2 * 19
8700 = 2^2 * 3 * 5^2 * 29
8712 = 2^3 * 3^2 * 11^2
8800 = 2^5 * 5^2 * 11
8892 = 2^2 * 3^2 * 13 * 19
8928 = 2^5 * 3^2 * 31
8960 = 2^8 * 5 * 7
9100 = 2^2 * 5^2 * 7 * 13
9108 = 2^2 * 3^2 * 11 * 23
9300 = 2^2 * 3 * 5^2 * 31
9324 = 2^2 * 3^2 * 7 * 37
9540 = 2^2 * 3^2 * 5 * 53
9702 = 2 * 3^2 * 7^2 * 11
9800 = 2^3 * 5^2 * 7^2
9984 = 2^8 * 3 * 13
9996 = 2^2 * 3 * 7^2 * 17
10140 = 2^2 * 3 * 5 * 13^2
10164 = 2^2 * 3 * 7 * 11^2
10332 = 2^2 * 3^2 * 7 * 41
10350 = 2 * 3^2 * 5^2 * 23
10400 = 2^5 * 5^2 * 13
10620 = 2^2 * 3^2 * 5 * 59
10656 = 2^5 * 3^2 * 37
10692 = 2^2 * 3^5 * 11
10764 = 2^2 * 3^2 * 13 * 23
10780 = 2^2 * 5 * 7^2 * 11
10836 = 2^2 * 3^2 * 7 * 43
10890 = 2 * 3^2 * 5 * 11^2
10980 = 2^2 * 3^2 * 5 * 61
11100 = 2^2 * 3 * 5^2 * 37
11172 = 2^2 * 3 * 7^2 * 19
11466 = 2 * 3^2 * 7^2 * 13
11484 = 2^2 * 3^2 * 11 * 29
11616 = 2^5 * 3 * 11^2
11628 = 2^2 * 3^2 * 17 * 19
11808 = 2^5 * 3^2 * 41
11844 = 2^2 * 3^2 * 7 * 47
11900 = 2^2 * 5^2 * 7 * 17
12060 = 2^2 * 3^2 * 5 * 67
12150 = 2 * 3^5 * 5^2
12168 = 2^3 * 3^2 * 13^2
12276 = 2^2 * 3^2 * 11 * 31
12300 = 2^2 * 3 * 5^2 * 41
12348 = 2^2 * 3^2 * 7^3
12384 = 2^5 * 3^2 * 43
12636 = 2^2 * 3^5 * 13
12740 = 2^2 * 5 * 7^2 * 13
12780 = 2^2 * 3^2 * 5 * 71
12900 = 2^2 * 3 * 5^2 * 43
13050 = 2 * 3^2 * 5^2 * 29
13056 = 2^8 * 3 * 17
13068 = 2^2 * 3^3 * 11^2
13140 = 2^2 * 3^2 * 5 * 73
13300 = 2^2 * 5^2 * 7 * 19
13356 = 2^2 * 3^2 * 7 * 53
13524 = 2^2 * 3 * 7^2 * 23
13536 = 2^5 * 3^2 * 47
13572 = 2^2 * 3^2 * 13 * 29
13600 = 2^5 * 5^2 * 17
13950 = 2 * 3^2 * 5^2 * 31
14076 = 2^2 * 3^2 * 17 * 23
14080 = 2^8 * 5 * 11
14100 = 2^2 * 3 * 5^2 * 47
14196 = 2^2 * 3 * 7 * 13^2
14220 = 2^2 * 3^2 * 5 * 79
14300 = 2^2 * 5^2 * 11 * 13
14508 = 2^2 * 3^2 * 13 * 31
14592 = 2^8 * 3 * 19
14652 = 2^2 * 3^2 * 11 * 37
14868 = 2^2 * 3^2 * 7 * 59
14940 = 2^2 * 3^2 * 5 * 83
14994 = 2 * 3^2 * 7^2 * 17
15200 = 2^5 * 5^2 * 19
15210 = 2 * 3^2 * 5 * 13^2
15246 = 2 * 3^2 * 7 * 11^2
15264 = 2^5 * 3^2 * 53
15372 = 2^2 * 3^2 * 7 * 61
15732 = 2^2 * 3^2 * 19 * 23
15900 = 2^2 * 3 * 5^2 * 53
16020 = 2^2 * 3^2 * 5 * 89
16100 = 2^2 * 5^2 * 7 * 23
16224 = 2^5 * 3 * 13^2
16236 = 2^2 * 3^2 * 11 * 41
16524 = 2^2 * 3^5 * 17
16640 = 2^8 * 5 * 13
16650 = 2 * 3^2 * 5^2 * 37
16660 = 2^2 * 5 * 7^2 * 17
16758 = 2 * 3^2 * 7^2 * 19
16884 = 2^2 * 3^2 * 7 * 67
16940 = 2^2 * 5 * 7 * 11^2
16992 = 2^5 * 3^2 * 59
17028 = 2^2 * 3^2 * 11 * 43
17052 = 2^2 * 3 * 7^2 * 29
17248 = 2^5 * 7^2 * 11
17316 = 2^2 * 3^2 * 13 * 37
17325 = 3^2 * 5^2 * 7 * 11
17340 = 2^2 * 3 * 5 * 17^2
17460 = 2^2 * 3^2 * 5 * 97
17568 = 2^5 * 3^2 * 61
17664 = 2^8 * 3 * 23

• oh wow you're quick, never mind! So here is my question to this, how are all these numbers going to be describe? Granted they all have tau=24, but how can all these be generalized? – user424890 Nov 7 '17 at 0:55
• @JasonBennett you asked for 36 divisors, which is what these do. How about if you write out why there are 36 divisors for the first five successes, 1260, 1440, 1800, 1980, 2016. – Will Jagy Nov 7 '17 at 0:59

There are an infinity.With distinct arbitrary primes $p_1,p_2,p_3,p_4$ we have the possibilities $$p_1p_2p_3^2p_4^2\\p_1p_2p_3^8\\p_1^{35}\\p_1^3p_2^8\\p_1^2p_2^{11}\\p_1^2p_2p_3^8$$ Maybe others I don't see now.