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...
24 Divisors:Find the least positive integer with $24$ positive divisors.
ANSWERED:
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.
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$\begingroup$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.
$\endgroup$ 6 $\begingroup$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 * 23There 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.
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