Number of factors in the smallest number possible.

Question

I was working with LCMs and GCFs, and I came across this problem:

"What is the smallest positive integer n having exactly 12 factors?"

So far, I have tried listing the first numbers 2-13 and finding the LCM, but I got the problem incorrect. I also tried making a number that is a power of 2, but I also got the wrong answer. I was hoping someone could help and tell me how to set up and solve this problem.

Thanks!

Answer 1

I also tried making a number that is a power of 2, but I also got the wrong answer.

If $2^{12}$ didn't work, then the question is probably asking for the product of the first 12 prime numbers: $2*3*5*7*11*13*...$

Answer 2

Links:

https://en.wikipedia.org/wiki/Primorial#Table_of_primorials

https://en.wikipedia.org/wiki/Highly_composite_number

Hence, for $n$ to be a highly composite number, the k given prime numbers $p_i$ must be precisely the first $k$ prime numbers (2, 3, 5, ...); if not, we could replace one of the given primes by a smaller prime, and thus obtain a smaller number than $n$ with the same number of divisors (for instance 10 = 2 × 5 may be replaced with 6 = 2 × 3; both have four divisors);

the sequence of exponents must be non-increasing, that is $c_1 \geq c_2 \geq \cdots \geq c_k;$ otherwise, by exchanging two exponents we would again get a smaller number than n with the same number of divisors (for instance 18 = 21 × 32 may be replaced with 12 = 22 × 31; both have six divisors).

Also, except in two special cases n = 4 and n = 36, the last exponent $c_k$ must equal 1. It means that 1, 4, and 36 are the only square highly composite numbers. Saying that the sequence of exponents is non-increasing is equivalent to saying that a highly composite number is a product of primorials.

I said this in comments but I see no response. The number of divisors is found by writing the prime factorization of the number, then, for each $p^k,$ multiply the number of divisors by $(k+1).$ So, the number of divisors of $60 = 2^2 \cdot 3^1 \cdot 5^1$ is $\tau(60) = (2+1)(1+1)(1+1) = 12.$

Now, the phrase "a product of primorials" refers to a number in which the prime factors are consecutive primes beginning with $2,$ and the exponents are non-increasing. The first types of examples: $$ 2^a, \; \; \mbox{with} \; \; a \geq 1 \; \; \mbox{and} \; \; \tau(2^a) = (a+1),$$ $$ 2^a 3^b, \; \; \mbox{with} \; \; a \geq b \geq 1 \; \; \mbox{and} \; \; \tau(2^a 3^b) = (a+1)(b+1),$$ $$ 2^a 3^b 5^c, \; \; \mbox{with} \; \; a \geq b \geq c \geq 1 \; \; \mbox{and} \; \; \tau(2^a 3^b 5^c) = (a+1)(b+1)(c+1),$$ $$ 2^a 3^b 5^c 7^d, \; \; \mbox{with} \; \; a \geq b \geq c \geq d \geq 1 \; \; \mbox{and} \; \; \tau(2^a 3^b 5^c 7^d) = (a+1)(b+1)(c+1)(d+1),$$ and so on. The smallest number with a certain target number of divisors will always be a product of primorials. For example, the smallest number with exactly $11$ divisors would be $2^{10} = 1024.$

The products of primorials with exactly $12$ divisors are, not in numerical order, $$ 2^{11}, \; \; 2^5 3^1, \; \; 2^3 3^2, \; \; 2^2 3^1 5^1. $$ I think those are all. I will check again later.

Yep

jagy@phobeusjunior:~$ 
60   2^2 3 5
72   2^3 3^2
96   2^5 3
2048   2^11

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Examples of numbers that are products of primorials include $n!$ and $\operatorname{lcm} \{1,2,3,4, \ldots, n \}.$ Also included are the highly composite numbers and the superior highly composite numbers. So are the superabundant and colossally abundant numbers. However, a highly abundant number, such as $18,$ may not be the product of primorials.