The smallest natural number with "n" divisors(factors)

[M.RezaMathing]

Hello, what is the solution to this problem? "The smallest natural number with "n" divisors"
I have seen problems like this with different numbers but I wanted to know the general solution.
Thanks in advance.

 

[Deleted member 4993]

Hello, what is the solution to this problem? "The smallest natural number with "n" divisors"
I have seen problems like this with different numbers but I wanted to know the general solution.
Thanks in advance.

To get a definitive answer, please post an example-problem/s.

 

[M.RezaMathing]

For example, what is the smallest number with 12 divisors(or factors)

 

[JeffM]

For example, what is the smallest number with 12 divisors(or factors)

The clue to this problem is the Fundamental Theorem of Arithmetic: every number can be expressed uniquely as a prime or a product of primes.

 

[Dr.Peterson]

For example, what is the smallest number with 12 divisors(or factors)

Please show an attempt at that specific problem. (Or maybe something smaller, like 6 divisors.)

Of, if you have trouble with that, how would you find the number of divisors of, say, 360?

 

[pka]

For example, what is the smallest number with 12 divisors(or factors)

Here is a way for you to investigate.
$540=(2^2)(3^3)(5)$ so $(2+1)(3+1)(1+1)=24$ so that $540$ has $24$ divisors.

 

[M.RezaMathing]

Please show an attempt at that specific problem. (Or maybe something smaller, like 6 divisors.)

Of, if you have trouble with that, how would you find the number of divisors of, say, 360?

Firstly, based on other replies and what I saw on other sites, to find the number of divisors of a number we first write it as a product of prime numbers, then we add 1 to the exponents in the prime factorization for the number and then multiply them. As for 360, $360=2^3.3^2.5$ and $(3+1)(2+1)(1+1)= 4*3*2=24$ so 360 has 24 divisors. Secondly, for the smallest number which has 6 divisors, the exponent(s) +1 multiplied together should be 6. So it can be $2^5=32$ but there are different ways to get 6 and we can use different bases, so how should I find the smallest number?

 

[Dr.Peterson]

Firstly, based on other replies and what I saw on other sites, to find the number of divisors of a number we first write it as a product of prime numbers, then we add 1 to the exponents in the prime factorization for the number and then multiply them. As for 360, $360=2^3.3^2.5$ and $(3+1)(2+1)(1+1)= 4*3*2=24$ so 360 has 24 divisors.

Right.

Secondly, for the smallest number which has 6 divisors, the exponent(s) +1 multiplied together should be 6. So it can be $2^5=32$ but there are different ways to get 6 and we can use different bases, so how should I find the smallest number?

For 6 divisors, we can factor n as 6 alone, or as 3*2, so the exponents can be 5 alone, or 2 and 1; the smallest numbers you can make with those exponents are 2^5 = 32 (with divisors 1, 2, 4, 8, 16, and 32), and 2^2*3^1 = 12 (with divisors 1, 2, 3, 4, 6, 12), so the latter is the winner.

The task you've set for yourself is to see if there is a general "formula" for any n.

It may be that the only way to find the smallest number for any given n is to do exactly what I just did for 6, and compare the numbers you get. Or there may be a pattern that always works, such as listing the prime factors of n in descending order and assigning each of those, minus 1, as the exponent on each successive prime as a base, which worked for 6. Maybe you can prove or disprove this.

 

[M.RezaMathing]

The task you've set for yourself is to see if there is a general "formula" for any n.

It may be that the only way to find the smallest number for any given n is to do exactly what I just did for 6, and compare the numbers you get. Or there may be a pattern that always works, such as listing the prime factors of n in descending order and assigning each of those, minus 1, as the exponent on each successive prime as a base, which worked for 6. Maybe you can prove or disprove this.

So for n=12: 12=2*2*3, 2^2*3^1*5^1=60 (with divisors 1,2,3,4,5,6,10,12,15,20,30,60). for n=15: 15=3*5, 2^4*3^2=144. Am I right?

 

[Dr.Peterson]

So for n=12: 12=2*2*3, 2^2*3^1*5^1=60 (with divisors 1,2,3,4,5,6,10,12,15,20,30,60). for n=15: 15=3*5, 2^4*3^2=144. Am I right?

Those both follow the guess I made; to be sure that these are the smallest possible numbers, we have to compare with other possibilities.

For n=12, we could factor that as 12, 6*2, 4*3, or 3*2*2; resulting numbers are 2^11 = 2048, 2^5*3^1 = 96, 2^3*3^2 = 72, and 2^2*3^1*5^1 = 60. Clearly 60 is the smallest.

For n=15, we could factor it as 15 or 5*3, resulting in 2^14 = 16384 and 2^4*3^2 = 144, and again 144 is the smallest.

So you're right; the next thing to do is to look for a proof that this pattern will always lead to the smallest possible number. Or, you could think about what it might take to find a counterexample, where some other number in our list will be smaller.

So I would first just try a few more small numbers, perhaps with higher powers involved. Let's try 8, a cube:

For n=8, we can factor it as 8, 4*2, or 2*2*2, resulting in 2^7 = 128, 2^3*3^1 = 24, and 2^1*3^1*5^1 = 30. I would call this a counterexample, since this time the smallest number doesn't come from a product of separate primes. (And in fact, that fits what I was expecting, though it came sooner than I thought: Since primes spread out as you go along, there should come a point where using more primes in the product costs more than using higher exponents.)

So, unless you're satisfied having to try all possibilities, which seems to be the only way to be sure, you would need to propose a different sort of pattern.