A certain positive integer has exactly 20 positive divisors. What is the smallest number of primes that could divide the integer?

c)  I believe the answer to this is 240....to see why......

As per my explanation to  (b) above, the largest numer of primes that could factor such a  number is 4.....

Note that  2,3 ,5 and 7   are the smallest primes......then...using the reasoning from (b), above.... we are looking for four exponents, that, when 1 is added to each and all are multiplied together, would equal 20.

But no such integers  k, l, m and n  exist such that (k + 1)(l + 1)(m + 1) (n + 1)  = 20  ...  where k, l ,m and n ≥ 1......so.....this number, whatever it is, can't have 4 prime factors

Let's drop 7 out of the mix and suppose it has just 3 prime factors...2, 3 and 5....again  we are looking for three exponents,  that, when 1 is added to each and all are multiplied together, would equal 20.  Put another way, we are looking for k, l and m ≥ 1   such that (k + 1)(l + 1)(m + 1) = 20

Note that the  only possibility here  is when we have  2 *2 *5  = 20.....and the smallest possible product would be 2^(4 )* 3^(1) * 5^(1) =  2^4 * 3 * 5 = 240

Now......the only remaining posibility is  that this number is composed of the two smallest primes, 2 and 3,  and we are looking for  some k  and l ≥ 1 such that (k + 1)(l + 1) = 20.....clearly, the only possibilities  are when k = 4 and l = 5, or vice-versa

So this number would factor as either 2^3 * 3^4   = 648   or 2^4 * 3^3 = 432....and both are > 240......

(a) One.  The number 2^19 has twenty divisors, all of the form 2^n where n is an integer going from 0 to 19.  The only prime involved is the number 2.

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A) is 1,  just like Alan said.

if all the 20 divisors are the first 20 powers of 2 there will be one prime divisor (That is 2).  This number would be 2^19

Any prime to the power of 19 would work just as well I think.

I do not have a pen and paper or a computer. I just have my phone. So I am going to put down my rambling thoughts.

B) Let me think.....

1×2      2factors

1 x2x3.    3+2c2=4factors

1x2x3x5.  4+3c2+3C1=4+3+1=8factors

1×2×3×5×7=

5+4C2+4C3+4C4

=5+6+4+1=16factors

mmm I can't use any more primes....there would be too many factors. 

So maybe 4 primes is the most? .....

1×2×3×5×7×2

How many factors does this have?

All the 16 from before plus 12,20,28,60,84,140,420

That makes 16+7=23 factors

thats to many. :(

maybe 3 primes is the mosts.

At this point in time I really need a pen and paper and/or computer and some combinatory mathematics.

But do you get where I am going here?

Like I said.  I am really just thinking out loud. There can not be many primes. :(

This question is definitely not answered  

b) .....I believe the answer to this is 4.....to see why......

Note that.....the number of possible divisors for a^k (where a is prime, and k ≥ 1)  is just (k + 1)

For instance, 3²  = 9, has  (2 + 1) = 3 possible divisors...... 1, 3 and 9

And the number of possible divisors for a^k * b^m    where a, b are prime, and k, m ≥ 1 .....  is just   (k + 1) (m +1)

So.....consider a positive integer factored as a^k * b^m *cⁿ * d^p  where a, b, c, d are prime and k, m, n and p  = 1

Then, the possible number of divisors is just (k + 1)(m + 1)(n + 1)(p + 1)  = (2)(2)(2)(2)  = 16

Note that, if another prime (also to the first power)  factored the number, we would have (2)(2)(2)(2)(2)  = 32 possible divisors.....and that's too  many.....!!!!

Edit...this answer is incorrect......Melody has the correct solution, below.....!!!!!

Chris, I do not see how you have answered the question.

(Although i did get something out of your algebra logic.)

You have found - like i did - that the product of 4 unique primes can give an answr with 16 factors.

BUT the question calls for EXACTLY 20 factors.  Do you think that such a number does not exist?  

b)

Perhaps I have the answer - using your algebra Chris :)

The factors of 20 are 1,2,4,5,10,20

Now if we are going to have the largest number of primes to make 20 factors we need the most integers greater than 1 that can multiply to give 20.

they are 2*2*5

So the answer is 3 unique primes (6 primes altogether).  This is the maximum number of prime numbers that can divide this composite number with 20 factors.

a * b * c^4      

c) So the smallest positive integer that has exactly 20 positive divisors will be

3*5*2^4

$${\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{5}}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{{\mathtt{4}}} = {\mathtt{240}}$$

Thanks Chris - this was a real joint effort.   Are you happy with the answer ?   

Sorry Chris I did not realize you had already answered - I only saw your first answer.

You must have posted the other s when Iwas off doing things and I did not realize that they were there. 

Anyway, we came to the same conclusion  :/

Melody, I agree with your answer to (b).....I was assuming unique primes.....but clearly there could not be 4 unique primes that would produce a positive interger with exaxctly 20 divisors.....so .....your answer of 3 is correct.....

Yep Chris, we are both brilliant     LOL   :))

Thanks Mellie for making us think.  We can always rely on you for that :)