(a) A certain positive integer has exactly 20 positive divisors. What is the smallest number of primes that could divide the integer?
(b) A certain positive integer has exactly 20 positive divisors. What is the largest number of primes that could divide the integer?
(c) What is the smallest positive integer that has exactly 20 positive divisors?
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. :(
b) .....I believe the answer to this is 4.....to see why......
Note that.....the number of possible divisors for ak (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 ak * bm where a, b are prime, and k, m ≥ 1 ..... is just (k + 1) (m +1)
So.....consider a positive integer factored as ak * bm *cn * dp 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.....!!!!!
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......
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 :/