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(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?

 Jun 28, 2015

Best Answer 

 #5
avatar+129849 
+14

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......

 

 

 Jun 30, 2015
 #1
avatar+33657 
+11

(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.

 .

 Jun 29, 2015
 #2
avatar+118667 
+11

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. :(

 

 Jun 29, 2015
 #3
avatar+118667 
+5

This question is definitely not answered  

 Jun 30, 2015
 #4
avatar+129849 
+11

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.....!!!!!

 Jun 30, 2015
 #5
avatar+129849 
+14
Best Answer

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......

 

 

CPhill Jun 30, 2015
 #6
avatar+118667 
+10

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?  

 Jun 30, 2015
 #7
avatar+118667 
+10

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 ?   

 Jun 30, 2015
 #8
avatar+118667 
+5

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  :/

 Jun 30, 2015
 #9
avatar+129849 
+10

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.....

 

 Jun 30, 2015
 #10
avatar+118667 
+5

Yep Chris, we are both brilliant     LOL   :))

 

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

 Jun 30, 2015

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