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What is the smallest positive integer that has exactly 6 perfect square divisors?

 Jul 1, 2015

Best Answer 

 #1
avatar+128079 
+11

(1^2) * (2^2) * (3^2)  * (2^2)  will give us what we need = 144

 

Notice that the "2^2" at the end combined with the other 2^2  will give us 16  = 4^2 

 

Likewise, it will produce 6^2 = 36 when combined with  the "3^2"

 

 

  

 Jul 1, 2015
 #1
avatar+128079 
+11
Best Answer

(1^2) * (2^2) * (3^2)  * (2^2)  will give us what we need = 144

 

Notice that the "2^2" at the end combined with the other 2^2  will give us 16  = 4^2 

 

Likewise, it will produce 6^2 = 36 when combined with  the "3^2"

 

 

  

CPhill Jul 1, 2015
 #2
avatar
+5

Question is unclear as to whether the divisors can be the same number (as shown in answer #1) and we need SIX , not FOUR as provided in answer #1....

If you CAN use the same number twice, why not use it three...or even six times

THEN the answer would be    1^2  *  1^2  * 1^2  * 1^2   *  1^2  *  1^2  = 1

If you can't use the same number more than once, then the answer would be:

1^2  *  2^2  * 3^2  * 4^2  *  5^2  * 6^2 =  518400

ASSUMING you can't use NEGATIVE numbers.   IF you CAN use negatives it would be :

1^2  *  -1^2  * 2^2  * -2^2    * 3^2  *  -3^2  =  1296

 Jul 2, 2015
 #3
avatar+128079 
+6

Sorry, Anonymous....I'm sticking with my answer......the question doesn't specify how many prime factors we can use (or even any restictions relating to the powers on those primes).....it just wants to know the smallest possible positive integer that has 6 perfect square divisors.....

 

Of course.....I suppose that we might claim that " 1 " could have as many perfect square divisors as we wished, but that answer seems trivial.......I'm assuming that all the perfect square divisors are unique....

 

And 144 is certainly < 518400 or 1296   !!!!

 

 

 Jul 2, 2015
 #4
avatar+118587 
+6

Thank you anon - I really like it that you have gotten involved.   

 

Yes, I think Chris is right

$${factor}{\left({\mathtt{144}}\right)} = {{\mathtt{2}}}^{{\mathtt{4}}}{\mathtt{\,\times\,}}{{\mathtt{3}}}^{{\mathtt{2}}}$$

 

1^1=1

2^2=4

3^2=9

(2*2)^2=16

(2*3)^2=36

(2^2*3)^2=(4*3)^2=144

 Jul 2, 2015
 #5
avatar
+6

I'll concede you are right......the question does NOT specify the divisors multiplied together must equal the result, just that they must be divisors.

 Jul 2, 2015

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