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Of all numbers between 1 and 15!(factorial), how many of them are BOTH perfect squares and perfect cubes for the same number(such as 64)? Any help would be appreciated. Thank you.

 Jan 23, 2019
 #1
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+1

The total number of perfect squares and perfect cubes for the same number would be: (15!)^(1/6) =104 such numbers. And these numbers will consist of ALL prime numbers up to: 10^[(log(15!))/ log(6)]=~105. The last prime number to be included would be 103.
However, all these prime numbers must be raised to a MINIMUM power of 6 and multiples of 6. So, the first of these numbers would look like this:


2^6,  2^12,  2^18,  2^24,  2^30,  2^36(which is < 15!). Then, they will continue with the next prime: 3^6,  3^12,  3^18....etc. Then you will have a combination of powers such as: 2^6  x  3^6  x  5^6 and multiples thereof.
The computer-generated numbers will begin with:


(1, 64, 729, 4096, 15625, 46656, 117649, 262144, 531441, 1000000........etc......and end in: 1000000000000,  1061520150601, 1126162419264, 1194052296529, 1265319018496), for a grand total = 104.

 Jan 23, 2019
 #2
avatar+21191 
+6

Of all numbers between 1 and 15! (factorial),

how many of them are BOTH perfect squares and perfect cubes for the same number(such as 64)?

 

lcm(2,3) = 6  (least common multiple)

 

So the numbers that are going to be perfect squares and perfect cubes are precisely the 6th powers of the integers:
\(\begin{array}{rcll} 1^6 &=& 1 \\ 2^6 &=& 64 \\ 3^6 &=& 729 \\ 4^6 &=& 4096 \\ 5^6 &=& 15625 \\ ... \\ n^6 &<& 15! \\ \end{array}\)

 

So n are the numbers between 1 and 15! (factorial)

\(\begin{array}{|rcll|} \hline n^6 &<& 15! \\ n &<& \sqrt[6]{15!} \\ n &<& \sqrt[6]{1~307~674~368~000} \\ n &<& 104.57228596314317\ldots \\ \mathbf{n} & \mathbf{=} & \mathbf{104}\\ \hline \end{array}\)

 

104 of them are BOTH perfect  squares and perfect cubes.

 

laugh

 Jan 24, 2019

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