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.
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.
+26376 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.
