5. Determine the largest positive integer $n$ with $n<500$ for which $6048\left(28^{n}\right)$ is a perfect cube (that is, it is equal to $m^{3}$ for some positive integer $m$ ).
5. Solution 1
We start by finding the prime factorization of $6048\left(28^{n}\right)$.
Note that $28=4(7)=2^{2} 7^{1}$.
Therefore, $28^{n}=\left(2^{2} 7^{1}\right)^{n}=2^{2 n} 7^{n}$.
Also, $6048=8(756)=8(4)(189)=8(4)(7)(27)=2^{3}\left(2^{2}\right)\left(7^{1}\right)\left(3^{3}\right)=2^{5} 3^{3} 7^{1}$.
Thus, $6048\left(28^{n}\right)=2^{5} 3^{3} 7^{1}\left(2^{2 n} 7^{n}\right)=2^{2 n+5} 3^{3} 7^{n+1}$.
This is a perfect cube exactly when the exponent of each of the prime factors is a multiple of 3 . Since the exponent on the prime 3 is a multiple of 3 , the problem is equivalent to asking for the largest positive integer $n$ less than 500 for which each of $2 n+5$ and $n+1$ is a multiple of 3 .
If $n=499$, these are 1003 and 500 ; neither is divisible by 3 .
If $n=498$, these are 1001 and 499; neither is divisible by 3 .
If $n=497$, these are 99 and 498 ; both are divisible by 3 .
Thus, $n=497$ is the largest positive integer less than 500 for which $6048\left(28^{n}\right)$ is a perfect cube.
Solution 2
We note that $6048=28(216)$.
Therefore, $6048\left(28^{n}\right)=216(28)\left(28^{n}\right)=216\left(28^{n+1}\right)=6^{3} 28^{n+1}$, since $6^{3}=216$.
Since $6^{3}$ is a perfect cube, then $6048\left(28^{n}\right)$ is a perfect cube exactly when $28^{n+1}$ is perfect cube. Since 28 is itself not a perfect cube, then $28^{n+1}$ is a perfect cube exactly when $n+1$ is a multiple of 3 .
Thus, we are looking for the largest positive integer $n<500$ for which $n+1$ is a multiple of 3 . Since 500 and 499 are not multiples of 3 , but 498 is a multiple of 3 (since $498=3(166)$ ) we
have $n+1=498$ or $n=497$.
Therefore, $n=497$ is the largest positive integer less than 500 for which $6048\left(28^{n}\right)$ is a perfect cube.
ANSWER: 497