Find a positive integer m such that
1/2m is a perfect square and
1/3m is a perfect cube. Can you
find a positive integer n for which
1/2n is a perfect square,
1/3n is a perfect cube and
1/5n is a perfect fifth power?
To find a positive integer m such that 1/2m is a perfect square and 1/3m is a perfect cube, we can start by considering the prime factorization of m.
Let's write m = 2^a * 3^b * k, where k is a positive integer without factors of 2 or 3.
Now, we know that 1/2m is a perfect square, which means that 2m = 2^(a+1) * 3^b * k must be a perfect square. Similarly, 1/3m = 2^a * 3^(b-1) * k should be a perfect cube.
Since 2m and 1/3m should be perfect squares and perfect cubes respectively, their powers of 2 and 3 must be even.
From the equations above, we can determine that a + 1 and b must be even. Let's say a + 1 = 2x and b = 2y, where x and y are positive integers.
Now let's simplify the equations:
2m = 2^(2x) * 3^(2y) * k = (2^x * 3^y)^2 * k
1/3m = 2^(2x - 1) * 3^(2y - 1) * k
So, we can see that if we choose x = 1 and y = 1, then 2m will be a perfect square and 1/3m will be a perfect cube. This means that:
m = 2^(2x - 1) * 3^(2y - 1) * k
= 2 * 3 * k
= 6k
Therefore, m must be a multiple of 6.
Now, let's move on to finding a positive integer n such that 1/2n is a perfect square, 1/3n is a perfect cube, and 1/5n is a perfect fifth power.
Using a similar approach, we can write n as n = 2^c * 3^d * 5^e * m, where m is a multiple of 6.
To make 1/2n a perfect square, c+1 must be even, making c even. Similarly, d and e must also be even to satisfy the conditions for 1/3n and 1/5n.
Considering these conditions, let's choose c = d = e = 2 and substitute the values:
n = 2^c * 3^d * 5^e * m
= 2^2 * 3^2 * 5^2 * (6k)
= (2 * 3 * 5 * k)^2 * (6k)
= (30k)^2 * (6k)
= (180k^2) * (6k)
= 1080k^3
Therefore, the positive integer n that satisfies the given conditions is n = 1080k^3, where k is a positive integer.
Note: This is just one possible solution, and there can be other values of n satisfying the given conditions.