What's the least positive integer,
n, having the following properties:
 n = (a^2)/2
 n = (b^3)/3
 n = (c^5)/5
(where a, b, and c are integers)
Here's what I got. For the number to be as small as possible, n can only contain factors of 2, 3, and 5.
Let's analyze each requirement separately. For n to be an integer,
a=2^x
b=3^y
c=5^z
Each line then says that n must be the following forms
2^(2x1)
3^(3y1)
5^(5z1)
The smallest n would be a product of each of the three forms. So n=2^(2x1)*3^(3y1)*5^(5z1).
The find the appropriate x, y, and z. We need to consider 2n being a perfect square, 3n being a perfect cube, and 5n being a perfect 5th power.
Plugging the form of n into those requirements gives:
3^(3y1)*5^(5z1) = perfect square
2^(2x1)*5^(5z1) = perfect cube
2^(2x1)*3^(3y1) = perfect 5th power
The first line says that y and z are odd. But it's more convenient to say that y is in the form (2k+1)/3 and z is in the form (2k+1)/5.
The second line says that x is of the form (3k+1)/2 and z is of the form (3k+1)/5.
The last line says that x is of the form (5k+1)/2 and y is of the form (5k+1)/3.
NOTE: Not all the k's work and k is arbitrary.
In order for x to be of the forms (3k+1)/2 and (5k+1)/2, it must be of the form, (15k+1)/2. The smallest k that works is 1, making x=8.
In order for y to be of the forms (2k+1)/3 and (5k+1)/3, it must be of the form, (10k+1)/3. The smallest k that works is 2, making y=7.
In order for z to be of the forms (2k+1)/5 and (3k+1)/5, it must be of the form (6k+1)/5. The smallest k that works is 4, making z=5.
So x=8, y=7, z=5. And n=2^(2x1)*3^(3y1)*5^(5z1).
n=2^(15)*3^(20)*5^(24), which is a very large number.

Posted by np_rt
on 20041127 18:43:38 