What's the least positive integer, n
, having the following properties:
(where a, b, and c are integers)
- n = (a^2)/2
- n = (b^3)/3
- n = (c^5)/5
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,
Each line then says that n must be the following forms
The smallest n would be a product of each of the three forms. So n=2^(2x-1)*3^(3y-1)*5^(5z-1).
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^(3y-1)*5^(5z-1) = perfect square
2^(2x-1)*5^(5z-1) = perfect cube
2^(2x-1)*3^(3y-1) = 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^(2x-1)*3^(3y-1)*5^(5z-1).
n=2^(15)*3^(20)*5^(24), which is a very large number.
Posted by np_rt
on 2004-11-27 18:43:38