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Exponential Difficulties 2 (Posted on 2004-11-27) Difficulty: 4 of 5
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)

See The Solution Submitted by SilverKnight    
Rating: 4.0000 (5 votes)

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Solution | Comment 3 of 9 |

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^(2x-1)
3^(3y-1)
5^(5z-1)

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
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