Sieve

Project Euler 124: Determining the kth element of the sorted radical function

Project Euler 124: Determining the kth element of the sorted radical function

Problem 124 of Project Euler deals with radical functions, something I have never heard of before. So that always makes it interesting. It reads

The radical of n, rad(n), is the product of distinct prime factors of n. For example, 504 = 23 x 32 x 7, so rad(504) = 2 x 3 x 7 = 42.
If we calculate rad(n) for 1 ≤ n ≤ 10, then sort them on rad(n), and sorting on n if the radical values are equal, we get:

Unsorted
 
Sorted

n

rad(n)

n

rad(n)

k
1
1
 
1
1
1
2
2
 
2
2
2
3
3
 
4
2
3
4
2
 
8
2
4
5
5
 
3
3
5
6
6
 
9
3
6
7
7
 
5
5
7
8
2
 
6
6
8
9
3
 
7
7
9
10
10
 
10
10
10

Let E(k) be the kth element in the sorted n column; for example, E(4) = 8 and E(6) = 9.

If rad(n) is sorted for 1 ≤ n ≤ 100000, find E(10000). Continue reading →

Posted by Kristian in Project Euler, 7 comments
Project Euler 72: How many elements would be contained in the set of reduced proper fractions for d ≤ 1,000,000

Project Euler 72: How many elements would be contained in the set of reduced proper fractions for d ≤ 1,000,000

I don’t think it is a coincidence that this exact problem comes right now as we shall see in the solution of the problem. Problem 72 of Project Euler reads

Consider the fraction, n/d, where n and d are positive integers. If n<d and HCF(n,d)=1, it is called a reduced proper fraction.

If we list the set of reduced proper fractions for d ≤ 8 in ascending order of size, we get:

1/8, 1/7, 1/6, 1/5, 1/4, 2/7, 1/3, 3/8, 2/5, 3/7, 1/2, 4/7, 3/5, 5/8, 2/3, 5/7, 3/4, 4/5, 5/6, 6/7, 7/8

It can be seen that there are 21 elements in this set.

How many elements would be contained in the set of reduced proper fractions for d ≤ 1,000,000?

The first thing to note is that in here HCF is the same as the greatest common divisor or GCD as I usually use. The second thing to note is that we are probably looking for a very large number, but let us take a look at the problem. Continue reading →

Posted by Kristian in Project Euler, 15 comments