# level 5: k-Primes (2019-11-23)

###### tags: `Codewars` `python`

A natural number is called k-prime if it has exactly k prime factors, counted with multiplicity. For example:

k = 2 --> 4, 6, 9, 10, 14, 15, 21, 22, ...
k = 3 --> 8, 12, 18, 20, 27, 28, 30, ...
k = 5 --> 32, 48, 72, 80, 108, 112, ...
A natural number is thus prime if and only if it is 1-prime.

Task:
Complete the function count_Kprimes (or countKprimes, count-K-primes, kPrimes) which is given parameters k, start, end (or nd) and returns an array (or a list or a string depending on the language - see "Solution" and "Sample Tests") of the k-primes between start (inclusive) and end (inclusive).

Example:
countKprimes(5, 500, 600) --> [500, 520, 552, 567, 588, 592, 594]
Notes:

The first function would have been better named: findKprimes or kPrimes :-)
In C some helper functions are given (see declarations in 'Solution').
For Go: nil slice is expected when there are no k-primes between start and end.
Second Task (puzzle):
Given a positive integer s, find the total number of solutions of the equation a + b + c = s, where a is 1-prime, b is 3-prime, and c is 7-prime.

Call this function puzzle(s).

Examples:
puzzle(138) --> 1 because [2 + 8 + 128] is the only solution
puzzle(143) --> 2 because [3 + 12 + 128] and [7 + 8 + 128] are the solutions

my code:
<pre>
def primes(n):
primfac = []
d = 2
while d*d <= n:
while (n % d) == 0:
primfac.append(d)
n //= d
d += 1
if n > 1:
primfac.append(n)
return len(primfac)

def count_Kprimes(k, start, nd):
return [i for i in range(start, nd+1) if primes(i)==k]

def puzzle(s):
a, b, c = count_Kprimes(1, 2, s), count_Kprimes(3, 8, s), count_Kprimes(7, 128, s)
return sum(i + j + k == s for i in a for j in b for k in c)
</pre>