# Task
 The sequence of `Chando` is an infinite sequence of all Chando's numbers in ascending order.

 A number is called `Chando's` if it is an integer that can be represented as a sum of different positive integer powers of 5.

 The first Chando's numbers is 5 (5^1). And the following nth Chando's numbers are:
 ```
 25  (5^2)
 30  (5^1 + 5^2)
 125 (5^3)
 130 (5^1 + 5^3)
 150 (5^2 + 5^3)
 ...
 ...
 ```

 Your task is to find the Chando's nth number for a given `n`.

# Input/Output


 - `[input]` integer `n`

  `1 <= n <= 7000`


 - `[output]` an integer

  nth Chando's number

Source: codewars | Difficulty: EASY
Tags: Number theory

Starter code: def nth_chandos_number(n):
	

Solutions:
def nth_chandos_number(n):
	return int((bin(n) + '0')[2:], 5)


from itertools import combinations, chain
from math import log

def nth_chandos_number(n):
	l = [0] + [5 ** i for i in range(1, int(log(n, 5)) + 10)]
	m = chain(*[combinations(l, ni) for ni in range(1, len(l) + 1)])
	ls = sorted(list(set([sum(i) for i in m])))
	return ls[n]


ARCHIVE = [sum((5 ** (i + 1) for (i, s) in enumerate(reversed(bin(n)[2:])) if s == '1')) for n in range(1, 7001)]

def nth_chandos_number(n):
	return ARCHIVE[n - 1]


from itertools import combinations as c

def nth_chandos_number(n):
	(comb, generate, i) = ([1], [], 1)
	while len(generate) <= n:
		generate.extend(sorted([j for k in range(1, len(comb) + 1) for j in c(comb, k) if j[0] == i]))
		i += 1
		comb.insert(0, i)
	return sum((5 ** i for i in generate[n - 1]))


nth_chandos_number = c = lambda n: n and c(n // 2) * 5 + n % 2 * 5


lst = [5]
for i in range(2, 14):
	x = 5 ** i
	lst.append(x)
	temp = lst[:]
	for j in temp[:-1]:
		lst.append(j + x)

def nth_chandos_number(n):
	return lst[n - 1]


def nth_chandos_number(n):
	(a, i, j, m) = ([], -1, 1, 5)
	while len(a) < n:
		m = 5 ** j
		a.append(m)
		while j > 1 and i < 2 ** (~-j) - 1:
			a.append(m + a[i])
			i += 1
		(i, j) = (0, j + 1)
	return a[~-n]


def nth_chandos_number(n):
	b = '{:b}'.format(n)
	r = 0
	for d in b:
		r = r * 5 + int(d) * 5
	return r


def nth_chandos_number(n):
	(t, ans) = (1, 0)
	while n:
		t *= 5
		if n % 2:
			ans += t
		n //= 2
	return ans