I need x^y == integer (input) cant figure out how to do so I CANT USE MATH IMPORT tried to do something like this:
a = int(input("Please Enter any Positive Integer:"))
power = 1
i = 1
while(i <= a):
power = power * a
i = i + 1
for example the input is 8 i need the program to find x^y==8 in this case the output needs to be: x=2 y=3 hope its clear thanks in advance.
You can find prime factors and take the greatest common denominator (gcd) of prime counts.
For example 216000's prime factors are 2^6, 3^3, 5^3 so the power will be 3. For each of the primes keep count/3 as the power to compute the base: 2^2 * 3^1 * 5^1 = 60. So 216000 = 60^3
def primeFactors(N): # returns dictionary of {prime:count}
result = dict()
p = 2 # p is a candidate prime factor
while p*p<=N: # prime candidates up to √N (remaining N)
while N%p == 0: # count prime factors
result[p] = result.get(p,0)+1
N //= p # by removing factors, only primes will match
p += 1 + (p&1) # next potential prime
if N>1: result[N] = 1 # anything remaining after √N is a prime
return result
def gcd(a,b=0,*c): # gcd using Euclid's algorithm
if c: return gcd(gcd(a,b),*c) # for multiple values
return a if not b else gcd(b,a%b) # Euclidian division version
def findPower(N):
counts = primeFactors(N) # {prime factor: count}
power = gcd(*counts.values()) # power is gcd of prime counts
base = 1
for f,c in counts.items(): # compute base
base *= f**(c//power) # from remaining prime powers
return base,power
output:
print(findPower(216000)) # (60,3)
print(findPower(8)) # (2,3)
print(findPower(81)) # (3,4)
print(findPower(371293)) # (13,5)
print(findPower(29**7)) # (29,7)
print(findPower(1522756**5553)) # (1234, 11106)
print(findPower(12345**12345)) # (12345,12345)
Here is a solution:
def findfactors(number):
factors = []
search = number
for n in range(2, number//2):
while search%n==0:
factors.append(n)
search = search // n
if search != 1:
factors.append(search)
return factors
def findxpowery(number):
res = findfactors(number)
diffactors = set(res)
#print("factors: ", res)
#print("different factors", diffactors)
res2 = [res.count(x) for x in diffactors]
#print("count the different factors", res2)
rr = 1
for r in diffactors:
rr = rr*r
#print("multiplicate all the different factors", rr)
if len(set(res2))==1:
#print("solution is", rr, "^", list(set(res2))[0])
return (rr, list(set(res2))[0])
else:
#print("no solution")
return None
number = (9**2)
r = findxpowery(number)
print("solution is", number, "=", r[0], "^", r[1])
Output is :
solution is 81 = 3 ^ 4
I think I understand what you are trying to achieve, this should work, I'm sure someone else can give you a better solution though.
a = int(input("Please Enter any Positive Integer:"))
max_ = round(a**(1/2)+1)
for x in range(2, max_):
for y in range(2, max_):
p = x**y
if p == a:
print(x,'power',y)
print(p)
elif p > a:
pass
To check if a number a can be a perfect power of x you may check
x ** int(round(math.log(a, x))) == a
Then to find out if there it is a perfect number of anything, loop from 2, to the square root of the value
from math import sqrt, log, ceil
def find_perfect_power(a):
for value in range(2, ceil(sqrt(a)) + 1):
if value ** int(round(log(a, value))) == a:
return True, value, int(round(math.log(a, value)))
return False,
Example
for i in range(1, 30):
r = find_perfect_power(i)
if r[0]:
print(i, r)
2 (True, 2, 1)
4 (True, 2, 2)
8 (True, 2, 3)
9 (True, 3, 2)
16 (True, 2, 4)
25 (True, 5, 2)
27 (True, 3, 3)
num = int(input("enter number: "))
# if you have x and y values for example x = 5 and y = 5
x = 5
y = 5
if x ** y == num:
print("True")
else:
print("False")
