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# Project Euler #243: Resilience

# Project Euler #243: Resilience

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when p = 1/10, the answer is near 1.6516447 * 10^103 I hope this infomation will help you

phew!! at last after 4 months my code worked!! learned a lot with this challenge

is there a way to see the test cases? dont understand why I cant pass more than two

this is the possible answer ig:

def lcm(c,d):

def gcd(c,d):

a=int(input())

for i in range(a):

i=3

z=[]

while True:

x=sum([1 for j in range(1,i) if(lcm(j,i)==(i*j))])

## print("x:",x)

val=x/(i-1)

## print(val)

## print(b[0]/b[1])

if(val < b[0]/b[1]):

i=i+1

Accordinng to the problem description if we calculate the resilience of d = 9, we see that R(9) = 2/8 < 4/10. If so then, d = 9 should be the smallest denominator for which R(d) < 4/10 But in the example given in problem description d = 12 is considered as smallest denominator having a resilience R(d) < 4/10. Am I interpreting the problem wrongly ?