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  3. Project Euler #37: Truncatable primes
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Project Euler #37: Truncatable primes

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22 Discussions

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  • Kirahvi
     + 0 comments
    n            = int(input())
trunc_primes = [23, 37, 53, 73, 313, 317, 373, 797, 3137, 3797, 739397]
my_primes    = [i for i in trunc_primes if i < n]
print(sum(my_primes))

  • nkosi9976
     + 0 comments

    100 points 

    # Enter your code here. Read input from STDIN. Print output to STDOUT
def is_prime(n):
    if n < 2:
        return False
    for i in range(2, int(n**0.5) + 1):
        if n % i == 0:
            return False
    return True

def is_truncatable(n):
    s = str(n)
    for i in range(1, len(s)):
        if not is_prime(int(s[i:])) or not is_prime(int(s[:i])):
            return False
    return True

limit = int(input())
trunc_primes = []
for p in range(10, limit):
    if is_prime(p) and is_truncatable(p):
        trunc_primes.append(p)
print(sum(trunc_primes))

  • alhassanchoubas1
     + 0 comments

    Python

    N = int(input())
primes = {2, 3, 5, 7}
truncatable_primes = []

def is_prime(n):
    if n in primes:
        return True
    elif n == 1:
        return False
    elif n % 2 == 0:
        return False
    else:
        # Check for divisibility up to the square root of n
        for i in range(3, int(n**0.5) + 1, 2):
            if n % i == 0:
                return False
        return True

def is_truncatable_prime(n):
    num_strL = num_strR = str(n)
    
    while len(num_strL) > 0:
        curr = int(num_strL)
				
        if is_prime(curr):
            primes.add(curr)
            num_strL = num_strL[1:]
        else:
            return False
    
    while len(num_strR) > 0:
        curr = int(num_strR)
				
        if is_prime(curr):
            primes.add(curr)
            num_strR = num_strR[:-1]
        else:
            return False
            
    return True

for i in range(11, N):
    if is_prime(i):
        primes.add(i)
        if is_truncatable_prime(i):
            truncatable_primes.append(i)

# print(truncatable_primes)
print(sum(truncatable_primes))

  • Venkateshgurram5
     + 0 comments

    Easy logic. 100 points

    def is_prime(n):
    if n==2:
        return True
    if n%2==0 or n<2:
        return False
    if all([n%i for i in range(3,int(n**0.5)+1,2)]):
        return True
    return False
n=int(input().strip())
s=0
for i in range(10,n+1):
    if is_prime(i):
        t=i
        while t>0:
            if not is_prime(t):
                break
            t=t%(10**(len(str(t))-1))
        if t==0:
            t=i
            while t>0:
                if not is_prime(t):
                    break
                t=t//10
            if t==0:
                s+=i
                
print(s)

  • ankith_kumar68
     + 0 comments

    code in c:

    int is_prime(int num){

      if (num == 1) return 0;
  for (int i= 2; i<=sqrt(num); i++) {
     if(num%i == 0) return 0;
  }
  return 1;

    }

    int main() {

    int n;
scanf("%d",&n);
unsigned long long int sum = 0;
for (int i = 11 ; i<n; i++) {
     if(is_prime(i)){
         int digit = 1+log10(i),flag = 0;
         for (int j = 1; j<digit; j++) {
              if(is_prime(i/pow(10,j)) && is_prime(i%(int) pow(10,digit-j))) continue;
               else {
                   flag = 1;
                   break;
               }
         }
         if(flag == 0)  
             sum += i;
     }
}
printf("%lld\n",sum);
return 0;

    }