Thread: Getting a different answer to the one expected/published

Truncatable primes - The number 3797 has an interesting property. Being prime itself, it is possible to continuously remove digits from left to right, and remain prime at each stage: 3797, 797, 97, and 7. Similarly we can work from right to left: 3797, 379, 37, and 3. Find the sum of the only eleven primes that are both truncatable from left to right and right to left. NOTE: 2, 3, 5, and 7 are not considered to be truncatable primes.

The following is my effort to sole this. the answer i get is 573 which makes sense until you consider there is only supposed to be 11 and the example given is bigger than the 11 i found

Code:

#include <stdio.h>
#include <stdlib.h>

#define Countto 1000000

int FindMod( int *, int );
void LoadPrimes( int *Primes );
int CheckPrime( int *, int );
int Truncatable( int *, int, int, int );

int main()
{
    int NumSieve[Countto] = { 0 };
    int Sum = 0, Counter = 0;

    LoadPrimes( NumSieve );

    for (int i = 11; i < Countto; i += 2)
    {
        int Mod = 10, Limit = FindMod( &Mod, i );

        if ( CheckPrime( NumSieve, i % Mod) )
            if ( Truncatable(NumSieve, Mod, Limit, i ) )
            {
                printf("%d\n", i);
                Sum += i;
                Counter++;
            }
        if ( Counter == 11 )
            break;
    }

    printf("Sum of the 11 truncatable primes is: %d\n", Sum);
    return 0;
}

int FindMod( int *Mod, int Num )
{
    int Counter = 0;

    while ( Num % *Mod < Num )
    {
        *Mod *= 10;
        Counter++;
    }

    return Counter;
}

void LoadPrimes( int *Primes )
{
    for ( int i = 4; i < Countto; i += 2 )
        Primes[i] = 1;

    for ( int i = 3; i < 1000; i += 2 )
    {
        if ( Primes[i] == 0 )
            for ( int j = i * 2; j < Countto; j += i)
                Primes[j] = 1;
    }
}

int CheckPrime( int *Primes, int Index )
{
    return Primes[Index] == 0; //Index is prime
}

int Truncatable( int *NumSieve, int Mod, int Limit, int i )
{

    int TruncNum[Limit];

    for ( int j = 0; j < Limit; j++ )
    {
        Mod /= 10;
        if ( !CheckPrime( NumSieve, i % Mod ) )
            return 0;

        TruncNum[j] = i % Mod;
    }

    for ( int j = Limit - 1; j >= 0; j-- )
    {
        if ( !CheckPrime( NumSieve, ( i - TruncNum[j] ) / Mod  ) )
            return 0;

        Mod *= 10;
    }


    return 1;
}

where am i going wrong

so is 3 and 7 but they count those on 3795

Originally Posted by stahta01

Wow, you are truly ignorant about math.

Goodbye.

Tim S.

nothing to do with mathS i can speak and understand plain English

ok am i understanding this correctly? in the function is_truncatable_prime you test if the number passed is not prime ie if a number is prime the if statement fails and the function continues. However, function is_prime tests weather the number is even or not and if it is returns 2. Does this not cause the !is_prime to fail and therefore allow even numbers through.

that's it im going to the ophthalmologist

Originally Posted by cooper1200

that's it im going to the ophthalmologist

Just look at my previous post.