Goldbach Conjecture, Objective-C

Goldbach Conjecture

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

int IsPrime(int number) {
    // Check if a number is prime
    int i = 0;
    
    for (i = 2; i < number; i++) {
        if (number % i == 0 && i != number) { return 0; }
    }
    return 1;
}

int IsEven(int number) {
    // Check if a number is even
    
    if (number % 2) {
        // Number is odd
        return 0;
    } else {
        // Number is even
        return 1;
    }
}

int IsRange(int number) {
    if (number >= 2) {
        return 1;
    } else {
        return 0;
    }
}

int ArrayLength(int array[]) {
    int size = (sizeof array)  / (sizeof array[0]);
    return size;
}

int HowManyPrimes (int number) {
    int numberOfValues = 0, i = 0;
    for (i = 2; i < number; i++) {
        if (IsPrime(i) == 1) { numberOfValues++; }
    }
    
    return numberOfValues;
}

int main(void)
{
    // 1. Input Number
    
    int number = 0;
    printf("Please enter a number from 2 to plus infinite that is even:\n");
    //scanf("%d", &number);
    
    // 2. Check if number meets criteria
    
    if (number == 2) {
        printf("The answer is:\n\n");
        printf("   --> (1,1) ");
        printf("Although 1 is not a prime number, it was considered as such when Goldbach made his conjecture!\n");
        printf("\nThere was a total of 1 unique answer!\n");
        return 0;
    }
    
    if (IsEven(number) == 0 || IsRange(number) == 0) {
        printf("The number is wether not within range, or not even. Please, enter a number between 2 and +infinite that is even!\n");
        return 0;
    }
    
    // 3. Generating an array of prime number at max as big as the number
    
    int nbPrimes = HowManyPrimes(number), primes[nbPrimes], i = 0, j = 0, numberOfPair = 0;
    
    printf("The function is calculating! This may take a while...\n");
    for (i = 2; i < number; i++) {
        if (IsPrime(i) == 1) { 
            primes[j] = i;
            j++;
        }
    }
    
    // 4. Proceed numbers in array to check the possibilities and printing results
    
    printf("The computation is complete, here are the answer(s):\n\n");
    for (i = 0; i < nbPrimes; i++) {
        int reminder = abs(number - primes[i]);
        if (IsPrime(reminder) == 1 && reminder != 1 && reminder >= primes[i]) {
            printf("   --> (%d,%d)\n", primes[i], reminder);
            numberOfPair++;
        }
    }
    printf("\nThere was a total of %d unique answer(s)!\n", numberOfPair);
    
    return 0;
}


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