C solution with gcd, short and easy with brief explanation
If anyone wants mathematical proof just comment below\n\nint gcd(int a, int b)\n{\n    if (a == 0) return (b);\n    return (gcd(b % a , a));\n}\n\nbool isGoodAr
If anyone wants mathematical proof just comment below\n```\nint gcd(int a, int b)\n{\n    if (a == 0) return (b);\n    return (gcd(b % a , a));\n}\n\nbool isGoodArray(int* nums, int numsSize)\n{\n    if (numsSize == 1) return (nums[0] == 1);\n    int flag = gcd(nums[0],nums[1]);\n    for(int i = 2 ; i < numsSize ; i++) flag = gcd(flag, nums[i]);\n    return (flag == 1);\n}\n//if there exists one integer greater than 1 that can divide all the numbers, return false\n//else return true\n```
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