@ tushar just one modification to you code would make the things
correct.makin if (a % 2*prime[b] == 0) inspite of if (a % prime[b]
== 0) would take care of even things
hope i am correct..
thanx! for the reply
On Jul 5, 10:47 pm, Tushar Bindal <tushicom...@gmail.com> wrote:
> If my interpretation is right, following should be the code.
>
> int main()
> {
> int userInteger = 0;
> cout << "Enter A Number "<<endl;
> cin >> userInteger; // Ask For a number from the user
> if (userInteger > 0) // Is the number valid?
> {
> int result = 0;
> int prime[5] = { 2, 3, 5, 7, 11 };
> int a,b, count = 0;
> *for(a=1;a<=userInteger;++a) // Looping to user's input
> {
> for(b=0;b < 5;++b) // Looping the prime numbers array
> {
> if (a % prime[b] == 0) //if divisible by any number, just end it there
> break;
> }
> //break or end of loop will bring the control here
> if(b==5) //a was not divisible by any of the first 5 prime numbers
> {
> result+=a;
> ++count;
> }}*
>
> cout << "Numbers Not evenly divisible by 5 prime numbers: " << count
> << endl;
> cout << "The Sum of numbers not evenly divisible by 5 prime numbers: "
> << result << endl;}
>
> getch();
> return 0;
>
> }
>
> ---------------------------------------------------------------------------
> -----------------------------------------------------
>
> As per my solution,
> Test Cases:
> 1)
> userInteger = 20
> count = 4
> Numbers will be: 1, 13, 17, 19
> result = 50
>
> 2)
> userInteger = 35
> count = 7
> Numbers will be: 1, 13, 17, 19, 23, 29, 31
> result = 133
>
> Even numbers can never be there in this list as they are all divisible by 2.
> Bfore 169, only prime numbers can be included.
> Hope my interpretation of your question was correct
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