# Euler Q1 C++ code

Andrew Mathenge mathenge at gmail.com
Thu Jul 24 00:04:05 BST 2008

```Kinuthia's analysis is correct. That's the answer.

For example, when n = 15.

15 is divisible by 3 and 5 so it will be added to sum  twice.

Andrew Mathenge.

On Wed, Jul 23, 2008 at 4:38 AM, kinuthiA muchanE <muchanek at gmail.com> wrote:
> This email is from a ubuntu-ke subscriber to other subscribers inluding you:)
> Eric,
> Firstly, I think you are using an old version of c++, but that would not
> prevent you from getting the correct answer ;). Secondly, the mistake
> you are making is that you are incrementing sum twice, the question is
> if an integer is a multiple of 3 OR 5. So, if you modify the condition
> in the "while" loop to:
>
> <code>
>        if (n%3 == 0 || n%5 == 0) {
>                sum += n;
>                }
> </code>
>
> ... all should work well. The reason it is working with 10 is because
> you are iterating over very few numbers, but as size of the number
> increases things change!:-)
>
> Does this help?
> Kinuthia...
> On Wed, 2008-07-23 at 09:41 +0300, Eric Kivuti wrote:
>> This email is from a ubuntu-ke subscriber to other subscribers inluding you:)
>> This is the code I used (cpp attachment available):
>>
>
>> 0 mod 3 = 0
>> 1 mod 3 = 1
>> 2 mod 3 = 2
>> 3 mod 3 = 0
>>
>> so for all x mod 3 = 0
>> */
>> #include <iostream.h>
>> void main(){
>>  int n=1, sum=0;
>>  while (n<10){
>>   if((n%3)==0){
>>    cout<<n<<", ";
>>    sum+=n;
>>   }
>>   if((n%5)==0){
>>    cout<<n<<", ";
>>    sum+=n;
>>   }
>>   n++;
>>  }cout<<"\n\nsum = "<<sum<<endl;
>> }
>>
>> Output:
>>
>> 3, 5, 6, 9
>>
>> Sum = 23
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```