CODEWITHSUNDEEP
c++mathlogic
user768417
user768417

Reputation:

Find the sum of all multiples of 3 or 5 up to 1000

I'm doing the problems on Project Euler in C++, but I'm not getting the right answers to the first one.

Here's my code:

#include <iostream>

using namespace std;

int main()
{
    int b;
    int c;

    for (int a = 0; a <= 1000;)
    {
        a = a + 3;
        b = a + b;
    }
    cout << b << "\n";

    for (int a = 0; a <=1000;)
    {
        a = a + 5;
        c = a + c;
    }
    cout << c << "\n";

    b = b + c;
    cout << b << "\n";
    return 0;
}

My output is:

167835
101505
269340

Where's the error in my logic?

Upvotes: 0

Views: 4824

Answers (6)

Shamkumar Rajput
Shamkumar Rajput

Reputation: 166

Consider, Find the sum of all multiples of 3 up to 20?

Ans : =>

3, 6, 9, 12, 15 this are multiples of 3 up to 20

Sum of all multiple of 3 up to 20 is => [3 + 6 +9 + 12 + 15]

(3 + 6 +9 + 12 + 15) you can rewrite in following way

3 (1+ 2+3 +4+5 ) = > 3 (15) => 45

sum of sequence can be calculated using following formula

K(K+1)/2 = > here K is 5 => 5 (5+1)/2 = >15

In general, We can say that multiple of any number (N) within given range R

K = R/N;

N* (K (K+1))/2

In our case R =20 and N =3

int sumDivisibeBy(int R, int N)
{
   int K = R / N;
   int  SEQSUM =  ((K*(K + 1)) / 2));
   return (N*SEQSUM)
}

In your case you need to call this function thrice =>

sumDivisibeBy(1000,3) + sumDivisibeBy(1000,5)-sumDivisibeBy(1000,15)

Upvotes: 2

Austin Salonen
Austin Salonen

Reputation: 50245

While others have posted exactly where you've erred, you should be trying to figure out how you got the wrong answer as well.

In your case, you could have written all the values you determined to be multiples of 3 and multiples of 5; then you could have analyzed the 333 multiples of 3 you should've seen and the 199 multiples of 5 you should've seen.

I don't want to give away the keys to finding the actual solution (despite the fact that others have already) but part of the problem solving at PE is debugging.

Upvotes: 0

Leon
Leon

Reputation: 1141

wouldn't bother incrementing by 3 and by 5, you can increment by 1 and check whether numbers are divisible by 3 or by 5. Computers are designed for number crunching.

int sum = 0;
for (int i = 0; i < 1000; i++)
{
    if (i%3 == 0 || 
        i%5 == 0)
    {
        sum += i;
    }
}
cout << "SUM:" << sum << endl;

Upvotes: 0

Whoopska
Whoopska

Reputation: 149

Along with double counting multiples of 15, your increments are in the wrong order. If you increment a first, you will have values above 1000. Also I'm not sure about c++ initializing ints, but maybe set them equal to 0, at least for readers.

Upvotes: 0

tskuzzy
tskuzzy

Reputation: 36476

You're double counting numbers that are multiples of 3 and 5 (i.e. multiples of 15).

Upvotes: 2

Sven Marnach
Sven Marnach

Reputation: 602745

You are adding all values that are both multiples of 3 and 5 (i.e. multiples of 15) twice. Additionally, you will also include 1002 and 1005, which probably isn't intended.

Upvotes: 6

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