Find the sum of all the multiples of 3 or 5 below 1000

Question

If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23. I have the following code but the answer do

Answer 1

You might start by iterating from 3 to 1000 in steps of 3 (3,6,9,12,etc), adding them to the sum like,

int i = 3, sum = 0;
for (; i < 1000; i += 3) {
  sum += i;
}

Then you could iterate from 5 to 1000 by 5 (skipping multiples of 3 since they've already been added) adding those values to the sum as well

for (i = 5; i < 1000; i += 5) {
  if (i % 3 != 0) sum += i;
}

Then display the sum

printf("%d\n", sum);

Answer 2

#include<stdio.h>
#include<time.h>
int main()
{
    int x,y,n;
    int sum=0;
    printf("enter the valeus of x,y and z\n");
    scanf("%d%d%d",&x,&y,&n);
    printf("entered   valeus of x=%d,y=%d and z=%d\n",x,y,n);
    sum=x*((n/x)*((n/x)+1)/2)+y*((n/y)*((n/y)+1)/2)-x*y*(n/(x*y))*((n/(x*y))+1)/2;
    printf("sum is %d\n",sum);
    return 0;
}
// give x,y and n  as 3 5 and 1000

Answer 3

Rather than using range/loop based solutions you may wish to leverage more math than brute force.

There is a simple way to get the sum of multiples of a number, less than a number.

For instance, the sum of multiples of 3 up to 1000 are: 3 + 6 + 9 + ... + 999 Which can be rewritten as: 3* ( 1 + 2 + 3 + ... + 333)

There is a simple way to sum up all numbers 1-N:

Sum(1,N) = N*(N+1)/2

So a sample function would be

unsigned int unitSum(unsigned int n)
{
    return (n*(n+1))/2;
}

So now getting all multiples of 3 less than 1000 (aka up to and including 999) has been reduced to:

3*unitSum((int)(999/3))

You can do the same for multiples of 5:

5*unitSum((int)(999/5))

But there is a caveat! Both of these count multiples of both such as 15, 30, etc It counts them twice, one for each. So in order to balance that out, you subtract once.

15*unitSum((int)(999/15))

So in total, the equation is:

sum = 3*unitSum((int)(999/3)) + 5*unitSum((int)(999/5)) - 15*unitSum((int)(999/15))

So now rather than looping over a large set of numbers, and doing comparisons, you are just doing some simple multiplication!

Answer 4

Using the stepping approach, you can make a version:

#include <stdio.h>

int main ( void ) {

    int sum = 0;

    for (int i = 0; i < 1000; i += 5) {
        sum += i;
    }
    for (int i = 0; i < 1000; i += 3) {
        if (i % 5) sum += i;  /* already counted */
    }
    printf("%d\n", sum);
    return 0;
}

which does a whole lot fewer modulo computations.

In fact, with a counter, you can make a version with none:

#include <stdio.h>

int main ( void ) {

    int sum = 0;
    int cnt = 6;

    for (int i = 0; i < 1000; i += 5) {
        sum += i;
    }
    for (int i = 0; i < 1000; i += 3) {
        if (--cnt == 0) cnt = 5;
        else sum += i;
    }
    printf("%d\n", sum);
    return 0;
}

Answer 5

Perhaps you should do

sum += i // or sum = sum + i

instead of

sum = sum + 1

Additionally, be careful when printing long unsigned ints with printf. I guess the right specifier is %lu.

Answer 6

int main(int argc, char** argv)
{
    unsigned int count = 0;
    for(int i = 1; i < 1001; ++i)
        if(!(i % 3) && !(i % 5))
            count += i;
    std::cout << i;
    return 0;
}