Help please: Find the sum of all the multiples of 3 or 5 below 1000. (I'm using a website called Project Euler.net to help me practice code. I'm already stuck on the first problem.)

Question

Can you help me with this problem?

Answer 1

something like this?
public int findSum(){
int sum=0;
for(int i=0;i<1000;i++){
// if i is a multiple of 3 OR 5
// (divided by either with ZERO remainder)
// add i to the total sum
if(i%3 == 0 || i%5 == 0)
sum+=i;
}
System.out.println(sum);
}

or am i misunderstanding the question?

Answer 2

i'll go through the code line by line

sum = 0;
this variable will hold the final sum

for(int i=0;i<1000;i++)
this is the loop which goes through all the numbers from 0 to 999 (1000 is not included)
each iteration of the loop the variable 'i' which was initialized to 0 (int i=0) will incremented (i++) as long as its value is LESS than 1000 (i<1000)

inside the loop (remember that each loop cycle the value of i is increased by 1)
if(i%3 == 0 || i%5 == 0)
this condition checks if i can be divided by 3 with no remainder (i%3 == 0)
OR (|| in the code)
if i can be divided by 5 with no remainder (i%5 == 0)
if one of the conditions is met, add the value of 'i' to the sum.

a little explanation about the operator % (modulus):
this is a very handy operator which can be used in many different ways in programming.
in this code it is used to determine whether the value 'i' can be divided fully by 3 or 5 in a given loop cycle (iteration).
the way this works is as follows:

modulus of variable A on variable B (A%B) will give you the remainder, or in other words, you can say it like this: how many times can i put B inside A and what do i have as leftovers?
the leftovers is the remainder
few examples with numbers:

7%5 -> i can put 5 inside 7 one time and that is it, but i still have 2 as leftovers, so that is the remainder.

3%6 -> i can put 6 inside 3 ZERO times since it is bigger than 3, so the remainder is 6(as a rule of thumb, if the right side variable is bigger than the left, then the modulu is the right side variable)

28%5 -> i can put 5 inside 28 five times and my remainder is 3 (28-25=3)

and finally
when a number A is a multiple of B there will be NO remainder left

33%11 = 0 because 11*3 == 33

and that is exactly what was done inside the loop.

Answer 3

then it was all worth it :)

Answer 4

public class sumNumber
{
Public static void main ( String[] args)
{
Int num; numb;
Int sum1; sum2;
Int sumTotal;

for( num=1; n<=1000;num++)
{
sum1+=num*3;
}

for(numb=1; numb<=1000;num++)
{
sum2+= num*5;
}

sumTotal = sum1+sum2;
System.out.println("the total of the sum is" + sumTotal);
}
}

Answer 5

Hey, do u know what is the answer of the question? Is it 4004000?

Answer 6

@Burey thank you! Can you explain how you got the right answer? I still don't completely understand.

Answer 7

@Burey Thanks for taking the time to explain! This was very helpful and now I understand it!

Answer 8

@Vicky Li the answer is 233168

In Eclipse I put in:

public final class Problem1{
public static void main(String[ ]args){
System.out.println(new Problem1( ).run( ) );
}
public String run( ){
int sum= 0;
for (int i= 0; i< 1000; i++){
if (i% 3== 0 || i% 5== 0)
sum += i;
}
return Integer.toString(sum);
}
}
//output: 233168

Answer 9

sum <- 0
for (i in 1:1000){
if (i%%3 ==0){
sum = sum + i
}
}
print(sum)