find all multiples? .:Resolved:.

[voidflux]

Hello everyone, I again ran into a simple algebra 1 concept that i can't figure out. I have to write a program to add up all the multiples of 3 between 1 and 100; This should be cake once i find out how you find a mutliple of a number. Anyone care to give me the run down on how that works? I'm sure I did it before, I got a final grade of 98 in calc which is sad because I forgot basic algebra concepts. Maybe thats why i scored so low on my SAT's heh anyways, thanks for listening

[manavo11]

A loop like this maybe?

Dim Total As Integer
Dim i As Integer
 
Total = 0
For i = 1 To 100
    If i Mod 3 = 0 Then
        Total = Total + i
    End If
Next
 
MsgBox Total

[voidflux]

I didn't even look at ur code because I want to do it on my own (coding) I don't want you to tell me how to code it, I just need the logic behind finding multiplies of 3, what does find a multiple mean? Like I need an example, what is a multiple? Thanks for the responce.

[Darkwraith]

A multiple of a number m is a number, n, such that n = m * k where k is some integer.

[DiGiTaIErRoR]

I'll explain his source code...

The Mod operator returns the remainder of a division.

So any number that is a multiple of 3(i.e any integer * 3) divided by 3 will return a remainder of 0, thus no remainder.

Simple huh.

[Osiris]

Finding multiples goes like this...
9 is a multiple of 3 because when you multiply 3 × 3 you get 9
10 is not a multiple of 3 because you cannot multiply any whole number with 3 and get 10...

now if you divide 9 into 3 your answer will be 3 with a remainder of 0
if you divide 10 into 3 your answer will be 3 with a remainder of 1

all multiples of a number when divided by that number always have a remainder of 0... so thats what manavo11 was using in his code, he was checking all the numbers from 1 to 100 to see if they had a remainder of 0 when they are divided into 3...

hope this helps

[voidflux]

Excellent! Thank you everyone for your replies!
Here is what I came up with:

#include <iostream>
#include <iomanip>
using namespace std;


int main(void)
{
	//write a for statement to add all multiples 
	//of 3 between 1 and 100.
	int sum = 0;
	for(int i = 1; i <= 100; i++)
	{
		if((i%3) == 0)
		{
			sum += i;
			cout << setw(5) << i;
		}

	}
	cout <<"\nThe sum of the multiplies is: "
		 << sum << endl;
	return 0;
}

    3    6    9   12   15   18   21   24   27   30   33   36   39   42   45   48
   51   54   57   60   63   66   69   72   75   78   81   84   87   90   93   96
   99
The sum of the multiplies is: 1683
Press any key to continue

[Darkwraith]

You could make the loop faster if you counted off by 3's until your number is greater than 100.

 
for(int i = 3; i <= 100; i+=3)
{
     sum += i;
     cout << setw(5) << i;
}

Just a suggestion.

[voidflux]

That is true, and a good idea but it says It wanted you to test numbers 1-100, that loop would be testing numbers 3-100 even though you really don't need to test 1 and 2 i'm just following directions!

[Darkwraith]

Actually, its testing every third number from 3 to 100.

It makes me cry to see wasted processor cycles.

[twanvl]

Actually, since you are looking for the sum of the sequence:
a₀=0
a_n=a_n-1+3
for n=1...floor(100/3)=33
Which is:
(first+last)*number_of_terms/2 = (99+3)*33/2 = 1683

A program would look like:

Dim max As Long
max = 100
MsgBox (3+((max\3)*3))*max\6

This is the maths forum, after all