Programs that Find Numbers Divisible by 5 or 3

Question

i want to make a program to find how many number is divisible by 3 or 5 in a number for example 10 has 3 9 6 divisible by 3 and has 5 and 10 divisible by 5 so total is 5 and so on so i write my code

import math
n=float(raw_input())
div3=(n-2)/3
div5=(n-4)/5
f1=math.ceil(div3)
f2=math.ceil(div5)
sumss=f1+f2
print int(sumss)

but in some number it get wrong answer and the range of input number will be from 1 to 10^18 so i need to use math in it because the time limit for the problem test is 2 second any one have any efficiently equation to make that the loop cant make it it take very long time

Answer 1

This is probably a project Euler question. The issue is that some numbers can be shared by 3 and 5 . For instance 22 :

divisors of 3 : 3 6 , 9 , 12 , 15 , 18 , 21 .

divisors of 5 : 5 10 , 15 , 20

For both 15 occurs, so you did a double count.

The advantage is that 3 and 5 are relatively prime, so the only numbers that are shared are the ones dividable by 15 . So you simply need to undo the double counting:

n=int(raw_input())
div3=n//3
div5=n//5
div15=n//15
sumss=div3+div5-div15
print sumss

In case you allow double counting ( 15 should be counted twice), you can simply use:

n=int(raw_input())
div3=n//3
div5=n//5
sumss=div3+div5
print sumss

Note that the programs omitted the floating point arithmetic: this will result in both faster and more precise programs since floating point numbers work with a limited mantisse and thus can fail to represent a large number correctly (resulting by small errors). Furthermore in general integer arithmetic is faster.

Project Euler #1

Now the problem statement of Project Euler is a bit different: it asks to sum up these numbers. In order to do that, you have to construct an expression to sum up the first k multiples of l :

 k
---
\
/     l*i
---
i=1

Using Wolfram Alpha , one gets this expression . So you can calculate these as:

def suml (k,l) :
    return k*(k+1)*l/2

n=int(raw_input())
div3=n//3
div5=n//5
div15=n//15
sumss=suml(div3,3)+suml(div5,5)-suml(div15,15)
print sumss

This program gives 119 for n=22 which - you can verify above - is correct if you count 15 only once.

Answer 2

I am not sure whether I got the question right, but here is some idea:

n=float(raw_input())
div3=int(n/3)
div5=int(n/5)
div15=int(n/15)
sumss=div3+div5-div15
print sumss

EDIT: Ah, found the project Euler.

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.

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

That is a different task then the question posted here. It says bellow the number and to find the sum .

I am not sure whether it would be the right thing to post the solution here, so I am rather not doing that.

EDIT2: from Project Euler:

We hope that you enjoyed solving this problem. Please do not deprive others of going through the same process by publishing your solution outside Project Euler. If you want to share your insights then please go to thread 1 in the discussion forum.