Finding multiples of 3 or 5 below 1000

Question

I am just starting to try my hands in the problems of Project Euler and i have got stuck in the first code itself. Can someone please tell me in the below code where i am going wrong.

The problem statement is as follows:

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.

And my solution is as follows:

multiple1 = 5
multiple2 = 3
index = 2
sum = 0
while multiple1 < 1000 or multiple2 < 1000:
    if multiple1 < 1000:
        sum = sum + multiple1
        multiple1 = 5 * index
    if multiple2 < 1000:
        sum = sum + multiple2
        multiple2 = 3 * index
    index = index + 1
    
print (sum)

I know someone has already posted the solution somewhere, but i just want to figure out where i am going wrong with the logic.

Asked By: Subhayan Bhattacharya

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Source

Answer 1

Your code will generate duplicate values. For example, 15 is multiple of both 3 and 5. Hence, you are adding duplicate values to the sum every time a number is multiple of both 3 and 5. The code below fixes your issue.

multiple1 = 5
multiple2 = 3
index = 1
sum = 0
while multiple1 < 1000 or multiple2 < 1000:
    multiple1 = 5 * index
    multiple2 = 3 * index
    if (multiple1 < 1000) and (multiple1 % 3 != 0):
        sum = sum + multiple1
    if (multiple2 < 1000):
        sum = sum + multiple2
    index = index + 1
print (sum)

Answered By: Deja

Answer 2

The simplest way to do this

total = 0
for i in range(1,1000):
    if i%3 == 0 or i % 5 == 0:
       total += i
print(total)

Answered By: Omkar

Answer 3

you can do without loop also

n=1000
remider=(n+2)//3 #get the reminder of 3  
remider2=(n+4)//5 #get the reminder of 5  
fsum = 3*((remider-1)*(remider))/2 #3*(N*(N+1)/2))  
ssum = 5*((remider2*(remider2-1))/2) #5*(N*(N+1)/2))  
first_list = list(range(3,3*(remider),3)) #list of mutiple of 3  
second_list = list(range(5,5*(remider2),5)) #list of mutiple of 5  
c_sum=sum(list(set(second_list) & set(first_list))) # sum of common elements  
print(fsum+ssum-c_sum)

Answered By: Shravan Jain

Finding multiples of 3 or 5 below 1000

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