CODEWITHSUNDEEP
pythonpython-2.7
Robbie
Robbie

Reputation: 1774

Project Euler Python

I am working on the Project Euler problems. I am on Problem 1 and for some reason I am getting the wrong answer for it. I don't know if its my math logic that's the problem or if the logic for the code is wrong. Here's the code:

def getSumMult3_5(n):
    count = 0
    i = 0
    i_1 = 0
    while i*5 < n:
        count += i*5
        i += 1
    while i_1*3 < n:
        count += i_1*3
        i_1 += 1
    return count

print getSumMult3_5(1000)

It's returning that

count = 266333

but that is not the correct answer. Count should equal 233168. Can anybody help me figure out why it's doing this?

Thanks!

Upvotes: 0

Views: 125

Answers (3)

Anirudh
Anirudh

Reputation: 1

Here's my code:

running_sum = 0





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

You're double-counting numbers like 15 that are multiples of both

Upvotes: 0

watwolves8855
watwolves8855

Reputation: 27

You are double counting multiples of 15. You can solve this by introducing a third while statement to decrease the count by 1 for each multiple of 15.

I suggest using an if/else statement:

def getSumMult3_5(n):
  s = 0
  for i in range(1, n+1):
    if i%3==0:
      s+=i
    elif i%5==0:
      s+=i
  return s

print getSumMult3_5(1000)

This is because that if the number is a multiple of 3, it does not check whether it is a multiple of 5; if it is not a multiple of 3, it checks whether it is a multiple of 5.

I also recommend using s as a variable name because it represents sum, which can't be used as it is a keyword. Count seems to refer to the number of times something occurs.

By only using one variable, and using an if/else statement, execution time will be less and it will be less confusing to read.

Good luck!

Upvotes: 0

AGreenman
AGreenman

Reputation: 225

You're double counting numbers that are multiples of both 5 and 3, such as 15.

Upvotes: 2

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