Reputation: 51
Beginner here- trying to make a simple python program that will compute/answer this problem:
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.
Currently this is what I have:
a = 0
b = 0
while a < 1000:
a = a + 3
print (a)
while b < 1000
b = b + 5
print (b)
This will print all the numbers being considered. I just need to add them together and that's my answer.
I would like one of two things to happen, instead of the code that I have written:
Upvotes: 4
Views: 19523
Reputation: 33
**with python**
print(sum([i for i in range(1,1000) if i%3==0 or i%5==0 ]))
Upvotes: 0
Reputation: 11
total=0
i=0
while i<1000:
if i%3==0 or i%5==0:
print(num)
total+=i
i+=1
print("Total is: ")
Upvotes: 0
Reputation: 1
def sum_multiply (n):
data = []
for num in range (1, n):
if num % 3 == 0 or num % 5 == 0:
data.append(num)
return sum(data)
sum_multiply(1000)
Upvotes: 0
Reputation: 18381
Actually this problem can be solved in O(1) instead of O(N) without using any loops or lists:
The required sum is the sum of all multiples of 3 plus sum of all multiples of 5 minus the sum of multiples of (3*5=15) below the given number 1000 (LIMIT=999). The sums are calculated as a sum of arithmetic series. It can be calculated in following way:
LIMIT=999
# Get the upper bounds for the arithmetic series
upper_for_three = LIMIT // 3
upper_for_five = LIMIT // 5
upper_for_fifteen = LIMIT // 15
# calculate sums
sum_three = 3*upper_for_three*(1 + upper_for_three) / 2
sum_five = 5*upper_for_five*(1 + upper_for_five) / 2
sum_fifteen = 15*upper_for_fifteen*(1 + upper_for_fifteen) / 2
# calculate total
total = sum_three + sum_five - sum_fifteen
# print result with Python 3
print(int(total))
The result is:
>>>
233168
Upvotes: 12
Reputation: 245519
While a for
loop would work, you could also use a generator expression and sum
:
sum(n for n in range(1000) if n % 3 == 0 or n % 5 == 0)
Upvotes: 0
Reputation: 994917
It is possible to do this in one line in Python using a generator expression:
print(sum(x for x in range(1000) if x % 3 == 0 or x % 5 == 0))
The range(1000)
produces all the integers from 0 to 999 inclusive. For each one of those integers, if it is divisible by 3 or divisible by 5, then it is included in the result. The sum(...)
function adds up all those numbers, and finally print(...)
prints the result.
Upvotes: 4
Reputation: 118031
I would use a for
loop to iterate over each number in your selected range
. Then you can check if the modulus %
is equal to 0, meaning it has no remainder when divided by those values, if so, add it to the total.
total = 0
for num in range(1000):
if num % 3 == 0 or num % 5 == 0:
print(num)
total += num
>>> total
233168
Upvotes: 1