CODEWITHSUNDEEP
pythonpython-3.x
mllnd
mllnd

Reputation: 575

Sum up all the integers in range()

I need to write a program that sums up all the integers which can be divided by 3 in the range of 100 to 2000. I'm not even sure where to start, so far I've got this tiny piece of code written which isn't correct.

for x in range(100, 2001, 3):
      print(x+x)

Any help is much appreciated!

Upvotes: 10

Views: 94326

Answers (6)

Ali A
Ali A

Reputation: 359

Let's do it in Recursive way .

Utilizing a range_sum function

def range_sum(array):
if array[0]==array[1]:
    return array[0]
else:
    return range_sum([array[0],array[1]-1])+array[1]
range=[1,4]
print(range_sum(range))

10

Upvotes: 0

Aristide
Aristide

Reputation: 3994

There is a closed formula for that.

If (u_i) is a sequence defined by its first term u_0 and its common difference r, then the sum of the n first terms of (u_i) is:

\frac{n(u_0 + u_{n-1})}{2}

EDIT: I have made this little video to explain it visually.

A popular anecdote attributes this formula to the young Johann Carl Friedrich Gauss.

In your case:

  • u_0 = 102
  • u_{n-1} = 1998
  • n = (1998 - 102) / 3 + 1 = 633

So, the sum is (633 * (102 + 1998)) / 2 = 664650.

As a general Python function with the usual range arguments start, stop, step:

def arithmetic_series(start, stop, step):
    number_of_terms = (stop - start) // step
    sum_of_extrema = start + (stop - step)
    return number_of_terms * sum_of_extrema // 2

In your case, the call would be:

arithmetic_series(102, 2001, 3)

The complexity is O(1) instead of O(n), so unsurprisingly:

%timeit sum(range(102, 2001, 3))
100000 loops, best of 3: 17.7 µs per loop

%timeit arithmetic_series(102, 2001, 3)
1000000 loops, best of 3: 548 ns per loop

Upvotes: 12

Deck
Deck

Reputation: 1979

There is a sum function 

>>> sum(filter(lambda x: x % 3 == 0, range(100, 2000)))
664650

But this is better:

>>> sum(x for x in range(100, 2000) if x % 3 == 0)
664650

Upvotes: 4

Inbar Rose
Inbar Rose

Reputation: 43477

Since you know the first number in this range that is divisible by 3 is 102, you can do the following:

Solution:

>>> sum(range(102, 2001, 3))
664650

To make it into a robust function:

def sum_range_divisible(start, end, divisor):
    while start % divisor != 0:
        start += 1
    return sum(range(start, end, divisor))

Using it:

>>> sum_range_divisible(100, 2001, 3)
664650

Note:

The advantage here is that you do not have to check each number in the whole range, since you are jumping by 3 each time.

Timing:

I have timed the different solutions, mine and aga's:

>>> import timeit
>>> timeit.Timer('sum(range(102, 2001, 3))').repeat()
[9.516391893850312, 9.49330620765817, 9.508695564438462]
>>> timeit.Timer('sum(x for x in range(100, 2001) if x % 3 == 0)').repeat()
[134.757627812011, 134.46399066622394, 138.34528734198346]

Conclusion:

My answer is faster by a factor of 14

Upvotes: 20

aga
aga

Reputation: 29434

Use generator expression and sum function here:

res = sum(x for x in range(100, 2001) if x % 3 == 0)

It's pretty self-explanatory code: you're summing all the numbers from 100 to 2000, inclusive, which are divisible by three.

Upvotes: 13

devil00
devil00

Reputation: 11

sum(filter(lambda l : l%3 ==0, range(100,2001)))

Upvotes: 0

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