Python -- Adding digits in a large number fails

Question

I'm working on a bit of code that adds up all the digits of a number and prints out the sum:

number = #some big 50+ bit number
sum = 0

while number != 0:
    sum += number % 10
    number = int(number/10)

print(sum)

I thought this was a pretty slick solution but when I added print statements to check, it wasn't dividing correctly when the number was very large. I thought that because I was using python 3.8 that I could have arbitrarily large numbers. Is this not the case or am I asking the wrong question?

I tried replacing number = int(number/10) to both number = math.trunc(number/10) and number = math.floor(number/10) and got the same results for both of those as well

Every version of python (as far as I know) has arbitrarily large ints. — xilpex, May 17 '20 at 03:36
You want to go for integer division (`numer//10`), the reason is: https://stackoverflow.com/questions/588004/is-floating-point-math-broken — Adam.Er8, May 17 '20 at 03:37
Do not call your variable `sum` - it is a built-in function. — DYZ, May 17 '20 at 03:42

GoodDeeds · Answer 1 · 2020-05-17T03:43:18.043

6

The / operator does float division. You need // for integer division. So, use:

number = #some big 50+ bit number
s = 0

while number != 0:
    s += number % 10
    number = int(number//10)

print(s)

Dividing as floats may not work since you lose precision.

edited May 17 '20 at 03:43

answered May 17 '20 at 03:37

GoodDeeds

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Still, do not call your variable `sum`. And there is no need for `int()`. – DYZ May 17 '20 at 03:42

score 3 · Answer 2 · answered May 17 '20 at 03:43

3

You can try this way. Convert number into string and loop through string

number = # big number
S = 0
for x in str(number):
  S += int(x)
print(S)

answered May 17 '20 at 03:43

Hirusha Fernando

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Python -- Adding digits in a large number fails

2 Answers2