Why does the following python digitSum code gives the wrong answer?

Question

digitSum=0
n=222222222222222222222222222222
while n!=0:
    digitSum=digitSum+(n%10)
    n=int(n/10)
print(digitSum)

The output should be 60, whereas the output is 86.

Add a `print(n)` in your loop, you'll see what is happening: after the first division, `n=22222222222222223739180810240`. — Kenney, Feb 17 '16 at 18:35

score 1 · Answer 1 · answered Feb 17 '16 at 18:44

1

This should work too.

digitSum=0
n=222222222222222222222222222222
stringValue = str(n)
length = len(stringValue)
digitSum = length * int(stringValue[0])
print(digitSum)

answered Feb 17 '16 at 18:44

Nick

70
1
5

score 0 · Accepted Answer · answered Feb 17 '16 at 19:11

Use a // b for integer division, not int(a / b).

On the first iteration of your loop, your code is effectively doing int(222222222222222222222222222222 / 10). This results in a floating point value which you expect to be 22222222222222222222222222222.2 but in reality is 22222222222222223739180810240.0 due to limitations in floating point precision. This value propagates to the next iteration of your algorithm, where it obviously causes your wrong answer.

On the other hand // in Python performs integer division, which results in the truncated result you want. Because Python has unlimited integer range, this will work correctly even with very large numbers. So 222222222222222222222222222222 // 10 correctly results in the same number with the last digit removed.

Why does the following python digitSum code gives the wrong answer?

2 Answers2