integer-division

This question already has an answer here: Integer division: How do you produce a double? 10 answers if I have something like: long x = 1/2; shouldn't this be rounded up to 1? When I print it on the screen it say 0. It's doing integer division, which truncates everything to the right of the decimal point. Integer division has its roots in number theory. When you do 1/2 you are asking how many times does 2 equal 1? The answer is never, so the equation becomes 0*2 + 1 = 1, where 0 is the quotient (what you get from 1/2) and 1 is the remainder (what you get from 1%2). It is right to point out that

How could I go about finding the division remainder of a number in Python? For example: If the number is 26 and divided number is 7, then the division remainder is 5. (since 7+7+7=21 and 26-21=5.) Uku Loskit you are looking for the modulo operator: a % b for example: 26 % 7 Of course, maybe they wanted you to implement it yourself, which wouldn't be too difficult either. The remainder of a division can be discovered using the operator % : >>> 26%7 5 In case you need both the quotient and the modulo, there's the builtin divmod function: >>> seconds= 137 >>> minutes, seconds= divmod(seconds, 60)

I need to find whether a number is divisible by 3 without using % , / or * . The hint given was to use atoi() function. Any idea how to do it? Subtract 3 until you either a) hit 0 - number was divisible by 3 b) get a number less than 0 - number wasn't divisible -- edited version to fix noted problems while n > 0: n -= 3 while n < 0: n += 3 return n == 0 MSalters The current answers all focus on decimal digits, when applying the "add all digits and see if that divides by 3". That trick actually works in hex as well; e.g. 0x12 can be divided by 3 because 0x1 + 0x2 = 0x3. And "converting" to hex