division

I am not really trying to optimize anything, but I remember hearing this from programmers all the time, that I took it as a truth. After all they are supposed to know this stuff. But I wonder why is division actually slower than multiplication? Isn't division just a glorified subtraction, and multiplication is a glorified addition? So mathematically I don't see why going one way or the other has computationally very different costs. Can anyone please clarify the reason/cause of this so I know, instead of what I heard from other programmer's that I asked before which is: "because". CPU's ALU

Division in sqlite return integer value sqlite> select totalUsers/totalBids from (select (select count(*) from Bids) as totalBids , (select count(*) from Users) as totalUsers) A; 1 Can we typecast the result to get the real value of division result? Just multiply one of the numbers by 1.0 : SELECT something*1.0/total FROM somewhere That will give you floating point division instead of integer division. In Sqlite the division of an integer by another integer will always round down to the closest integer. Therefore if you cast your enumerator to a float: SELECT CAST(field1 AS FLOAT) / field2

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

I am just wondering, if I want to divide a by b, and am interested both in the result c and the remainder (e.g. say I have number of seconds and want to split that into minutes and seconds), what is the best way to go about it? Would it be int c = (int)a / b; int d = a % b; or int c = (int)a / b; int d = a - b * c; or double tmp = a / b; int c = (int)tmp; int d = (int)(0.5+(tmp-c)*b); or maybe there is a magical function that gives one both at once? On x86 the remainder is a by-product of the division itself so any half-decent compiler should be able to just use it (and not perform a div again

Every positive integer divide some number whose representation (base 10) contains only zeroes and ones. One can prove that: Consider the numbers 1, 11, 111, 1111, etc. up to 111... 1, where the last number has n+1 digits. Call these numbers m 1 , m 2 , ... , m n+1 . Each has a remainder when divided by n, and two of these remainders must be the same. Because there are n+1 of them but only n values a remainder can take. This is an application of the famous and useful “pigeonhole principle”; Suppose the two numbers with the same remainder are m i and m j , with i < j. Now subtract the smaller