What is Modulus Division in computer science?
538012
modulus division is division that returns the remainder of a division\n\nSo for example, \n5/3 = 1 with remainder 2, so 5 mod 3 = 2\n\nit is basically returning the remainder for a division
866735
I believe you use the remainder theorem for this, which states that the remainder when a polynomial f(x) is divided by (x-k) is\nf(k)\n\nThus the remainder would be f(-1) = 11
676773
Since your examples are of the form "x^y mod N",\nI would recommend that you look at the page:\n\nhttp://en.wikipedia.org/wiki/Exponentiation_by_squaring\n\nand, in particular, the section titled:\n\nFurther applications\n\nwhere it discusses the most efficient method for\n"fast computation of large exponents modulo a number".\nFor your examples, it isn't as critical, but if you're\nexploring cryptography, you'll end up having to\ndeal with much larger exponents.\n\nFor your specific examples, you could use a "bignum"\npackage and just brute-force compute 123^7 and take\nthe remainder after division by 187.
806367
The best way to handle creating a program of that sort is to use the Mod function to find whether a number comes out evenly i.e. 7 Mod 3 returns 1 because it gives the remainder after 7 is divided by 3. Use a For...Next statement that take the number (intNum) that the user inputs and "Mod's" it by every number below it. This will give you some of the factors so you can find out the perfect, abundant, and deficient of numbers.
150391
It is a strings proccessing alg which uses a failure function to improve the run time rather than bm alg\nrun time:m+n\nm:the length of the pattern\nn: the length of the text\nwe use tis algorithm to find the pat string in the text string..\nmore details \nmail me at ziko@math.com\ni can provid a pascal source for it
559200
By order of precedence of operations:\n\ny = 2^1/2  is the same as\ny = (2^1)/2  which is just\ny = 2/2 which yields\n\ny = 1 <<<<<<<<<<<<<<<<<<<<<<
312034
writing it as a quotient means writing it as a division, as a fraction:\n\n(n*1)/5
885409
A number that can be written as a fraction.
964841
Figure it out - when you divide 60 by each of those numbers, which one leaves a remainder?
173044
The period is pi/2.\n\nsin(x+pi/2)=sin(x)cos(pi/2) + cos(x)sin(pi/2)=cos(x)\ncos(x+pi/2)=cos(x)cos(pi/2) - sin(x)sin(pi/2)=-sin(x)\n\n\nAnd obviously, g(x)=mod(sinx)+mod(cosx) has a period of pi/2.\n\n- - - - - - -  - - - - \n\nmod is short for modulus, which means absolute value.
1137788
Nope, 0/0 could be 1 but it could also be anything else.   Division is the opposite of multiplication and 0 times anything at all, even infinity, is still zero.
241809
Quote: "The binary arithmetic operators are +, -, *, /, and the modulus operator %.", more abstractly an arithmetic operator is an operator which is applied to an arithmetic type.  Hope this helps.
1033912
It's actually pretty easy.  When you take a number, jumble around it's digits and then subract the new number from the old, you are always going to get a difference divisible by 9.  Since 9 divides the sum of the digits of all numbers divisible by 9, then nine divides the sum of the digits of the difference.  When you give all but one of the digits (and knowing that the last one is not 0), the computer just takes the sum of the digits you gave it, and subtracts it from a multiple of 9 (so that the remainder is between 1 and 9).  The number that you picked will always be the remainder.
1142175
One property of fibonacci series is the following\n Take two consecutive fibonacci numbers and apply to them the euclidean division algorithm.\nThen they will get the maximum number of steps( or the maximum number of divisions).\n So a computer will require more time for division of big numbers if you use euclidean division algorithm.
882691
Its an old arithmetic trick used when adding two large numbers.  You sum the first numbers digits, then the second numbers, ... etc, then take the remainder when you divide that sum by 9.  If the sum of the total's digits does not yield the same remainder, you made a mistake. For example:\n\n213+756+1148=2417\n2+1+3=6\n7+5+6=18/9 = 2 r 0\n1+1+4+8=14/9=1 r 5\nAdding remainders, we have 11/9=1 r 2.  But the sum of our digits in 2+4+1+7=14/9=1 r 5.  So something is wrong. Turns out the 4 should be a 1, 2117.  So its just an error correction method.\n\nSteve
1091280
There are many division in physics which division are you asking for?
95584
I rember doing something like this: you had to take the int you store the input in, and divide it by ten with the mod operator (%), and store it in a new int.  Since the mod operator divides the two numbers, and then returns the remainder, that will separate your last number.  For example:\n\nSay the input is 4321, and you have it stored in int input;\n\nint first_num = input % 10;\n\nthis takes your value in input, divides it by 10, then takes the remainder and stores it in first_num.  In this case, first_num should now equal 1.  \n\nAfter that, you have to shorten your input from 4321 to 432 by dividing (regular dividing this time) it by ten, and storing it in another new int.  (Since int can't have any decimals, dividing by ten just shortens it by the first number).  Then you do the whole thing over until you have all four numbers in separate ints, and then output them to separate lines.  \n\nHope that's clear enough--it's hard to explain this stuff so it's easy to understand.  Good Luck!