i want by any language thje code of prime factorization for prime number ,first i want traning:we input the number of sample and put it in array,in this array must used the number by this method 2*2=4 2*3=6 2*5=15 ...... let example we have input n=125\nthen we have 8 traning and 7 testing ,from traning we find the weight from algorthim ex:batch back or stoastic ....\nand we find the weight from test sample and get the output and compare with the value in array to find the accurcy thanks\n\niam sory for my bad english language\nm.n.a
200471
look at resource below: (code in Python language)\n\n(By the way is it your school assignment? you have asked this question in many categories!!!, so if it is your assignment, i hope this will not be in Python, but try to grasp on the algorithm not on the code: if u know the algo better, u can easily convert into code with little effort)
399305
with n a perfect square  represent n as x^2 + p with prime p = 1 mod 4 and write p = y^2 + z^2. This yields the representation n = x^2 + y^2 + z^2. \n the largest n for wicht this is not possible is 9634\n\nHow serious are you with your questions ?\nsee http://answers.yahoo.com/question/index;_ylt=AvivP2vdn_qSdnT5M_Siyhfsy6IX?qid=20061106081029AAr4yHJ\n\nyour chooses answer is not an answer
494044
How about 45360 = 2^4 * 3^4 * 5 * 7\n# factors = (4+1)(4+1)(1+1)(1+1) = 5*5*2*2 = 100\nThe number of factors of the number:\nx = p(1)^r(1) * p(2)^r(2) * p(3)^r(3) * ... * p(n)^r(n), where this is the prime factorization, is:\n(r(1) + 1)(r(2) + 1)(r(3) + 1)...(r(n) + 1)\n\nThe question did not ask for just prime factors but all factors.\nFor instance, if 2 and 3 are factors then so is 6.\n45360 has exactly 100 factors.\nThey are:\n1,  2,  3,  4,  5,  6,  7,  8,  9,  10,  12,  14,  15,  16,  18,  20,  21,  24,  27,  28,  30,  35,  36,  40,  42,  45,  48,  54,  56,  60,  63,  70,  72,  80,  81,  84,  90,  105,  108,  112,  120,  126,  135,  140,  144,  162,  168,  180,  189,  210,  216,  240,  252,  270,  280,  315,  324,  336,  360,  378,  405,  420,  432,  504,  540,  560,  567,  630,  648,  720,  756,  810,  840,  945,  1008,  1080,  1134,  1260,  1296,  1512,  1620,  1680,  1890,  2160,  2268,  2520,  2835,  3024,  3240,  3780,  4536,  5040,  5670,  6480,  7560,  9072,  11340,  15120,  22680,  45360
526238
You mean LCM?  LCD is lowest common denominator.\n\nFor each number that you are trying to find the LCM for, break the number down into its prime factors (recall that the primes are 2, 3, 5, 7, 11, 13, 17, etc.), then gather the highest prime factor of each prime.\n\nAn example will work better here.  Let's try this:  Find the LCM of 9, 12 and 150.\n\nSTEP 1:  Get prime factors of each:\n9 = 3*3 = 3²\n12 = 2*2*3 = 2² * 3\n150 = 2*3*5*5 = 2 * 3 * 5²\n\nSTEP 2:  Take the highest prime factor of each prime:\n2 >> highest is 2² from 12\n3 >> highest is 3² from 9\n5 >> highest is 5² from 150\n7 >> we don't have any primes for 7 or higher, so we can stop\n\nSTEP 3:  Multiply the highest prime factors from step 2 together to get the LCM:\n\nLCM = 2² * 3² * 5² = 900\n\nHope that helped a little!\n\n~ ♥ ~
408554
Why do you need to implement it yourself? Why not use the Crypt::Rijndael_PP module from CPAN?\n\nIf anything, you could use the module as a working implementation to study ( or copy from ).\n\nEDIT: Changed module name.
567366
The $7,938,000 answer is INTEGER.\n\n9 x 14 x 20 x 5 x 7 x 5 x 18 = 7,938,000.\n\nBecause the prime factors of 7,938,000 are {2,2,2,2,3,3,3,3,5,5,5,7,7}, you can eliminate letters with values based on other prime numbers that are not included in this solution set.  In other words, the following letters could NOT be in the word you are trying to find:  K {11}, M {13}, Q {17}, S {19}, V {22}, W {23}, and Z (26}.\n\nAlso, it would have been very unlikely for P {16} to have been a letter because it would have forced ALL remaining letters to have odd values (as all four 2's from the prime factorization would have been used up in the one letter "P").\n\nEliminating M, Q, S, and V were significant clues.  It eliminated any derivations of the words "Multiply," "SuM," "Quotient", "SQuare," "Subtract," and "diVide."\n\nSo, when "rectangle" and "triangle" included acceptable letter combinations but were just a little too big, I noticed that these 2 words both had T, N, G, E, and R.  Shortly thereafter, INTEGER fell into place.\n\nI hope you get some really worthwhile extra credit for this one.
370958
To see what conjecture they are interested in, it really is necessary to consider a number of examples, exactly as they said. So lets set up a table. We will call the product in question N.\nRemember that we are only interested in prime numbers for p.\n\np               N                N/p\n---------------------------------\n2                    2               1\n3                    3               1\n5                  25               5\n7                721           103\n11      3628801     329891\n13  479001601 36846277\n\nNotice that the numbers N so far have all been divisible by p. So, if we represented  them on a p-clock, they would all be p o'clock (or 0 o'clock if you prefer). That is the conjecture they are interested in. You are supposed to conjecture that the value of 1*2*3*...*(p-1)+1 in p-hour arithmetic is always p o'clock.
311031
Primes less than 100:\n2\n3\n5\n7\n11\n13\n17\n19\n23\n29\n31\n37\n41\n43\n47\n53\n59\n61\n67\n71\n73\n79\n83\n89\n97\n\nFrom this list we keep only:\n3\n7\n13\n17\n23\n31\n37\n43\n47\n53\n67\n71\n73\n79\n83\n97\n\nGood luck...
384646
First, start listing the prime #'s starting with the number 2.\nAnd let's only go up to 50, because 2 x 50 = 100 and we need 3 #'s.\n\nPRIME #'S LESS THAN 50:\n\n2\n3\n5\n7\n11\n13\n17\n19\n23\n29\n31\n37\n41\n43\n47\n53 ........oops!  STOP, too far.\n\nOkay.  Now we want the product of the 3 #'s needs to be 100.\n\n100 is an even #, so one of the numbers must be even.  \nThat eliminates, quite a lot.\n\nActually, that eliminates all of the #'s but the number 2.\nSo, 2 is the 1st prime number less than 100.\n\nYa, know. . . .\n \nI gave it a try.\n\nAnd I can't seem to find the other two numbers.\n\nYou got me stumped.\n\nPerhaps I read the question wrong.\n\nDid you mean:\nWhat three different prime numbers, multiplied together, have a product of 100?\n\nIf so, you stumped the professor!  :o)
551836
pseudo code sort of, based on C++... easiest for me... I am going to do this with an array of integers because the number of integers read into the program can be different than just three.  This can be easily changed though...\n\nstart program:\n\nint arrayInt[3];  //  places to store three integers\nint largestNumber = 0;  // place to store largest number\ncout << "enter three numbers:  ";   // request for three numbers\nfor (int i = 0; i < 3; i++)\n{     cin >> arrayInt[i]; }    \n// takes the three values from the user's input and stores them in an element in the arrayInt array.\n// so... now you need to compare all of the elements in arrayInt\nfor (int i = 0; i < 3; i++)  // do this three times\n{\n     if ( arrayInt[i] > largestNumber )  \n// comare largestNumber to the current element in the array\n// if it is larger, give largestNumber the value of that element\n     {\n         largestNumber = arrayInt[i] );\n     }\n}\n\n// display the largest number\ncout << "the largest number is:  " << largestNumber << endl;\n\n ******************************************************************\n\n1. initialize variables for the three input integers and a variable to store the largest integer (set this one equal to zero).\n2. ask the user for the three numbers\n3. read the user's input into the variables to store them\n4. compare the values stored in the three variables by either doing it individually (if an array is not used) or with a for loop.  Go through each of the three variables that store the user's input and compare the stored value to the largestNumber variable.  The largestNumber variable should be set equal to 0 since we know that all the numbers will be non negative, therefore 0 is the lowest any of the input numbers can be.  If the value of a variable that stores a user given integer is larger than the largestNumber variable, store the value of the input variable in the largestNumber variable ( largestNumber = inputValue2 ).  As you check all of the three input variables, the largest of all of the values checked will be stored in the largestNumber variable. \n5. tell the user the largest variable
72413
{determinses whether a number is prime}\nvar number, j : integer;\n    prime : boolean;\nbegin\n     writeln('type your number');\n     readln( number );\n     prime := true;\n     for j := 2 to abs(number) -1 do{allows neg nos.}\n         if number mod j = 0 then\n            prime := false;\n     if prime then\n        writeln( number , ' is a prime number')\n     else\n         writeln( number , ' is not a prime number');\n     readln\nend.
1990
There are some efficient ways of finding prime numbers:\n\nThe naive way to see if a number n is prime is to divide it by all the integers from 2 to the square root of n.  If none of these divide n evenly, n is prime.\n\nThere are more sophisticated ways linked below - if you want to find all primes in a given range, the Sieve of Eratosthenes (link below) is a good way to do that.  There are also probabilistic tests - i.e. you can be 99.9% (or however much you want less than 100%) sure n is prime.  These have the general advantage of being much faster.  The Fermat primality test (link below) is a good example of these, and relatively simple to implement.
566249
Prime numbers are those that are divisible  by itself while compound numbers are those that are divisible by many of its multiples. eg of primary number--1,2,3,5,7,11,13 etc\neg of compound number--4,6,8,9,12 etc\nquotient is the result obtained when a number is divided by another number.
206321
the simple program given above if fine.. in terms of the conncept.. but there's a bug... \nthe problem is with looping.. coz when u enter a number such as 80301  or 55302 ( a no. having 0 in between), the loop breaks at the point when zero is encountered.\n\n so the output in the first eg. is  8 0 ,sum= 8 (instead of correct 8 0 3 0 1 , sum=12\n\n\nso the logic of handling the number is absolutely correct but the looping is a comparatively difficult job in this case!
484008
To get prime factors, keep dividing by numbers and divide those until you can't divide them no more!  For example,\n\n34 \n17 * 2\n\n(Can't divide 17 any more and can't divide 2 -- both of these are prime numbers)\n\nAnother example,\n\n120\n12 * 10\n2 * 6 * 2 * 5\n2 * 2 * 3 * 2 * 5 (<-- all prime numbers here)\n\nTry to get the rest yourself and post the answers, we'll tell you if you got them right!  ;-)
350162
Go for it... ask some questions and I'll try and help.  However, be forewarned, I like to give people the *method* to solve a problem, so they can do it later on their own.  You will rarely find me just giving out answers.  Hope that is okay.\n\nIn your example:\n64 - 40ab\n\nThe best you can do is factor out a common factor from 64 and 40.  Both of these numbers are divisible by 8 (8 x 8 and 8 x 5)... that means you get:\n\n8(8 - 5ab)\n\nSince 8 and 5 share no further common factors, you can't factor them any further.  Also, a and b are only part of the second number, so you can't factor those...  the answer is:\n8(8 - 5ab)
288856
Now, the basic definition of P(n,k) is n!/(n-k)!\n\nUsing this, we can try to re-express the formula. Notice that the numerator can be written as (n-k)!(n-k+1)(n-k+2)(n-k+3)..., with the last term being (n-k+k)\n\nNow, substitute this back into the original formula and try to carry out some cancellation. Observe that you are left with a product, and not a fraction. Study this product. How many terms are there? And how is this relared to "a factorial-type computation using exactly k factors beginning with n"?