Find prime numbers p so that ppp X pp=ppppp, p may be different prime numbers.Each step should  have p only.?
553854
I came up with eight possible answers.  I'm assuming what you are saying is that each digit must be a prime number (e.g. in the set {2,3,5,7}) It's pretty obvious that the product will end in 5, and neither of the multipliers can end in 2.\n\nUsing an Excel spreadsheet I came up with the following answers:\n325 x 73 = 23725\n337 x 75 = 25275\n355 x 77 = 27335\n375 x 73 = 27375\n575 x 57 = 32775\n735 x 35 = 25725\n737 x 75 = 55275\n775 x 33 = 25575
853126
Hay here goes your code:\n//comment to make you more clear..\n\n\nlimit = 100;       // arbitrary search limit\n\nfor (i = 2; i <= limit; i++) { \n  is_prime[i] = true   // assume all numbers are prime at first\n}\n\nfor (n = 2; n <= sqrt (limit); n++) {\n   if (is_prime[n]) {\n      // eliminate multiples of each prime, starting with its square\n      for (i = n^2; i <= limit; i += n) { is_prime[i] = false }\n   }\n}\n\nfor (n = 2; n =< limit; n++) { \n  if is_prime[n] then print n\n}\n\nIts algorithm for prime number modify as per ur programing language\ngoog luck
346216
4p+9+(-7p)+2\n\nGroup the p's and the numbers\n\n(4p - 7p) + (9 + 2)\n\nSimplify\n\n(-3p) + (11)\n\nANSWER : -3p + 11
203743
SpawnOfSpock, the fact that there are an infinite amount of numbers does not imply there are also an infinite amount of primes. It is a serious enough question to have been proven in myriad ways by some of the greatest mathematicians.\nThe following is a common one. Suppose there were a finite amount of primes. Then the product p of all these primes is divisible by each prime. Consider the number p+1. Since p+1 was not in the set, it must be divisible by a prime number. Let that prime be s. But (p+1)/s = q + 1/s for some integer q and thus p+1 must also be prime. Thus our assumption that the set of all primes is finite must be wrong.
738875
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.
175248
Unfortunately I don't know of a quick method but I can help you a little here. \n\nFirst, as you probably already know, prime numbers can only be evenly divided by 1 and the number itself.\n\nSecond, since all numbers are a product of prime numbers you can save time in by not testing division by non-prime numbers like 4 and 6 and 9 and so on.\n\nThird, you only have to test those prime number that are smaller than the square root of the number.\n\nIn your example, 117, you need only test the prime numbers less than 10.81.... since that is the square root of 117. So do a trial divide of 117 by 2, 3, 5, and 7.\n\nOf course there are a few other rules that can help you eliminate the first three choices. If the number is odd, it can't be evenly divided by 2. If the sum of digits isn't divisible by 3 then the number isn't either. And all numbers that are divisible by 5 end in 5 or zero.\n\nFor 117, the number is odd so that eliminates 2. The sum of 1 and 1 and 7 is 9 which is divisible by 3 so 117 is divisible by three and therefore not prime.\n\nSo now you may notice that 117 is divisible by 3 and 39 but I said you only need to test numbers smaller than the square root. What about 39? Well, since the number is divisible by three that alone proves the number is not prime. If you want the full prime factorization, you need to keep dividing.\n\nGood luck.
355476
here's a tip. p*p = p^2\n\nnow do your own homework. How do you expect to pass the test?! Y!A won't be there for you then.
238182
p1    p2\n--- = ---\nq1    q2\n\n 5     2\n--- = ---\n 2    q2\n\nCross multiply,\n\n4/5
901836
I disagree with Peter because when I solved for p in both equations the resulting expressions are different.\n\nHere are my steps and my solutions:\n\nIn the second equation, multiply both sides by (p + 1)\np = r (p + 1)\n\nDistribute r through:\np = rp + r\n\nSubtract rp:\np - rp = r\n\nFactor out a p:\np(1 - r) = r\n\nDivide both sides by (1 - r):\np = r / (1 - r)  \n\nIn the first equation, multiply both sides by r:\np = r(r - 1)\n\nThese are different...
867162
5| 2 -9 -7 13\n..... 10 ..5 -10\n-----------------\n....2 1 ..-2 3\n\nSo P(5) = 3
904379
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
620710
A prime number cannot be divided to simpler terms, you cannot multiply two numbers to get to it, a fraction with that number in the numerator or denominator cannot be simplified... its relatively illogical and impossible to use an equation to find a number that can't be divided, multiplied, or simplified. \n\nIn simpler terms, in order to find a solution to ANY answer the solution has to have clues, probabilities, a paper trail, something....
352698
7 & 103\n\nYou can go here to find the prime factors for any number:\nhttp://www.acme.com/software/factor/
1085354
This a variation on the question of whether or not there are an infinite number of twin primes (obviously there cannot be any primes between twin primes p,p+2). This is an open question. It is not even known that every even natural number is the difference of two primes even without the restriction that they be consecutive primes. Obviously, it's true for the even number 666 since you gave an example, but it is clearly a problem for sufficiently large even numbers.  The solution to this is related to the Reimann hypothesis, as are many significant number theory problems like Goldbach's conjecture: "every even number can be expressed as the sum of two primes in at least one way."  However, if Pn+1 and Pn are two consecutive primes such that (Pn+1)-Pn = 2k then there are rigorous upper bounds for the number of such consecutive primes less than a given number X, and this number is less than C(k)* X/(logX)^2, where C(k) is a computable constant depending on k.\n\nThere is an excellent book: The New Book of Prime Number Records (by Paulo Ribenboim) which is very topically comprehensive.\n\nA good link:\nhttp://mathworld.wolfram.com/topics/PrimeNumbers.html
499816
Factor each number into their prime factors.\nTake the greatest number of occurrences of each\nprime and use them and only them  as the factors of your answer.\n\n  For example 4 and 10. Primes for each are 2,2 and\n2,5. 2 occurs at most two times and 5 occurs at most\nonce. So the factors in your answer are 2,2,5\nand the answer is 20. \n  Concise but complete. Exhaustive but exclusive.\nThat's elegance.
939960
If P is an odd prime, then half of the time we have P = 1 mod 4, and the other half of the time we have P = 3 mod 4.\n\nThis is the same as saying that half of the odd primes are of the form 4n + 1, and the other half are of the form 4n + 3.
1165002
This is a wild guess.\n\n3 x 5 x 7 x 11 x 13 x 17 x 19 x 23 x 29 x 31 x 37 x 41 = the number.\nThe answer will be the product of the first 12 primes that aren't 2. (If 2 were a factor, the answer would be even, not odd.)\n\nFor the rest, use a calculator. And check to make sure all those numbers are prime. Sorry this is such a whack answer.