I'm not stupid, but I can't work it out for myself. Eg how do you prove there is an infinite series of primes? How do you prove that the series of primes is unpredictable? I'm not sure any of these examples have been proved yet, so feel free to pick other examples.\n\nThanks
135249
Here's a pretty good link, see below
219486
All primes (ignoring 2) are odd.\n\nDivide odd numbers into 3 groups: those of the form 6n - 1, those of the form 6n + 1, and those of the form 6n + 3.\n\nIgnoring 3 itself, all primes are of the form 6n + 1 or 6n - 1, as all numbers of the form 6n + 3 or 6n - 3 are divisible by 3 and not prime. \n\nYour proposition follows from that fact.
523476
This is Goldbach's conjecture and has NOT been proven\ntrue or false.  Good luck solving it!!\n\nI looked at the above answers, adn they dont seem to understand the question.  You shouldl have said:\n\nAny even number greater than 2 CAN be written as the sum of two primes. Prove true or false.
365517
The flaw in that sort of reasoning is due to the untrue assumption that the solution of all infinite series is infinite.  For example, the sum of the series \n\n1/2 + 1/4 + 1/8...\n\nhas infinite terms, but a finite solution of 1.  Applied to time and distance, your problems are solved.\n\nGary H
652234
I’m not seeing how this is solvable.  I’m getting two possible answers.\n\nBut here’s how you should work the problem.  \n\nFirst of all, I think you pretty much have to assume that the x,y < 10.\n\nSo we know that Bill can’t tell what his two numbers are just by looking at the product.  As it turns out, there are very few products that allow this to happen: they are 12,16,18,24,36.  So you factor out all the number pairs that would make those products, and the sums are 7,8 (twice), 9, 10(twice), 11(twice), 12, and 13.\n\nSo we know that Albert is seeing one of those sums.  If he sees 7, then he knows that the only way for his numbers to add up to 7, and multiply up to either 12, 16, 18, 24, or 36 is for the numbers to be 3 and 4.  But he says nothing.  Therefore, he isn’t seeing 7.  Neither is he seeing 9, 12 or 13, for the same reason.  So, now we know the sum must be either 8,10, or 11.  And now Bill knows that.\n\nSo, if Bill were seeing 12 as the product, he would know that the numbers were 2 and 6, since they add up to 8.  But he says nothing, so he’s not seeing 12.  Likewise, if he were seeing 18, he would know that the numbers have to be 2 and 9.  But he doesn’t, so he’s not seeing 18, and he can’t be seeing 36 either.  So that means he must be seeing either 16 or 24.\n\nAnd now Albert knows this, so either Albert is seeing a sum of 8, and knows that the numbers are 4 and 4, or he’s seeing 11, and the numbers are 3 and 8.  I just don’t know how to eliminate either of those options.
250572
This site has a table of primes p where 1 < p < 1,000,000,000,000\n\nIf you want to know N, you'll have to count them.
98692
i think this will work...\n\nint prime=43;\nint isprime=0;\nfor(int i=2 ;i<prime;i++)\n{\n   if((prime%i)==0))\n   {\n       isprime=1;\n       break;\n   }\n}\nif(isprime==1)\n   printf("Not prime");\nelse\n   printf("Prime !");\n\n\n=====\nGot it na??
33371
Let's say that we want 1024 bit key. We want to generate two primes that are 512 bit long.\n It is known that the number of primes less than n is about n/ln(n) when n is sufficiently big. So the probability of a number near n being prime is about 1/ln(n). n is 512 bit long, so the probability is about 1/ln(2^512) = 1/(512*ln(2)) = 1/350.\n Since we know that even numbers are not primes(except for 2), we only need to check for odd numbers. This doubles the probability to about 1/180. Therefore, we only need few hundred trials to generate RSA key that is 1024 bit long.\n\nAs you can see, generating a RSA key is actually simple. On the other hand, factoring takes forever. This is why RSA is very secure(for now).
112548
The answer is quite simple! By using logarithmic series.\n\nFor any number x+1 , the natural log can be expressed as\n\nln(1 + x) = x - x^2 / 2  +  x^3/3  -  x^4/4  + ..... (infinity)\n\nNow you must be amazed that who the heck would calculate this infinite series! This series may seem infinite but it has a finite answer. It is analogous with summation of infinite geometric series (i hope you know this term).
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You can prove it by induction.\n\nIt is obviously true when n = 1.\n\nInduction pypothesis:  Suppose it is true for n\n\nWe need to show  is true for (n+1).\n\nDoes 24 divide (5^(2*(n+1))-1))?\n\nWe can rewrrite this as:\n\n25*(5^2n)-1  Subtract 24 from this to get 25*(5^2n)-25 \n\nThis is just equal to 25*(5^2n)-1).\n\nThis has to be divisible by 24 -- since one of the factors is (5^2n)-1) and our induction hypothesis says that it is.  \n\nAdd 24 to it and we get something that is also divisible by 24 -- but it is (5^(2*(n+1))-1)) -- so our induction is done.
169943
The primes in the 1st answer aren't consecutive.\n\nHere's a list just  using the prime numbers up to 1,000\n\n149-139\n191-181\n251-241\n293-283\n347-337\n419-409\n431-421 --- two in a row here\n557-547\n587-577\n641-631\n701-691\n719-709\n797-787\n821-811\n839-829\n929-919\n\nAnother interesting note, the highest difference between the primes from 1-1,000 was only 18.  And I think that only came up once.
438155
We sometimes see that 0!=1. Why is that? It would seem that we cannot do the multiplications from one "up to" zero. And, if we go backwards (1x0), we get zero.\n\nExperimenting with factorials, we come up with n!=n(n-1)!. For example 17!=17x(16!):\n\n16!=1x2x...x16\n17!=(1x2x...x16)x17\n\nThat equation (n!=n(n-1)!) just dictated to us where to put the parentheses. By making n=1, we can find 0!:\n\n1!=1(0!)\n0!=1\n\nAnd, it turns out that 0!=1 works very well in many situations (in probability, for example).\n\n\n--------------------------------------------------------------------------------\n\nAddendum:\n\nThe above proof that 0!=1 is based upon n!=n(n-1)!, which is in turn based upon the definition of factorial. So, it would seem to be a valid proof. But, 0! cannot be defined directly from the definition of factorial. So, mathematicians like to define 0! as 1, without proving it. So, the proof just amounts to a demonstration that defining 0! as 1 is consistent with the definition of factorial.\n\nI received email saying that my proof that 0!=1 is invalid because 0! is a constant, and you cannot solve for a constant. Wrong. I can find 4! (in a number of ways), and 4! is a constant. I can solve for the square root of 7, probably using my calculator, and it is a constant, too. There are numerous ways of finding pi, and it too is a constant.\n\nn! is also the number of permutations (ways of arranging) exactly n things. It makes sense to say that there is one way to arrange zero things. So again, 0!=1.
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Consider the numbers\nA1,\nA1+A2,\nA1+A2+A3\n..\nA1+A2+A3+...+An.\n\nDo the computations modulo n. If none of these is divisible by n, then we have n numbers and n-1 non-zero residue classes (mod n), so there are two of these numbers which are equivalent (mod n) by the pigeon hole principle. But then subtracting gives a sum of the required form that is 0 (mod n), in other words, so that n divides that sum.\n\nLooking at the proof, there is no need for the sequence to be monotonic.
15662
There are MANY faults in this reasoning.\n\nA) A finite number can only be divisible by integers that are SMALLER than the given number. In the first example, 720 is NOT divisible by 721, 722, etc. In the second example, 1,000,000,000,000,000,000! is not divisible by 1,000,000,000,000,000,000! + 1.\n\nB) A much easier way to find out the factors of a number is to factor it: for example: \n720 = 72 x 100 = 9 x 8 x 10 = 3x3 x 2x2x2 x 2x5, then you can use a method in combinatorics to find out how many different numbers you can make by multiplying the factors of 720 in different ways to see how many different numbers go into 720.\n\nC) Combining arguments A and B, you CAN show that 1,000,000,000,000,000,000! IS divisible by all integers smaller than 1,000,000,000,000,000,000. (I do intend to leave out "!")\n\nD) Finally, and most importantly, if you are going to show something is true for all integers you must either sit down for all time testing the validity of your hypotheses on EVERY NUMBER or you can use Mathematical induction.\n\nInduction means that you show your hypothesis is true for some initial number n (e.g. show that 720 is divisible by 1). Then you must show that IF your hypothesis is true for any n, THEN it must be true for the next number, n+1. You must carry out this argument WITHOUT REPLACING n WITH A NUMBER!\nYou should try to prove that1,000,000,000,000,000,000! is divisible by all positive integers smaller than 1,000,000,000,000,000,000 by showing if it's true for n, then it's true for n-1.
422316
the prime factorization can be figured out by repeatedly dividing the number by a prime number.\n for example 12\n 12 divided by 2 = 6,   6 divided by 2= 3,  3 divided by 3= 1\n the answer is 1.\ni hope this was helpful.
18132
Only one side is correct, the other is FALSE (for both sum and product)!\n\nYou can have n irrational numbers (n>1) with a rational sum:\nx1=x2=...=x(n-1)=sqrt(2), xn=-(n-1)*sqrt(2)\nx1+x2+...+xn=0\n\nYou can also have n irrational numbers (n>1) with a rational product:\nx1=x2=...=xn=2^(1/n)\nx1*x2*...*xn=2\n\nThe correct side (when all numbers are rational, their sum and product are rational) is trivial, you can prove it through induction.
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We assume that Sum and Prod know what we know.\n\nFrom the rules, xy is not prime.\nFrom Prod's first statement, x and y are not both prime (although one of the two could be)\nFrom Sum's first statement we know that x+y is a sum none of whose possible components are both prime (x+y is not even), but from her last statement, it must have more than one possibility - this rules out any x+y > 36.\nThis shaves x+y down to 11,17,23,29, or 35.\n\nChecking 11, we see that if x=2 and y=9, Prod wouldn't be able to initially deduce the numbers from knowing xy = 18, thinking they could be 2,9 or 3,6. This solution runs into a problem when we realize that Sum could still be thinking that x,y could be 3,8.\n\nNow we have to exclude all the other possibilities. All the other numbers possible for x+y have multiple solutions except for x+y = 35\n\nAt first Sum is thinking 2*33=66 or 3*32=96. Either way Prod can't deduce on her own. Once Prod hears Sum's statement Prod is able to narrow it down. Prod could not have 66, or Prod would still think the Sum could be 2+33=35 or 6+11=17. Therefore Prod must have 96 allowing them both to deduce the numbers 3 and 32.