Proof that  3 is a prime number?
467453
This can be proven by exhausing all the possibilities.  A prime number is defined as a number whose only factors are 1 and itself.  For the number 3, we need only check 1 number (the number 2), as 1 and 3 are already factors.  Dividing 2 into 3, we quickly find that the answer is 1.5, which shows that 2 is not a factor.  Therefore, the only factors of 3 are 1 and 3, which, according to the definition of a prime number, means that 3 is prime.
545916
4, 9, 25, 49\nPerfect squares such that the square root is a prime number.
892074
Because 3 and 23 are prime, the number will be a multiple of (3 x 23) = 69.\n\nNext, the two digit number in the thousand and hundreds place must be *just before* the two digit number in my tens and ones place.\n\nThe answer is 2829.\n\nYou can confirm this by checking the list below:\n2001\n2070\n2139\n2208\n2277\n2346\n2415\n2484\n2553\n2622\n2691\n2760\n2829 <--\n2898\n2967\n3036\n3105\n3174\n3243\n3312\n3381\n3450\n3519\n3588\n3657\n3726\n3795\n3864\n3933\n4002\n4071\n4140\n4209\n4278\n4347\n4416\n4485\n4554\n4623\n4692\n4761\n4830\n4899\n4968\n5037\n5106\n5175\n5244\n5313\n5382\n5451\n5520\n5589\n5658\n5727\n5796\n5865\n5934\n6003\n6072\n6141\n6210\n6279\n6348\n6417\n6486\n6555\n6624\n6693\n6762\n6831\n6900\n6969\n7038\n7107\n7176\n7245\n7314\n7383\n7452\n7521\n7590\n7659\n7728\n7797\n7866\n7935\n8004\n8073\n8142\n8211\n8280\n8349\n8418\n8487\n8556\n8625\n8694\n8763\n8832\n8901\n8970\n9039\n9108\n9177\n9246\n9315\n9384\n9453\n9522\n9591\n9660\n9729\n9798\n9867\n9936
1121426
I don't know if this is what you are looking for but simplified this fraction would be 24/43 (divided by 3). And you cannot simplify it any further because 43 is a prime number, as far as I remember...
850647
Too long to print but here's the website\n\n(2^3021376)(2^3021377-1)\n\nThere are 1,819,050 digits\n\nhttp://www.calendarhome.com/prime/perfect4.html\n\nPlease note that the highest perfect prime number currently known, will continue to change when a higher one is found.
772421
This answer will explain how to calculate is a number is a prime number. Short and sweet...\n\nStart by trying to divide the number by 2, then 3, then 5, etc(prime numbers) until the number you are dividing when squared is greater than the number you are dividing\n\nSo with 37 you start by dividing by 2(which doesn't go), then 3(which doesn't go), then 5 etc until 7 as 7^2 =49 which is greater than 37.\n\nUnlike in one post which said you divide up to 17, which is a waste of time.\n\nIn this case 37 is a prime number
749230
Neither.  A prime number has exactly two distinct factors, and 1 has only one.  And obviously it isn't composite.\n\nSome people will say 1 is prime, but they are wrong.
857498
In math, it means "can be divided" by numbers other than themselves (3 is divisible but only by 1 and itself). Prime numbers like 1, 3, 5, 7, 11, 13, 17, etc. are unique in that you can't divide them by another number to end up with another whole number. Non-prime numbers, like 4, 6, 8, 9, 10, 12, 14 and 16 are divisible because you can divide them by other numbers like 2, 3 or 4 and end up with another whole number (no decimals). Surely, some mathematician will explain this better and in more correct terms, but I did what I could on such short notice.
216982
2, 3, 5, 7, 11, 13, 17, 19\n1 is NOT prime
685138
actually the way i see it there are two numbers in the space.\nThe difference between each number is a prime number(consecutive prime numbers)\ndifference between 1 and 3 is 2\ndifference between 3 and 6 is 3\ndifference between 6 and 11 is 5\ndifference between 11 and the next no. is 7 i.e 18\ndifference between 18 and next no. is 11 i.e. 29\ndifference between 29 and 42 is 13\ngot it?\nhope your question was wrong.. hehe\n\nWhat i'm trying to say is the two numbers are\n18 and 29
161312
"Mathematicians have tried in vain to this day to discover some order in the sequence of prime numbers, and we have reason to believe that it is a mystery into which the human mind will never penetrate." — Leonhard Euler\n\nWe don't know if there is a pattern, but if it is, it's pretty complicated. The formula x^2 + x + 41 gives primes up to 40, but that's as far as it'll go.\n\nThere is a possible pattern with the distribution of primes, but this has not been proven, and if you manage to prove the Reimann hypothesis, you'll be plenty rich!
231634
well i am trying to find out myself\n\n31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.\n\nthat makes 15\n\n2, 3, 5, 7, 11, 13, 17, 19, 23, 29.\n\nthats 10\n\nU have all 25 prime less than 100 !\n\nAlways remember 1 is not a prime number ! cuz its divisible only by 1 and not any other number. and prime number requires divisibility by 1 and the number itself. i hope u got everything !\n\nmodified....i messed up somewhat. i wrote 65 and forgot 71. its clear now im sure watever i've writeen is 100% correct.
18671
To show that if a natural number is divisible by three, then the sum of its digits is divisible by three, use inductive reasoning:\n\n(A)  Base case: 3 is divisible by 3, as any natural number is divisible by itself.  The sum of its digits is 3, which is, of course, divisible by 3.\n\n(B)  Assumption: Let k=d(n)d(n-1)...d(1) be a multiple of 3, where the d(i) is the ith digit from the right in k.  Assume that d(n)+d(n-1)+...+d(1) is divisible by 3.\n\n(C)  Induction: Now, consider k+3.  One of the following occurs:\n\n     (1)  3 is added to d(1) and no carries are involved.  The sum of the digits becomes d(n)+d(n-1)+...+d(1)+3, which is divisible by 3.\n\n     (2)  3 is added to d(1) and a carry to the tens place is involved.  After the carry, 10 is subtracted from d(1) and 1 is added to d(2).  Now, The sum of the digits becomes:\n\n  d(n)+d(n-1)+...+d(1)+3-10+1\n= d(n)+d(n-1)+...+d(1)-6,\n\nwhich is divisible by 3.\n\n     (3)  3 is added to d(1) and multiple carries are involved.  This occurs when a string of m-1 9s ending in the tens place exists.  After the carries, 10 is subtracted from d(1) and 1 is added to d(2) (making d(2) 10), 10 is subtracted from d(2) (making d(2) zero) and 1 is added to d(3), and so on, involving m carries.  Then, Now, The sum of the digits becomes\n\n  d(n)+d(n-1)+...+d(1)+3-10m+m\n= d(n)+d(n-1)+...+d(1)+9m-6,\n\n which is divisible by 3.\n\nHence, by induction, if a natural number is divisible by three, then the sum of its digits is divisible by three.  \n\nNow, you wanted the converse.  Similar proofs can show that for base cases 1 and 2, the sum of digits for k=3*p+1 and k=3*p+2 (where p is a natural number) will not be divisible by three, giving the proof that a natural number is divisible by three if and only if the sum of its digits is divisible by three.  QED
294107
>> Prove that for each positive integer n\n>> there exist n consecutive positive integers\n>> none of which is an integral power of a\n>> prime number.\n\nFor example, I challenge you with n=5 and\nyou respond 16, 17, 18, 19, 20 and I win\nbecause 16 = 2^4, 17 = 17^1, 19 = 19^1\nare all integral powers of a prime.  But if\nI challenge you with n=3 and you reply with\n33, 34, 35 then you win because none of\nthose are prime powers.\n\nA proof (or winning strategy for the second\nplayer) is that given n, you can choose 2n\nprimes greater than n and use the CRT\n(Chinese Remainder Theorem) to find a\npositive integer x that simultaneously\nsatisfies\n\nx == -1 modulo prime_1\nx == -1 modulo prime_2\nx == -2 modulo prime_3\nx == -2 modulo prime_4\n[...]\nx == -n modulo prime_(2n-1)\nx == -n modulo prime_(2n)\n\nThen each of the n consecutive positive\nintegers x + 1, x + 2, ..., x + n is\ndivisible by at least two primes and\nso is not a power of a prime.
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.
465400
2, 3, 5, 7, 11\n28\n\nFYI- 1 is not a prime number! It is only divisible by 1. To be a prime number, it must be divisible by 2 numbers.\n\nprime numbers up to 20-\n2, 3, 5, 7, 11, 13, 17, 19\n\nto find a prime number, start at the number 2. cross out every other number (that would be 4, 6, 8, even numbers). Then cross out every 3rd number, every 4th number, and so on. The remaining numbers are prime!\n\nHope this helps!
171079
http://winklerwordart.com/signalglyph (that ends with signalglyph) and http://winklerwordart.com/Primes.html (that ends with Primes.html -- why does Yahoo do that ... thing?) are not saying anything new about prime numbers.\n\nThe product of the first three primes -- 2, 3, and 5 -- is 30.  Every other prime number has to be relatively prime to 30.  That rules out 30n + k for all but 30(1 - 1/2)(1 - 1/3)(1 - 1/5) = 8 values of k in the range 0 .. 29.  I.e., no prime greater than five can be 30n + k for k = 0, 2, 3, 4, 5, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25, 26, 27, 28.  The primes greater than 5 (and their products amongst themselves) must all be of the form 30n + k for k = 1, 7, 11, 13, 17, 19, 23, 29.\n\nIf you go up to 2*3*5*7 = 210, then the primes greater than 7 can only fall into the 48 residue classes modulo 210 that are relatively prime to 210.\n\nAn even better way to view it is to go in the other direction.  All primes greater than 3 are either of the two forms 6n + 1 or 6n + 5.  Or go one more step: all primes greater than 2 are odd.  Woo hoo, do I get my own web site for that? :-)\n\n"Stopping" at the 8 residue classes modulo 30 that are relatively prime to 30 strikes just the right balance between too trivial (the single class 1 modulo 2 or the two classes 1 and 5 modulo 6) and too "complicated" (48 residue classes modulo 210, more for 2*3*5*7*11, etc.).