4) Let p_n denote the nth prime. For n>3, show that:?
360420
Bertrand's conjecture states that for every n > 1 there is always at least one prime p such that n < p < 2n.\n\nBasis of induction:\n7 = p_4 < p_1 + p_2 + p_3 = 2 + 3 + 5 = 10\n\nInduction step:\nSuppose that p_n < p_1 + p_2 + ...+ p_{n-1}.\nUsing Bertrand's conjecture for number p_n, we will find that\np_{n+1} <= p_{n+k} < 2*p_n < (p_1 + p_2 + ...+ p_{n-1}) + p_n.\n\nThus p_n < p_1 + p_2 + ...+ p_{n-1} for all n>3.
283463
p divides n+3 and p divides (n+3)+7 ---> necessarily p divides  7 so p=7
1163467
The YJ (Yablon-Jinydu) Conjecture is:\n\n lim (n->infinity) Mn ^ (1/n) == 1.5000 == 3/2 == M2/M1,\n where Mn is the nth Mersenne Prime,\n i.e. M1==2, M2==3, M3==5, M4==7, M5==13, M6==17, M7==19, \n      M8==31, M9==61, M10==89, etc.\nThis was discovered empirically by Yablon using a simple C-program to take nth roots of the first forty known Mersenne Primes. The approach to the limit of exactly 1.5000 == 3/2 was unmistakeable and is easily reproduceable.\n\nWith a proof of this conjecture, it is then easy to prove the Infinitude of the Mersenne Primes:\n\n There are an infinite number of Mersenne Primes !!!!!\n\n\nThe YJ Conjecture is proven as follows:\n\n(1) Prove lower and upper bounds on a constant K with\n\n   sqrt2 < 1.47 < 2^gamma < K < 2^gamma*ln3 < lg3 < phi < sqrt3 < 2\n   such that the conjecture holds for limit == K if\n   there is any such limit.\n(2) Prove K == 3/2 == 1.5000 using an epsilon-delta\n\n   simple calculus proof.\nThis proves the above assertions.
14699
Yes.\n\nFor any finite set of primes, there will always exist another prime not in that set.  So there can never be a "complete" finite set of all prime numbers.\n\nTake all known primes:\nP1, P2, P3,... Pn\nwhere Pn is the greatest prime.\n\n(P1 x P2 x P3 x.... x Pn) + 1 is not divisible by any of the known primes.\n\nIf it is not prime, then its prime factors are not in the original known set of primes.\n\nIf it is prime, then it itself is a prime that is not in the original known set of primes.
777706
You make the assumption that there are a finite number of primes.\nIf this is so, then it is possible to write them all out in a list.\nYou then multiply them all together, and add one to the product.\nThe result is not divisible by any of your existing prime numbers. So it is either the case that:\nIt is not divisible by any number, i.e. it is prime.\nIt is divisible by two or more prime numbers, which cannot be on your list.\nEither way, you have missed out at least one prime number. Therefore, not matter what your list of prime numbers, you have always missed out some, which means that there is an infinite number of primes.
136689
2*3*5*7*11*13 = 30030\n\nYou're correct.\n\n(I see questions regarding "1" as a prime - Google "prime numbers" and you'll find many links starting at 2. Just sayin'.....)
482620
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.
1067732
If the N numbers:\n\np, p + d, p + 2d, ..., p + (N-1)d\n\nare all prime, and if q is a prime, then either:\n\nq divides d and all elements of the sequence have the same residue modulo q (which would be nonzero unless N=1 and p=q)\n\nor\n\nany q consecutive elements of the sequence make up a complete residue set modulo q, i.e., their remainders modulo q will be the set {0, 1, 2, ..., 1-1} not necessarily in that order.\n\nthen note that unless 3 divides d that the first three taken modulo 3 form a complete residue set (i.e., the remainders modulo 3 will be the set {0, 1, 2} in some order).  Likewise the first five taken modulo five will have remainders modulo five of {0, 1, 2, 3, 4} modulo five.  So searches for N consecutive primes in arithmetic progression usually take N to be a multiple of all of the primes less than or equal to N.\n\nA search for\n\nprime number records consecutive arithmetic progression\n\nin Google,\n\nhttp://www.google.com/search?q=prime+number+records+consecutive+arithmetic+progression\n\nyields as its first link\n\nPrime Arithmetic Progression -- from Wolfram MathWorld\nhttp://mathworld.wolfram.com/PrimeArithmeticProgression.html\n\nwhich contains some sample sequences (smaller p and larger N, smaller N and larger p) and says it was last updated Friday, May 12, 2006 -- sample quote from that page:\n\n"\nFor example, 199, 409, 619, 829, 1039, 1249, 1459, 1669, 1879, 2089 is a 10-term arithmetic progression of primes with difference 210.\n"\n\n(note that 210 = 2 * 3 * 5 * 7 is the product of all primes less than or equal to ten).\n\nDan D
84784
You should really just draw out the factor tree.\n\nYou will get 2power4 times 3 in prime factorization.\n\nYou just find the lowest prime number and a multiple and branch down the tree in the same process. Also bring down the primes. Once you have all prime numbers you put them together them to get the prime factorization.\n\nHere is the tree:\n\n                             48\n                            /  &#92;\n                           2 * 24\n                          /   /  &#92;\n                         2 * 2 * 12\n                        /   /   /  &#92;\n                       2 * 2 * 2 *  6\n                      /   /   /    / &#92;\n                     2 * 2 * 2  * 2 * 3\n\n                             2^4 * 3
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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!
881190
Well, these numbers are each broken down in prime facorization like this:\n\n3: 3   (prime number)\n7: 7   (prime number)\n5: 5   (prime number)\n4: 2*2 (composite number)\n\nThe greatest amount of times 2 appears is twice. \nThree is once.\nFive once.\nSeven once.\n\nSo, 2*2*3*5*7= 420. You can check it on a calculator. \n(Above can also be written as: 2^2*3*5*7)
1139003
(n/(n - 3)) + (5/(2n - 6))\n(n/(n - 3)) + (5/(2(n - 3)))\nMultiply everything by 2(n - 3)\n(2n + 5)/(2n - 6)
367524
using principle of mathematical induction u can prove this\n check for n=1  2^2 -3 -1=0 which is div by 9\nlet this be true for n=m\n2^2m -3 m -1 =9k  ,where k is any integer\n\nnow for n=m+1 \n2^2(m+1) -3(m+1) -1\n=4.2^2m-3m-4\n=(2^2m-3m-1)+3.2^2m -3\n=9k+3(2^2m-1), now (2^2m -1) is an odd integer divisible by 3 for all m >0\ntherefore the above expression becomes 9k +3.3g ,where g is any integer\n=9(k+g)=9i  where i is an integer \ntherefore the condition for divisibility by 9 is true for n=m+1\nhence by principle of mathematical induction it is true for all n>=1
283645
if p is a factor of both, than p should also be a factor of the difference.\n\nn + 10 - (n + 3)\n\nThe resulting integer only has one factor greater than 1.  That's your answer!\n\n(I'll let you figure out the rest, don't want to spoil it for you.)
630943
1) Let n = the first integer, then\nn+1 = second integer\nThe product is n(n+1)\n2) Let n = the first even integer, then\nn+2 = second even integer and\nn+4 = third even integer.\nThe sum would be n+(n+2)+(n+4)
1017510
How do you come up with this:\n"Thus, in general, tau(sigma(A^2)) >= 3, but tau(p^b) = 1"\n\nI don't see that what you have done implies that tau(sigma(A^2)) >= 3, since the prime factor of  x^((2r+1)-1)/(x-1) could be the same as the prime factor of (y^(2s+1)-1)/(y-1).
813306
Start by dividing by 2 until you can't divide by 2. Then 3, then 5 etc..(prime numbers) until your number has been divided so many times that it equals 1.\n\nThe product is the number you divided into\n\nEg\nExpress 24 as product of primes\n\n24 / 2 =12\n12/2 =6\n6/2 =3\n3/3 =1\n\nThe prime productes are 2, 2, 2, 3\n\nSo you express it as 2^3*3\n\nNow do the same with 252