serious answers only.  Please note that this is not asking for a prime number, it is the composite.
213026
169 is 13 times 13.  Any smaller positive integer is either prime or divisible by 2, 3, 5, 7, or 11.
401716
1224=2*2*3*3*17\n4000=2*2*2*2*2*5*5*5\n6336=2*2*2*2*2*2*2*11\n3125=5*5*5*5*5\n\nNone. There is no prime # that is a factor of all; therefore, there is no #, prime/composite, that goes into all of them.
397942
It's quite complicated.  The link given goes into detail about an inductive method of obtaining this.  If you just want the answer it is -\n\n(n^4 - 6n^3 + 23n^2 -18n + 24) / 24
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.
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!  ;-)
182675
Do you mean the 26th Mersenne Prime? It has already been found, In 1978.\n\nIt is (2^23209) - 1, a number with 6,987 digits in it.\n\nThey are up to 43 Mersenne Primes now. All Mersenne Primes are of the form (2^n) -1, where n is prime, but only a minority of prime values of n generate a Mersenne number.\n\nIt takes many days of looking by supercomputers to find these very large numbers, the largest of which has over 9 million digits. That took 50 days to find and 5 days to check that it had no factors. \n\nThere is no formula, just systematic trial and error and dividing by all primes up to the square root of the candidate prime to see if they divide into it (modulo 0) exactly or not.\n\nThe 26th ordinary prime, counting from 2 upwards as the 1st prime number, is relatively small in comparison.\n\n2, 3, 5, 7, 11\n13, 17, 19, 23, 29\n31, 37, 41, 43, 47\n53, 59, 61, 67, 71,\n73, 79, 83, 89, 97 \n\nmaking the 26th prime 101
35614
it isn't prime: it's divisible by 7 (which is one of my favourite numbers).\nIf you add it to 668^2 you get 669^2 (but that's a trick you can do with every odd number).
448444
the order of any element, modulo p, divides p-1 (which is an even number for everything except the stupid case of p=2, which i'm omitting) and every number relatively prime to p, raised to the (p-1) power will be congruent to 1 (modulo p)\n\ntherefore b^(p-1) = 1 (mod p)\nso a^((p-1)/2) = (b^2)^((p-1)/2) = b^(p-1) = 1 (mod p)\ntherefore order(a) divides ((p-1)/2) and hence order(a) does not equal (p-1), which is the requirement for being a primitive root.\n\nhence, a can never be a primitive root
482412
The answer is 23475.\n\nLet's look at how I got there:\n\n\nThe first digit is a prime number:  2,3,5,7\n\nOne is not a prime, so anything starting with 1 is not actually an option.  There are two valid numbers, but one of them has repeating digits.\n\nGood.  That narrows it down a lot.\n\nThe third digit is twice the first digit:  Following from the numbers possible for the first digit, it's 4,6,10,or 14.  It obviously cannot be the last 2 since they are two digits.  That means it is either 4 or 6, and the first digit is 2 or 3.\n\nSo far we have\n2x4xx or\n3x6xx\n\nThe fourth digit is the third digit plus 3.  That's easy.  We now have\n2x47x or\n3x69x\n\nThe fifth digit is the difference between the first digit and the fourth digit.  7-2=5 and  9-3=6  We now have\n\n2x475 or\n3x696\n\nWait!  The second number has a repeated digit, 6.  It thus cannot be the number.\n\nLet's go on with option 1.  The second digit is the fifth digit minus the first digit.  5-2=3\n\nThe number is 23475.\n\n\nHope this helps.
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.
293574
the first step is basically to factor 120175.  the obvious factor to do first is 5.  this gives us 5*24035.  then factoring another 5 gives us 5*5*4807.  since 4807 obviously isn't divisible by 5 or 2, try 3.  doesn't work - so keep on trying the next biggest prime number (1,2,3,5,7,11,13,17...)  turns out that 4807 is divisible by 11.  this gives us 5*5*11*437.  keep on going and you get 5*5*11*19*23.  just checking, yes - 5*5*11*19*23=120175.\n\nnow converting these factors into letters, you get the available letters EEKSW.  this can be rearranged to be WEEKS.  however, remember that 5*5*11*19*23 can also be written as 25*11*19*23, since this also equals 120175.  this means the word could also have KSWY.  but i don't think you can rearrange those letters to make a word.\n\nso therefore the english word encoded as 120175 is WEEKS.  basically, you factored the number, converted them into letters, and solved the anagram.\n\n[[edit]]\n\ninuyasha's right - there could be an infinite number of "a"s in there too since 5*5*11*19*23 = 5*5*11*19*23*1*1*1*...\n\nhowever, this gives us WEEKS+AAA... so using the online anagram solver, found at http://wordsmith.org/anagram/ , there are no words that have "WEEKS" in them and "A"s.  therefore, the answer is still WEEKS.
33244
501/3=167\n\n167 is a prime number\n\nIs this what you were looing for?  If not, repost with more details.
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
625630
by any chance is this a question from your maths homework? lol :D\n\nI cant work it out myself because to put it simple, im crap at maths.
1418
1. It's one multiplied by ten to the twenty-third power.\n  -Otherwise ten to the twenty-third power.\n\n2. It is also an example of multiplicative identity.\n\nSorry if that is not what you are looking for.
518118
Try to factorise number 319.\nIt is 11 * 29\nSo you have 11 rows 29 people per row \nor 29 rows with 11 people per row.\n\nIf you wonder how to find it it is simple\nTry to divide 319 with prime numbers\n\n2,3,5,7,11,13,17\n\nOnly 11 is ok!
69527
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.