prime factorizations for 18 through 2500?
915493
18 = 2 x 3 x 3\n19 = 19\n20 = 2 x 2 x 5\n21 = 3 x 7\n22 = 2 x 11\n23 = 23\n24 = 2 x 2 x 2 x 3\n...\n2496 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 13\n2497 = 11 x 227\n2498 = 2 x 1249\n2499 = 3 x 7 x 7 x 17\n2500 = 2 x 2 x 5 x 5 x 5 x 5\n\nIs this for detention or something?  Is your teacher a masochist?
885353
Each zero at the end requires a 5 and a 2 somewhere in the factors.  Since there are more twos than 5s, all you need to do is count the number of 5s.  That's still an awfully big task.  Each number that is divisible by 5 has at least one 5 as a factor (e.g., 5, 10, 15...)  Of course there are 2000/5 = 400 of these.\n\nEach number divisible by 25 has at least 2 factors of 5.  There are 2000/25 of these or 80.  We already accounted for one of these 5s in the first count so this adds 80 more.\n\nEach number divisible by 5^3 (125) add another factor of 5 for 2000/125 = 16.\n\nNow, on to 5^4 this time the division is uneven: 2000/625 = 3.2 This means there are 3 numbers below 2000 divisible by 5^4\n\nOnce we get to 5^5 = 3125, it is bigger than 2000 so there are no more factors of 5.\n\nAll in all this comes to:  400 + 80 + 16 + 3 = 499 zeros.\n\nYou also asked about 1000!.  Using the same approach:\n\n200 + 40 + 8 + 1 = 249 zeros
873178
249 zeros end 1000!\n\nThere will be a zero for every factor of 10 multiplied in 1000!. Since 10 = 2*5 there will be as many factors of 10 as there are numbers of factors of 5.\nThere are 200 numbers from 1 to 1000 divisible by 5.  So 200 5's in these.\nThere is an additional 5 for every factor of 25=5*5.  There are 40 of these from 1 to 1000, so add another 40.\nThere is an additional 5 for every factor of 125 = 5*5*5, and there are 8 of these, so 8 more.\nAnd one more factor of 5 for 625 = 5*5*5*5.\nSo 200 + 40 + 8 + 1 = 249 factors of 5, thus 249 factors of 10 and 249 trailing zeros.\n\n2 is the first digit to the left of these zeros.\n\nMath proof too lengthy here and now.  http://www.justinwhite.com/big-calc/1000.html\n\nPretzels:  Good starting technique, however, in ignoring factors like 32 and 35 (because one has a 2 and one has a 5) you neglected to factor in the residue factors i.e., 32/2 = 16 and 35/5 = 7.  These will contribute to the last non-zero digit.  Also, the 3 in 30, the 4 in 40, th 6 in 60 etc. as well as the 3 in 300, the 4 in 400, the 6 in 600, etc. all contribute to this digit.\nThis is very complicated and will take a long boring time to type it all out.  Just use a computer/calculator to figure it.  It's 2.
812323
The prime factors of 1992 are 2 x 2 x 2 x 3 x 83.\n\nBy including 2 * 4 * 6, you have already covered 2 x 2 x 2 x 3, but you need to cover 83.  83 is prime, so the first time you have an even number that has 83 as a factor will be 166 (2 * 83).\n\nThat means you must figure the product of 2 * 4 * 6 * ... * 164 * 166 for it to be divisible by 1992.\n\nThe answer is largest even number in the product is 166.
914600
1272 = 2 x 2 x 2 x 3 x 53   ...prime factors\n2, 3, 4, 6, 8, 12, 24, 53, 106, 159, 212, 318, 424, 636,   ...all factors\noh, I forgot, and 1 and 1272.
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'.....)
118857
Ahkens provided the correct answer and Aseem provided the correct answer and the formulaic method of determining the relevant prime numbers. If you are confused, an easier way of obtaining the prime numbers for low numbers like 255 is simply to start with the lowest primes and divide them into 255 until you exhaust the possibilities. Firstly, the definition of a prime number is a number that can be divided, without a remainder, only by itself and by 1. (The number 1 is itself, however, NOT considered to be a prime - notice that Aseem correctly excluded it from his answer despite it appearing in the results of his formulae). \n\nUsing this definition, the prime numbers between 0 and 100 are:\n\n2,3,5,7,11,13,17,19,23,29,\n31,37,41,43,47,53,59,61,\n67,71,73,79,83,89,97\n\nSo 2 is the first prime that fits the above definition; but 2 doesnt divide into 255 so it doesn't fit into this answer.\n\nNext is 3; 255 / 3 = 85 so 3 is your first number.\nNext prime is 5; 85 / 5 = 17 so 5 is your next number.\n17 is a prime because it too fits the definition so 17 is your next number and as there are no remainders you have solved the problem. 3,5,17
780131
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.
955219
ok so prime factorization is basically factoring a number into it's multiple primes. for example, to prime factorize 12, u'd figure out what it's multiples are- 4 and 3 (or 6&2..1&12 u get the point), ok now break down 4 to prime numbers- 2 and 2. and since 3 is already prime u don't have to do anything to it. so the prime factorization of 12 is 2^2 and 3
835669
It's almost 2/3, but not exactly.\n\nInstead 667/1000 is already in lowest terms.\n\nThe prime factors of 667 are 23 x 29\nThe prime factors of 1000 are 2 x 2 x 2 x 5 x 5 x 5.\n\nThey have none in common so you can't reduce it further.
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.
46670
Factor trees are the way to go. Start with any composite number on the top of the page. Then think of any two factors of that number, then draw two branches downward from the first number to the two factors you determined. Then if either of the two numbers you chose are composite, find two factors of that number and draw two branches downward to the next level of factors. Keep doing this until the final row of numbers you get is all prime. Any prime factors that are repeated can be expressed with exponents.\n\nFor example, take the number 90:\nTwo factors of 90 are 9 and 10 (you can choose others if you wish).\nDraw two branches down from 90 to 9 and 10.\nBoth 9 and 10 are composites, so find factors for each of those. Factors of 9 are 3 and 3; factors of 10 are 5 and 2. Now our last line of prime factors is 3 x 3 x 5 x 2 or 3^2 x 5 x 2 (using exponents). And, BAM, there's our answer!\n\nVery crudely, your factor tree should look something like this (drawn without the branches):\n\n  90\n 10 9\n5 2 3 3
899443
39 can be devided by 3 people.  Scary.  \n\nBut you get the jist of what is a prime and a square from the other posts.
787255
factors of 250 - 1, 2, 5, 10, 25, 50, 125, 250\n(a trick to know if u forgot one of them is to try multiplying the last number with the first number. ex) 1x250, 2x125, 5x50, 10x25)\n\nLCM of 135 and 63 - 8505 (thats 135 times 63, usually there's a number b4 that will divide evenly, but in this case, its the numbers multiplied together)\n\nLCM is the least number that can be divided evenly by both 135 and 63, 3 would be the LCD (least common denominator)
806711
YES it is and here is a list of the firs 1000 prime numbers so just print it out and look at it while your homework.\nhttp://primes.utm.edu/lists/small/1000.txt
434001
i don't have the code but I guess you have to read something about 'numbers theory' and use the function 'mod'. once you understand the algorithm to generate prime numbers, write the code is gonna be easy.\nhope it helps
832927
Only perfect squares have an odd number of factors.  Generally factors pair up (e.g. factors of 12 are 1x12, 2x6, 3x4).  But in the case of perfect squares, one number "pairs up" with itself, making for an odd number of factors.\n\nIn your case there are lots of perfect squares greater than 300.  Assuming they want the first number with this characteristic, I'd pick 324.  (324 = 18²).\n\nThe prime factors of 324 are:\n2 x 2 x 3 x 3 x3\n\nThis means the factors of 324 are:\n1, 2, 3, 4, 6, 9, 12, 18, 27, 36, 54, 81, 108, 162 and 324\n\nSee how they pair up:\n1 x 324 = 324\n2 x 162 = 324\n3 x 108 = 324\n4 x 81 = 324\n6 x 54 = 324\n9 x 36 = 324\n12 x 27 = 324\nexcept for 18 which "pairs" with itself:\n18 (18 x 18 = 324)\n\nSo 324 has an odd number of factors.\n\n(Other possible answers are 361, 400, 441, 484, 529, 576, 625, 676, 729, 784, 841, 900, 961, 1024, 1089, 1156, 1225, 1296, 1369, 1444, 1521, 1600, 1681, 1764, 1849, etc.)\n\n(Note: if you want to prove it simply, pick 361... the square root is prime, so it has only three factors of 1, 19 and 361.)