what is the number of zeros in 100 factorial?
609644
100! = 100*99*98...3*2*1\n\nSources of zeroes are multiples of five and multiples of 10:\n10, 20, 30, ... (10 to 90 and 100, 11 zeroes)\n5*2, 25*2, ... (5, 15, 25... 95, will make 10 zeroes)\n25, 50, 75 (another 3 zeroes)\nSo I guess it should be total 24 zeroes.
597363
100\n\nhttp://en.wikipedia.org/wiki/Names_of_large_numbers
848063
488,895\n\nWrite every number as a six digit number with leading 0's if necessary.  i.e, 1 becomes 000001, 10 becomes 000010, etc.\nAlso, write 1,000,000 itself as 000000.  (We don't change anything by doing this.) We will subtract out the leading 0's later.\n\nTherefore, there are 1,000,000 6-digit numbers.  Each digit occurs just as many times as any other, so so far there are:\n1,000,000 * 6 / 10 or 600,000 zeros.\nNow subtract out all the leading zeroes.\n\nEvery 1-digit number has 5 too many.  There are 9 of these, so 5*9 = 45 too many zeroes.\nEvery 2-digit number has 4 too many.  There are 90 of these, so 4 * 90 = 360 too many zeroes.\nContinuing this way, there are another 3*900 = 2700 , 2 * 9000 = 18000 and 1 * 90000 = 90000 too many zeroes.\n\nThis leaves 600,000 - 45 - 360 - 2700 - 18000 - 90000\n= 488,895 zeros in total.
436598
Remember that a factorial is basically the product of that number, that number - 1, all the way down to 1.\n\n1. (n-1)(n-2)(n-3)(n-4)(n-5)...1\n---------------------------------\n(n-4)(n-5)...1\n\nCancel the (n-4) and everything after.\n\nANSWER : (n-1)(n-2)(n-3)\n\n2. (n+6)(n+5)(n+4)(n+3)...1\n---------------------------\n(n+3)(n+2)...1\n\nCancel the (n+3) and everything after\n\nANSWER : (n+6)(n+5)(n+4)\n\nEDIT : Lattisrectum, keep in mind that (n-4)! is (n-4)(n-5)(n-6)... all the way to 1, not (n-4)(n-3)... :)
657261
isn't it to add pairs of mumbers like (1+100) (2+99) (3+98).......(50+51)\n\nall the answers are 101, and there are 50 answers\n\nso 101 x 50 = 5050
1092588
actually, a googolplex is NOT 1 with 100 zeros. That's a googol. A googolplex is 10^(10^100) or 10^googol and can not be written as zeros as it would take more ink then there is matter in the universe to write.
797722
AARGHHH!!  I withdraw my earlier attempt.  It must be dementia, I'm afraid.  It's still asking about 1000! (1000 factorial), of course, but a careful accounting is needed, not only of the effect of all numbers ending in: \n\nA. 1, 2, 3, ... 9 (I did this by a variety of ways, always coming up with a final non-zero digit of 6 for the product of all such sets), \n\nbut also of the following two other types of number, which I regrettably neglected the first time around:\n\nB. 10, 20, 30 ...90 (as often as sets of numbers like that occur), and finally, of types:\n\nC. 100, 200, 300, ... 900 (this set occurs but once).  The final, unique 4-digit number (1000) only adds three zeros at the end and doesn't affect the final non-zero digit.\n\nO.K., here goes:\n\nNumbers of Type A.\n\n1 x 2 x 3 x ... x 9 = 362880; final non-zero digit is 8.\n\nThere are 100 such "basic sets" of type A altogether.  \nMultiplying them together, set after set, we get the \nfollowing repeated cycle of non-zero ending digits \n(top row is # of basic sets used so far; bottom row is the corresponding final digit.)\n\n1   2   3   4   5   6   7   8   9 ...\n8   4   2   6   8   4   2   6   8 ...\n\nNotice that after every four basic sets have been used, we end up with 6 as the final non-zero digit.  Since there are exactly 100 such sets (a multiple of 4), we end up with that same last non-zero digit for the product of ALL numbers of type A.  (But we're not done yet.) \n\nNumbers of Type B.\n\nThere are exactly 10 sets of numbers of Type B, i.e. ending with  exactly one (but not two or three) zeros.\n\nMultiplying just these sets together, we'll have the following analogous results:\n\n1   2   3   4   5   6   7   8   9  10  \n8   4   2   6   8   4   2   6   8     4   \n\nSo, the last non-zero digit in these "10's products" is 4 for this lot.\n\nNumbers of Type C.\n\nWe have just one set of "100's".  For this, the last non-zero digit is 8 (it's just our "basic set" with a large number of zeros following it); only the last non-zero digit is of interest.\n\nSo, to summarize, the different sets we've divided our numbers into, when multiplied together, give us final non-zero digits as follows:\n\nType A:   6\nType B:   4\nType C:   8 \n\nFinally, then, the last non-zero digit of the product of everything is the last digit of:\n\n6 x 4 x 8, or 2.  (PHEW!!)\n\nWell, it's now the value of 2 (as reported by zanti3 from the 100 factorial website at:\n\nhttp://www.justinwhite.com/big-calc/1000.html)\n\n(Finish off this URL with 1000.html, instead of 1000... The Yahoo site would not reproduce the full URL from my draft.  Alternatively, just go to Yahoo or Google with 1000 factorial 1000 - this worked for me.) \n\nBoy; what a problem!  Evidently the jury is still out on the question of my impending dementia.  (See zanti3's kind words, below about my earlier fiasco.) \n\nThanks again to zanti3 for bringing the existence of a (1000 factorial) web site to my attention.\n\nLive long and prosper, everyone.
982580
Use binomial coefficients to solve this.\n\n| n |           n!\n|    |   =  ----------\n| k |        (n-k)!k!\n\nn is the number of possibilities and k is the number of unordered outcomes.\n\nCombination of six numbers from ten possibilities.\n\n| 10 |       10!          3628800         \n|      | = ------------ = --------------- = 210\n|   6 |    (10-6)!6!     (24)(720)\n\nCombination of four numbers from ten possibilities.\n\n| 10 |\n|      |\n|   4 |\n\nYou should be able to figure out the second one on your own.\nHere's a link on factorials, just in case you don't know what they are.\n\nhttp://mathworld.wolfram.com/Factorial.html
254265
Your count starts with the point Zero.\n\n\nzero m  1\nTwo m   2  \nFour m  3\n\n100m    51\n\nhttp://www.futureaccountant.com/\n\nhttp://www.schoolingkids.com/
477858
There are 2 factors of the leading coefficient 2, and 3 factors of the constant 4, so 2 times 3 is 6.  Then there are 2 possible signs, so there are 12 possible zeros.  But that includes duplicates.  The possibilies are ±4/1, ±2/1, ±1/1, and ±4/2, ±2/2, ±1/2.  2/1 = 4/2 and 2/2 = 1/1, so if you remove the duplicates there are 8 possibilities.
707641
another zero would be the conjugate of the one you know,\nthen try the rational zeros test, or even syn div
987712
It stands for 1 followed by 100 Zeros
613818
0123456789\n6210001000\n\nHere you can see there are 6 zeros, 2 ones, 1 two, and 1 six.\n\nI suppose you were looking for something more like this though.\n1234567890\n2100010006\n\nFor those of you with the 1000.0008 solution, where is the 1 8?  I suspect that solution is incorrect.
658003
5050\n\nIf you add the numbers like this\n (1 + 100) + (2 + 99) + (3 + 98) + . . . . + (50 + 51) = ?\nthen each set adds up to 101 and there are 50 sets of numbers, so 101*50=5050
1072126
10000\n\nThat's counting 0000, 0001, 0002, all the way to 9999
358099
0.43 & 0.0062\n\nDivide by 100
359204
100 which is the binary equivalent of 4 if you are considering those 2 numbers to be added binary numbers