(1000)!, where !-factorial.Find the single digit  after adding the digits in it's value.?
613146
Nice problem. Here's the solution: 1000! is a multiple of 9 which is the same thing with saying that: S(1) = the sum of its digits is a multiple of 9 (Using the following known fact: a number is divisible by 9 if and only if the sum of its digits is divisible by 9). \n   Clearly S(1) < 1000! and 9 | S(1). Summing the digits of S(1) one obtains a number S(2) which is strictly smaller than S(1) and is a multiple of 9. And so on, until finally one arrives at a one digit number, say S(r) for some r  which is divisible by 9. But this can only be 0 or 9. Since S(r) is the sum of the digits of the previous number obtained by this algorithm, namely S(r-1), it can not be 0. So it must be 9! \n   Note that this type of proof works for any number divisible by 9, not only factorials of numbers, and in fact, with exactly the arguments displayed above it can be prooved that the single digit after adding the digits of an arbitrarly chosen number N is the digit d > 0 such that N-d is a number divisible by 9. (If you don't belive it, simply work out a computer programme that verifies this for all the numbers smaller than 1 000 000 for example).
53602
You need to round off the number to 4 digits and use the proper exponent to reflect the entire number:\n\n1.496 x 10^11  Your first answer is correct.
369365
move the decimal point to the left 9 spaces.\n.000000004.
732682
1110\n\nTo be divisible by 3, the total of the digits of a number has to be divisble by 3, so 1+1+1+0 = 3 (so you need at least 3 "1" digits).\nBut you need it to be divisble by 5 as well, and only numbers that end in 5 or 0 are divisble by five.\nSo, you take the 111 and add a 0, and that is your number.
165476
How about (a+c)/(b+d)  ? more interesting!!\n\nbtw this only work when a/b != c/d
698756
First conversion the numbers to binary, then make the OR operation.\n\nExp.:\n8 OR 3.\n\nis: 1000 OR 11\n\n= 1011.\n\nWhich mean: 11 (Dec.)\n\nYou can use the Windows Calc.
560969
23\nif intead of '=' there is +
445906
10 times 10 times 10 1000\nyou could use a calculator alot easier
138797
PI to "One Million" digits:\n\nhttp://3.141592653589793238462643383279502884197169399375105820974944592.com/index1.html
860441
I'm going to start by assuming that numbers starting with 0, other than zero itself, don't count.  Normally, since there are two choices for any individual digit, the number of numbers possible for any number of digits (d) is 2^d.\n\nHowever, since 0 can't be the starting digit for any but the one-digit numbers, your choices are cut in half, which would be 2^(d-1).\n\nThus, your answer would be the sum of:\n2 + 2^1 + 2^2 +...+ 2^9\n\nInterestingly enough, you'll notice that this answer ends up being 2^10.
171094
5000                  AHHHHHHHHHHHHHH!\nIt is 4100!\nGreat Question!
661659
(0! + 0! + 0! + 0!)!\n=(1+1+1+1)!\n=(4)!\n=24
361979
if the last # is in the tenths place put it over 10
280927
let the first 3 digits of your number be x = abc and the last four digits be y =defg\n\nwhere a-g are the digits\n\nThe sequence of operations can be rewritten as:\n\n((x*80 + 1) * 250 + 2 * y - 250)/2\n\nIf you distribute the first 250 you get\n\n(20000x + 250 + 2y - 250)/2\n\nCancel the 250's\n\n(20000x+2y)/2\n\nDivide both elements of the numerator by two (distribute the 2)\n\n10000x + y\n\nSince x is a three digit number abc, abc * 10000 = abc0000\n\nSince y is a four digit number defg, 10000x + y = abcdefg.  \nThus we see using algebra that you end up with the same 7 digit number you started out with.
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.
1157976
0000, 0001, 0002, 0003, 0004, 0005, 0006, 0007, 0008, 0009, 0010, 0011... etc, etc  untill u get to 9998, 9999\n\ntry it out, might take a long time but one combination will work :))
688360
Last few digits are zeros. For any positive integer 'n', factorial n i.e. n! has last digit equal to zero if n > 4. One question is how many zeros are there in 1000!? But you have not asked this question and so I am not going to give the answer.