If you have a message like "This is a test,"  You convert the message to numbers:\n\n20 08 09 19 09 16 01 20 05 19 20\n\nYou use a one-time pad, and line up the numbers:\n\n20 08 09 19 09 16 01 20 05 19 20 00 0\n61 23 87 89 41 12 30 90 23 71 29 08 7\n\nThe theory behind a one-time pad would be that you add the pairs of numbers to encrypt them.  So, like above, you'd have 20+61=81, 08+23=31, and so on.\n\nHere's the problem.  What do you do if the numerical value of a letter added to the data from the one-tme pad exceeds 100? In the above example, S=19, and if you get 89 from the one-time pad, the numbers add up to 108.  This makes the string of cyphertext an extra digit long, and thus makes the message impossible to decrypt.\n\nWhat would be the solution to this problem?  How would you get a three-digit sum reduced to two digits in such a way that a recipient could use the same one-time pad to accurately decrypt the message?
123832
Wrap-around, modular arithmetic.  If the sum exceeds 100, subtract 100 so the result is still 2-digit.  And when decoding, if the result goes negative, add 100.
287293
this is possible to figure out.  What it is a combination of numbers   i will do examples to keep it simple.   If your number was 00  or 000  then the answer was 9.  well its a pattern    if your number is 001 then the answer he came up with was 8.  well then if you had 002  then it was 7.  but also on the note if your number was 011 or some combination of this number then it was 8 also  because when you take 011 and add 0+1+1 you get 2 so it was 8.   the way it works is the numbers you type in it adds them together so there is a correlating number that goes with the answer that was added together.  so... if the numer added together was 0 then it was 9.  and the number added together was 1 then he came up with 8.  added together you got 2 then he said 7.  and just keep going ... 3 is 6,   4 is 5.    also on the same account    5 is 4,  6 is 3,   7 is 2,   8 is 1,  and 9 is 0.
233841
6666.  \nEliot could only figure out the code from the information\nthat Drew says is sufficient if all the digits in that number are\nthe same; otherwise, he could not know in what order the digits occur.\nFurthermore, since he cannot determine the number from her first\nstatement about it, the product cannot be unique and so the number\ncannot be 1111, 5555, 7777, 8888, or 9999.  Of the remaining\npossibilities only 6666 meets the requirement that the first digit\nof the square of the product of its digits should be odd:\n6x6x6x6 = 1296, squared = 1679616.
70337
There is no flaw as long as you recognize that .99999 repeats forever.  If you were to end it at some point and subtract, then as a previous poster said, you'd actually get 8.99....1.  But since it repeats, there is always another 9 and so it works.\n\nAnd consider this - if .9999999repeating did NOT equal 1, then there would be another number between them.  What would that number be?\n\n~~~~~~~~~~~~~~~~~~~~\nNo, the number in question is not infinitessimal, it doesn't exist.  You can never get to the end of the string of nines to try and squeeze another number in there :)
659308
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 :))
173240
This link will solve your problem, just download the utitlity.  \nNow you better rate my answer the best one since it does not include calling any one :)\n\nhttp://www.softpedia.com/get/Security/Decrypting-Decoding/Asterisk-Key.shtml
596777
Some simple tests to try out first:\n\n1. is the last digit is 5? (divisible by 5)\n\n2. do the digits sum to a multiple of 3? (divisible by 3)\n\n3. add all the digits in even positions (2nd, 4th, etc)\n    add all the digits in odd positions (1st, 3rd, etc)\n    subtract the 2 numbers\n    is this a multiple of 11?\n    e.g. given 9581: (9+8) - (5+1) = 17-6 = 11\n\nIf the answer to any of these is "yes", you have a composite number.\n\nIf not, tought luck! you will have to try dividing it by all odd numbers from 3 to half of the number (well, actually square-root of the number)\n\nAs an aside: try researching the term "recursion"
304655
How did you transmit that question to this web site on a machine that uses two bits of symbolic information; 0 and one?
450637
If you problem, or part of it, is:\n\nHow many 8 digit numbers can be formed from this list of digits?\n8! / (3! * 2!)\nThe number combinations of digits has tobe reduced by 2! because of the 2 "6"s.\nThe number combinations of digits has tobe reduced by 3! because of the 3 "2"s\n.\nBut since the number cannot exceed 65,500; all larger numbers must be eliminated. Subtract the number of 8 digit numbers that can be produced from the list that start with "9". Likewise "7". Likewise "69", "67", "66", "659", ""657", and "656".\n\n\n\n(Try to state your problem clearly and completely.)
574505
The number is too small for a real "pattern" to emerge.\nThe digits could be the first 10 values of some sequence given by some ( many ) recurrence relation, could be ...\n\nIn order for a proper pattern to emerge, it should be much larger...
408290
Firstly, it has to start with either a 1 or a 2, and end with a 2 or a 6. So we just take each case one at a time.\n1..2: that leaves 23556, so there are 5!/2! ways, ie 60 ways.\n1..6: that leaves 22355, so there are 5!/(2!2!) ways, ie 30 ways.\n2..2: that leaves 13556, so there are 5!/2! ways, ie 60 ways.\n2..6: that leaves 12355, so there are 5!/2! ways, ie 60 ways.\n\nNote that in each case there are 5! ways of arranging the other 5 numbers, but we need to divide by 2 if there is a pair of repeated numbers.\nThat gives a total of 210.\n\n(note that both of the other answers given so far are wrong - its not 360, because that person has forgotten they have to be even; and its not 120. They counted 2x2 possibilities for the first and last digits which is correct, but one of those cases already uses both 2s, so you *don't* want to divide by 2 in the end for that particular case (you're not counting them twice are you!)\nTheres really no other way of doing it than splitting them into cases like I did.
212787
not securely, some one has to know how to decrypt it. You only want the person with the correct key to be able to do so. Thats how public, private, and preshared keys work.  the end user knows the answer to ghe magic question, either by digital key or Preshared passphrase
469623
So call the number\nB = {b_0,b_1,b_2,b_3,b_4,b_5,b_6,b_7,b_8,b_9}\n(this reverse indexing makes things easier).\n\nYou must have b_0 != 0, since you were told it was a ten-digit number. So b_0 is between 1 and 9.\nSo b_0 = 0 - (number of the b_i which are non-zero)\n\nAlso note that Sum_i=0..9 (b_i) = 10 since B has ten digits, and each b_i has to have a single value between 0..9\nHence Sum_i=1..9 (b_i) = (10 - b_0) for any particular choice of b_0\n\nTo prove the number is unique, see below:\n\nIt's easy to show 6210001000 is one possible answer.\nb_0 = 6 since there are six 0's in B\nb_1 = 2 since b_2, b_6 =1\nb_6 = 1 since b_0 = 6\nAll other b_i are 0 since there are no other non-zero b_i.\n\nTo prove the number is unique, the easiest way seems to be to show that the other b_0 between 1..5 and 7..9 are impossible by (proof by contradiction).\n\nb_0 = 9 is impossible since then b_9 must be 1, which gives a contradiction giving '9000000001', which only has 8 zeros, also b_1 must be 1, et.c\n\nb_0 = 8 gives '8000000010', but then b_1 must be 1, and it has one less zero, so that becomes '7100000100'...\nbut that has two ones, so b_1++, and b_2 must be 1, and you have again one less zero so b_0=6, giving '6210001000', which finally works.\n\nHarder to show that b_0 = 1..5 are impossible.\n\nLet's start from the opposite extreme, b_0=1.\nThen Sum_i=1..9 (b_i) = (10 - b_0) = 9\nThis constraint means each b_i can only be 0, 1 or 2.\nSince we set b_0=1, this means we have one zero and eight ones e.g. B='11111111110', doesn't matter where you put the 0. But that doesn't work since clearly b_1 must >=8, thus Sum_i=1..9 (b_i) = 8+7>9 which is a contradiction.\n\nIt should be trivial to crank out the cases b_0=2..4, maybe there is a more compact proof.
432865
Most of it is pretty clear. One thing that might not be obvious are the "Received:" lines. As an e-mail message is passed from one server to another, the server adds a "Received:" line to the top of the e-mail message.\n\nWhat that means is that you can backtrack an e-mail message by the received lines. The top line is the information for the server that delivered it to you. The next line is a server that passed it along. The bottom Received line will be the information about the system that sent the e-mail to you.\n\nHowever, spammers and other scumsucking lowlifes can actually add an extra couple of Received lines to the top of an e-mail message trying to obfuscate where the message originally came from. You can eliminate some of them by understanding specifics of an IP address. For example, they'll use something like: 354.210.0.223 as an ip address but since each octet is a range from 0 to 254, you can see the first number cannot be valid so you can delete that line.\n\nPlus, since each server just adds a "Received:" line and nothing else, any received lines that fall after a non-received line:\n\nReceived:\nReceived:\nFrom \nReceived:\nFrom\n\nThat third received line is likely bogus.\n\nAnyway, other lines should be obvious as I said. Date, time, subject, from, lines, etc. Other lines might not be as clear and may be specific to an e-mail package or spam fighting package. Read the line and what the data is. It might become clear if you think about it.
525018
Most people use a base 10 system. Whatever you use, the base is that "Unreachable" number. For example, in base 10, there is no digit for 10, you start a new column representing how many 10's you've amassed. You have a 1's column a 10's column a 100's column and so forth. To get the numerical value you multiply (most people are so used to it, they just do it in their head). If the number is 823 you can express it as 3X1 plus 2X10 plus 8X100\n3x1 = 1\n2x10 = 20\n8X100=800\n-----------------\n1+20+800= 821\n\nSo 3 plus 20 plus 800. Make sense?\n\nIn binary "2" is the unreachable number, so you go 1, 10, 11, 100, etc. You have a 1's, 2's, 4's, 8's column.\nThe number 11011 would be 1Xone, plus 1xtwo, plus 0Xfour, plus 1xeight, plus 1x16 or 27 in base 10.\n\n1x 1 = 1\n1x 2 = 2\n0x 4 = 0\n1x 8 = 8\n1x16 = 16\n---------------\n1+2+8+16=27\n\nLet's look at your specific numbers 1011, that's \n\n1  x  1  =  1\n1  x  2  =  2\n0  x  4  =  0\n1  x  8  =  8\n--------------------\n1+2+0+8= 11\n\nYour top number is 11 (in base 10)\nYour bottom number:\n\n1 x 1 = 1\n1 x 2 = 2\n----------------\n1+2=3\n\nSo the problem states 11/3, that's pretty easy. 3.666(repeating)\n\nWith practice, it becomes easier. You learn to recognize the column values and mulitply in your head. \n\nHope that helps,\nOdd
206321
the simple program given above if fine.. in terms of the conncept.. but there's a bug... \nthe problem is with looping.. coz when u enter a number such as 80301  or 55302 ( a no. having 0 in between), the loop breaks at the point when zero is encountered.\n\n so the output in the first eg. is  8 0 ,sum= 8 (instead of correct 8 0 3 0 1 , sum=12\n\n\nso the logic of handling the number is absolutely correct but the looping is a comparatively difficult job in this case!
311059
1 - the initial 1\n11 - all numbers from phrase "previous has 1 digit 1"\n21 - all numbers from phrase "previous has 2 digit 1"\n1211 - all numbers from phrase "previous has 1 digit 2, then 1 digit 1"\n111221 - all numbers from phrase "previous has 1 digit 1, then 1 digit 2, then 2 digits 1"\n312211 - all numbers from phrase "previous has 3 digits 1, then 2 digits 2, then 1 digit 1"\n13112221 - all numbers from phrase "previous has 1 digit 3, then 1 digit 1, then 2 digits 2, then 2 digits 1"\n\nSo, the next will be\n\n1113213211 - all numbers from phrase "previous has 1 digit 1, then 1 digit 3, then 2 digits 1, then 3 digits 2, then 1 digit 1"\n31131211131221\n13211311123113112211\n11131221133112132113212221\nand so on