What kind of numbers are they and how does it work?
132568
Probably means the person has their number blocked.  That's all.
421609
I POSTED THE FIRST CORRECT ANSWER HERE AND WAS WRONGLY DELETED FOR A BOGUS VIOLATION REPORT\n\nI posted last night around 12pm, here was my (correct) answer:\n\nOk so let the tens digit be x, then the ones digit is (12-x). The value of the number is 10*x+(12-x) = 9x+12 which you are told = 11x 9x+12=11x 12=2x x=6 12-x=6 also So the number is 66. (Check x = 6*11 as claimed.)\n\n(There is some severely dishonest point-gaming going on around here)
430670
Actually -- it is a simple subtraction.\n\n-1,299
626766
Our familiar decimal number system developed over the centuries, basically because we have 10 fingers.\n\nComputers - deep down - use binary, because the most fundamental storage element is the bit, which can either be a 0 or 1 logically. In reality, it is usually represented by an electrical circuit, which is either on for 1 or off for 0.\n\nStrings of 1s and 0s are hard for people to use, so decades ago, computer scientists started cluster the bits together into groups of 3 of 4. Three bits map directy to an octal number system (0-7) and four bits map directly to a hexidecimal number system (0-9 then A-F).\n\nBinary doesn't map directly to decimal. 3 bits aren't enough, and 4 is too many, so another translation needs to be done, which is what's done under the scene on calculators and most other computers when numbers are shown to most users.
567878
555 is the natural number following 554 and preceding 556.\n\nIt is a sphenic number.\n\nIn base 10, it is a repdigit, and because it is divisible by the sum of its digits, it is a Harshad number. It also a Harshad number in binary, base 11, base 13 and hexadecimal.\n\nFive hundred and fifty five is:\n\n    * The telephone exchange for fictitious phone numbers in US movies - see 555 telephone number and North American Numbering Plan.\n    * A tongue twister in German. Try saying Fünfhundertfünfundfünfzig three times fast.\n    * The number of keyboard sonatas written by Domenico Scarlatti, according to the catalog by Ralph Kirkpatrick.\n    * The model number of the 555 timer IC, a classic integrated circuit (chip) implementing a variety of timer and multivibrator applications, and historically widely used in electronics.\n    * The years AD 555 and 555 BC.\n    * The Thai version of lol in a text conversation. "5" in Thai is pronounced "ha," so three of them would be "hahaha."\n    * The approximate height of the Washington Monument in feet.
444648
What type of information are you supposed to get? Numerical or categorical? A measure or a count?\n\nWell, you do understand the meaning of 'original,' right? You're supposed to come up with something. \n\nI'm still not sure what type of data you need. You need the response to be a number, like height? Or do you need to get yes/no and count the number of each that you get?\n\nI have had students come up with some pretty cool questions. Like comparing boys to girls on their answer to "When you take a shower, do you face toward the showerhead or away?"\n\nThere is a pretty cool segment from "This American Life" on superpowers. A guy asked people, "If you could choose between the following two superpowers, which would it be: flight or invisibility?" "What would you do with your power?"\n\nIf you need numbers, you could ask how many songs on their iPod. How many numbers stored in their cell phone.
28367
-1 works\n-1*3 + 1 = -2\n-2/2 = -1 (back to where we started)\n\nAddition> Now you say they should be pos! Jeez\n\n0 works though is it technically positive?\n\nWow, try a number like .1  - The numbers get REALLY big before coming back down to 1.  (5.00315450989997E+15)\n\n\n  Function IsOddNum(ByVal intNumber As Double) As Boolean\n    If (intNumber Mod 2 = 0) Then\n      Return False\n    Else\n      Return True\n    End If\n  End Function\n\n  Private Sub Button1_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Button1.Click\n    Dim num As Double = Convert.ToDouble(TextBox1.Text)\n    Dim count As Integer = 0\n    Literal1.Text = num.ToString() & "<BR>"\n    While num <> 1 And count <> 1000\n      If IsOddNum(num) Then\n        num = 3 * num + 1\n      Else\n        num = num / 2\n      End If\n      Literal1.Text &= num.ToString() & "<BR>"\n      count += 1\n    End While\n  End Sub\n\n\nAnother Note:  I just plotted the number of iterations for the numbers between 1 and 20,000.  The graph totally shows a pattern!  (Mathematicians love patterns)\n\nAnother Note, Doh, now the numbers have to be integers?  Jeez, there goes my repeating number answer.  Are you going to keep adding rules every time I get it right? LOL
514570
My trusty calculator (with BIG screen!) tells me 23,333.3333333.\n\nWould love to know why you need to know as they are some seriously big numbers.  Might have to ask a question to find out!
55999
Computers represent negative numbers in various fashions, but usually the most significant bit is set to 1 when the value is negative, and the remaining 7 bits represent the number, using one's or two's complement, which is a way of storing the information in the remaining 7 bits so that rules for addition and subtraction of negative and positive numbers remain constant. (for further information on 1's and 2's complement, see link).\n\nSo my guess is that the data type used to store your number is using some signed algorithm, and the first term that comes out negative is because you've superceded the max. number that can be stored...the value was added, the most significant bit became 1, and your computer interpreted it as a negative number and returned the result.\n\nThe solution is to use more bits to store your data. If you used 16 bits, you would not run into the negative numbers so fast. No matter how many bits you choose to store your data though, you'll eventually run out if your series continues endlessly - computers can represent numbers only up to a certain limit... programming languages do have data types that can represent substantially large numbers though - check the data types of the langauge you are using and something bigger...float/double/long??
19189
use bitwise operators. something like:\n\nnum_bits = 0;\nfor(i = 0; i < 32; i++) {\nif number & (1<<i) num_bits++;\n}\n\nnow num_bits holds the number of bits that were set in number. The "<<" operator does bit shifting. So "1<<i" equals a number that's all 0's except for a 1 at position i. If you bitwise and that with something, that tells you if that bit is set in the number. So we do this for each of the bits and add up the results.\n\nThis is assuming 32-bit integers. Adjust accordingly for 64-bit integers.
456088
HOLY SH*T..........it works......i did it three times.......i dont know how they did it......\n\nand i came up with a different number and i different symbol every time...
22961
the answer to the question "what kind of math is that?" actually does have an answer - it is called floating-point arithmetic.  and, well, congratulations.  you've just been exposed to a floating-point error, which arises because the rules for floating-point arithmetic and regular arithmetic are not the same.  (:\n\ncomputers represent numbers in memory by converting them to a binary number system, and then storing each 1 or 0 in a memory cell called a bit.  \n\nunfortunately this is a pretty limited way of storing numbers, because if you have a very large or very small number, you waste a lot of empty space.  it's just like writing large numbers in scientific notation:  i can write 6x10^23, or i can write 600000000000000000000000.  obviously the scientific notation is a lot easier and more compact.\n\nso computers deal with very large (or very small) numbers by representing them in what is essentially scientific notation.  half of the memory used for the number represents the "6" and the other half of the memory represents the "10^23".  \n\n(actually, because these things are binary, it is not a power of 10, but actually a power of 2.  but the idea is the same.)  \n\nin computer lingo this is called "floating-point" format because the decimal point is free to "float" around in the number by changing the exponent part.\n\na subtle kind of problem arises when converting certain numbers to binary form.  if i were to ask you to write down 1/3 in decimal form, you would probably say ".3" or ".3333" or ".3 repeating" because no matter how many additional 3's you add, you need an infinite number of digits to accurately write 1/3.\n\nwhen you convert numbers to binary, a similar problem arises, but it happens for different numbers than we are used to.  0.1 is a good (and rather infamous) example - if you try to write it in binary, it has an infinitely-long representation, just like 1/3 does in decimal.\n\nso, when you store a number like 0.1 in floating-point format, some of those digits have to get rounded away or discarded, because you'd need infinite amount of computer memory to accurately represent it.\n\nobviously, to any schmuck your problem of 1-.9-.1 should equal zero, but because of these rounding problems, in a floating-point machine some small fraction gets left over.  that small fraction is what google is returning to you (10^-17 is very, very small).  openoffice on the other hand, probably takes that 10^-17 and just rounds it down to zero.\n\nthese floating-point errors can wreak serious havoc on a computation-intensive algorithm.  they are a particular problem in scientific simulations, where a certain calculation may get repeated many many times.  \n\nif a small error crops up, that error can be quickly amplified by successive steps until it causes major problems.  so programmers who are writing numerical software have to be careful not to write code that introduces these errors (namely, subtracting two nearly-equal floating-point numbers, which is exactly what you did, is a serious problem).  so a good programmer will write code to ensure that the error stays negligible.\n\nyou can read more about floating-point errors here:\nhttp://en.wikipedia.org/wiki/Floating_point\nand some more advanced articles:\nhttp://support.microsoft.com/kb/q42980/\nhttp://docs.sun.com/source/806-3568/ncg_goldberg.html
540058
Consecutive numbers differ by 1.\nIf the first one is n, then the second one is n+1\nThat means the third one is the second number plus one (n+1)+1 = n+2\nThe fourth one is (n+2)+1 = n+3\nand the fifth one is (n+3)+1 = n+4\nAdd them altogether\nn + n+1 + n+2 + n+3 + n+4 = 5n + 10
328005
I agree with chapin4ever's answer.\nand the three digit number is the ASCII equivalent of the symbol.
72344
Multiplying by 12\nImagine your number to be pre-pended with a zero. Each digit (including the prepended zero ) but the last has a neighbor, i.e., the digit on its right.\nRule: to multiply by 12: Starting from the rightmost digit, double each digit and add the neighbor. This gives one digit of the result. If the answer is greater than 1 digit simply carry over the 1 or 2 to the next operation. Example: 0316 × 12 = 3,792:\nIn this example\nthe 3 is neighbor to the prepended zero.\nthe 1 is neighbor to the 3.\nthe 6 is neighbor to the 1.\nthe last digit 6 has no neighbor.\n6 × 2 = 12 (2 carry 1)\n1 × 2 + 6 + 1 = 9\n3 × 2 + 1 = 7\n0 × 2 + 3 = 3\nInstead of pre-pending a zero, you can think that the first digit only needs to be copied, adding any carry.\n\nMultiplying by 11\nRule: to multiply by 11, recopy the last digit. Then, two by two, add digits next to each other. Recopy the first digit.\n\nExample: 3,422 × 11 = 37,642\n\nRecopy 2\n2 + 2 = 4\n4 + 2 = 6\n3 + 4 = 7\nRecopy 3\n\n\nMultiplying by other numbers\nThe 'halve' operation has a particular meaning to the Trachtenberg system. It is intended to mean "half the digit, rounded down" but for speed reasons people following the Trachtenberg system are encouraged to make this halving process instantaneous. So instead of thinking "half of seven is three and a half, so three" it's suggested that one thinks "seven, three". This speeds up calculation considerably.\n\nIn the same way the tables for subtracting digits from 10 or 9 are to be memorized.\n\nMultiplying by 5\nRule: to multiply by 5: \nTake half of the neighbor \nAdd numbers up, two by two \nAdd 5 if number is odd \n[edit]\nMultiplying by 6\nRule: to multiply by 6: \nAdd half of the neighbor to each digit. \nIf the result is odd, add 5. \n\nMultiplying by 7\nRule: to multiply by 7: \nDouble each digit. \nAdd half of its neighbor. \nIf the result is odd, add 5. \n\nMultiplying by 8\nRule: to multiply by 8: \nSubtract last digit from 10. \nSubtract 9 from the other digits. \nAdd result to the neighboring digits on the right. \n\nMultiplying by 9\nRule: to multiply by 9: \nSubtract the last digit from 10. (Ex.: 10 - 3 = 7) \nSubtract the middle numbers from 9, and add to the number to the right. \nTake away 1 from the first number.
34285
I use the *67 trick.  Maybe it will not work with your provider, I have Sprint.  Test this:  From your cellphone, call a number for which you can see the called ID, i.e. your home phone.  You will probably see your cell phone number on the screen.  Then, call it again, but this time dial *67 before you dial the number.  What do you see on the screen?
369325
i dont know the exact number \n\nFind out your self