Can anybody solve these binary numbers questions?
150167
(a) 111100111101\n(b) 110001011
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.
403722
1,11,21,1211,111221....\n\n312211,13112221,1113213211\n\n\nBTW i love this riddle, i use it on for my chess team for brain warm-ups
910180
3ab(2ab^2)^4\n3ab(16a^4b^8)\n48a^5b^9\ntrust me!!!! i just got done doing this in class a week ago.
578454
I can do all that, but except for small numbers I wold never bother.\n\nThe "easy" way, such as it is, is to always go through binary on your way from decimal to hex. \nFor the others, list the bits and add/partition.
19517
Same way the decimal system or any other base number system works.\n\nReading from right to left, you take that digit and multiply it by the base raised to the column number (with column numbers counting from 0, by 1s). Then you add the values up. \n\nexample of why 714 = 714:\nyou take the sum of:\n4 x 10 ^ 0 \n1 x 10 ^ 1\n7 x 10 ^ 2   \n\nwhich becomes\n4 x 1   = 4\n1 x 10  = 10\n7 x 100 = 700\n\n\nAdd them up and you get 714\n\nSo, 1011001 in base 2 (binary) would be the sum of:\n1 x 2 ^ 0\n0 x 2 ^ 1\n0 x 2 ^ 2\n1 x 2 ^ 3\n1 x 2 ^ 4\n0 x 2 ^ 5\n1 x 2 ^ 6\n\nwhich becomes\n1 x 1   = 1\n0 x 2   = 0\n0 x 4   = 0\n1 x 8   = 8\n1 x 16  = 16\n0 x 32  = 32\n1 x 64  = 64\n\nAdd them up, and 1011001 represents 89 in base 10.
676828
Done it:\n\nA = 3\n\nB = 9\n\nc = 6\n\nyou're welcome, if you have doubts of how i did it... erminsho@yahoo.com.mx
938348
There are many combinations. One of them is 1,2,3,4,5,16. Is there some specific property you wanted these six numbers to have?
133517
the solutions is 7,8,3,2 or 7,6,5,2. The problem works with both sets of numbers.
359204
100 which is the binary equivalent of 4 if you are considering those 2 numbers to be added binary numbers
884616
Possible .X....Y\noutcome\nHHHH......3....3\nHHHT......3....2\nHHTH......2....2\nHTHH......2....2\nHHTT......2....1\nHTHT......2....1\nHTTH......1....1\nHTTT......1....0\nTHHH......2....3\nTHHT......2....2\nTHTH......1....2\nTTHH......1....2\nTHTT......1....1\nTTHT......1....1\nTTTH......0....1\nTTTT......0....0\n\nHence,\nP(X=3,Y=3)=1/16\nP(X=3,Y=2)=1/16\nP(X=3,Y=1)=0\nP(X=3,Y=0)=0\nP(X=2,Y=3)=1/16\nP(X=2,Y=2)=3/16\nP(X=2,Y=1)=2/16\nP(X=2,Y=0)=0\nP(X=1,Y=3)=0\nP(X=1,Y=2)=2/16\nP(X=1,Y=1)=3/16\nP(X=1,Y=0)=1/16\nP(X=0,Y=3)=0\nP(X=0,Y=2)=0\nP(X=0,Y=1)=1/16\nP(X=0,Y=0)=1/16\n\n.................JOINT PROBABILITY DISTRIBUTION OF x & y\n\n........0.1....2.3.....|P(X=x)   \n...&#92;Y|__________|\n..X&#92;.|..........................|\n..__|..........................|\n....0|1/16.1/16.0.0.....|2/16\n......|..........................|\n....1|1/16.3/16.2/16.0|2/16\n......|..........................|\n....2|0.2/16.3/16.1/16|6/16\n......|..........................|\n....3|0....0..1/16..1/16|2/16\n......|..........................|\nP(Y=y)|2/16.6/16.6/16.2/16|1\n\nfOR 2nd part we have to find \nP(X=3|Y=1)=P(X=3,Y=1)/P(Y=1)=0\nP(X=2|Y=1)=P(X=2,Y=1)/P(Y=1)=(2/16)/(6/16)=1/3\nP(X=1|Y=1)=P(X=1,Y=1)/P(Y=1)=(3/16)/(6/16)=1/2\nP(X=0|Y=1)=P(X=0,Y=1)/P(Y=1)=(1/16)/(6/16)=1/6\nHence\n      conditonal probability distritribution of X given Y=1\n......X...|0.1.2.3..|TOTAL\n.....___|_______|_____\nP(X=x)|0.1/3.1/2.1/6.|1\n\nForget about those (.........)s. They are provided so that the tables do not collapse.
21853
Whether you are looking for general information about binary numbers or looking to perform mathematical operations on them I would suggest you take a look at this article at Wikipedia, http://en.wikipedia.org/wiki/Binary_numbers I think it has all the information you are looking for.
593943
1) The sum of the positive 3-digit odd integers = 101 + 103 + 105 + ... + 995 + 997 + 999.  Adding the first of these to the last, we get 101 + 999 = 1100.  Adding the second of these to the next to last, we get 103 + 997 = 1100.  This pattern keeps repeating until 549 + 551 = 1100.  The question then becomes, how many of these 1100's do we have?  We have one for 101, 103, 105, ..., 549 (the later numbers share their 1100's with these).  How many numbers are in the set 101, 103, 105, ..., 549?  Well, there are 449 numbers in 101, 102, 103, ..., 549.  Since there is one more odd number than even number, we must have 225 odd numbers and 224 even numbers.  Thus, there are 225 odd numbers in the set 101, 103, 105, ..., 549.  Thus, we have 225 1100's to add together.\n\n225 * 1100 = 247500.\n\n\n\n2) This time we wish to add the numbers of the set 10, 11, 13, 14, ..., 94, 95, 97, 98.  Adding the first to the last we get 10 + 98 = 108.  Adding the second to the second to last we get 11 + 97 = 108.  The pattern keeps repeating until we get 53 + 55 = 108.  How many 108's do we have?  Well, we have one for 10, 11, 13, 14, 16, 17, ..., 52, 53 (again, the later numbers share their 108's with these).  How many numbers are in this set?  In the set 10, 11, 12, 13, ..., 52, 53, there are 44 numbers.  The multiples of three range from 12 to 51--from 3*4 to 3*17.  Thus, there are 14 multiples of three in this range.  This means that there are 30 numbers in the set 10, 11, 13, 14, 16, 17, ..., 52, 53.  Thus, we have 30 108's to add together.\n\n30 * 108 = 3240.\n\n\n\n3) Here, I am assuming that the raise of 10% also applies to the previous year's raise and not just to the "principal" of $18000.  Thus, I am assuming a compound raise instead of 5 simple raises.\n\nDuring the first year, you make $18000.  During the second year, you make 18000 * 1.1 = $19800.  During the third year, you make 19800 * 1.1 = $21780.  During the fourth year, you make 21780 * 1.1 = $23958.  Finally, during the fifth year, you make 23958 * 1.1 = $26353.80.\n\nGood luck.
81243
Using brain, of course.\n\nYou need to know the basic rules:\n0 + 0 = 0; 0 + 1 = 1; 1 + 0 = 1; 1 + 1 = 10.\nMoreover, 0 - 0 = 0; 1 - 0 = 1; 1 - 1 = 0; 10 - 1 = 1;\n1 - 10 = -1; 1 - 11 = -10.\n\nJust do the calculation your way using this relations.\n(Pen and paper are not bad companions.)\nElse, you can convert the binary number to decimal and calculate and then convert the answer back into binary.\n\nThe following link might help you. I saw a conversion table of number systems up there. Have a nice day.
40022
1 bit can have 2 values = 0 or 1\n1 byte has 8 bits = a total of 256 possible combinations\n1 Kb has 1024 bits\n1 KB has 1024 bytes\n1 MB has 1024 KB\n1 GB has 1024 MB\n1 TB has 1024 GB
424362
Problem #1:  Four consecutive integers have sum of -34.  Let the first integer be x.  The next three must be x+1, x+2, and x+3.  This gives you the equation\n\n-34 = x + x+1 + x+2 + x+3\n-34 = 4x + 6\n-40 = 4x\nx = -10\n\nSo, the 4 consecutive integers are -10, -9, -8, and -7.\n\nProblem #2:  Again, four consecutive integers can be written as x, x+1, x+2, and x+3.  This time the equation is:\n\n4(x + x+3) = 24 + 6(x+2)\n4(2x + 3) = 24 + 6x + 12\n8x + 12 = 6x + 36\n2x = 24\nx = 12\n\nSo, the 4 integers are 12, 13, 14, and 15.  \n\nIt pays to check your answers:  The sum of the first and fourth is 12+15=27.  Four times this sum is 108.  6 times the third integer is 6*14=84.  108 is 24 greater than 84.
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 :))