How do you add, subtract, multiply, and divide Binary numbers?
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.
843943
4, 0, -2, -2, 0\n\nFigure it out! Don't be lazy!
61382
oh crap i know this one    hang on     think think   an algorithm is a repeated process like converting binary to decimal and back.  but i can't think of one for finding the union sorry
97477
76063 in decimal\n0x1291F in Hexadecimal\n1001 0100 1000 11111 in binary
9961
0,1,1,2,3\n\nto get the next number add the previous 2 numbers
591222
0.\n...and the counting stops, because we are counting in binary....lol.
152638
It is the number 36 in binary.\n\nFrom right to left:\n0 = 0\n0 = 0\n1 = 4\n0 = 0\n\n0 = 0\n1 = 32\n0 = 0\n0 = 0\n\n32 + 4 = 36
1130664
1+1 = 10 if you are counting in binary.\n\n1+1=0 with 1 carry over if you are counting in urinary.\n\nanything bigger than binary = 2.\n\n1+1 = 1 if you are talking logical (OR) also.\n\n1+1 = 2 at first then it becomes 3 then 2 again if you are talking about marriage and divorce.
612486
This is pretty straighforward in C - I won't write the assignment for you, but the general proccess is as follows:  \n\n1.  Count the number of each coin - store that number in some variable (say, int num_dimes;  )  If you're reading from a file, it's a little more complicated, but if your user is entering something from a command line (as is the case with beginning programming assignments, usualy) it's just a matter of having them enter a number and using a function like get() to read the input.  \n\n2.  Multiply the number of each coins by the appropriate denomination.  For example - if you counted 6 dimes in the previous step, multiply 6 by .10.  Make sure that you store the result in a floating point variable, and it's generaly a good idea to ensure that you're multiplying two floats together in order to get a float answer.  Use something like this:\n  \nfloat dimes_total; \ndimes_total = (float)num_dimes * 0.10;   \n\nthe (float) in front of num_dimes casts that variable as a floating point number.  \n\n3.  After you've done this for each coin, simply add up the totals: \n\nfloat total; \ntotal = pennies_total + nickles_total  ... etc, etc.\n\n4.  print your answer to the screen - since dollars and cents use the same number base as arithmetic done by the compiler, nothing fancy is needed here.    In cases where you're dealing with different number bases, however, it gets to be a little more tricky.
121493
We are used to ten digits in our numbering system, but binary only has two (zero and one, or on and off) In our system when we get to nine, our ones spot is at the top and we have to use two columns for higher numbers. Same applies in binary. Column starts at zero and is full at one.\nOn our regular system our first two columns fill at nine nine - in binary the first columns fill at one one. nine nine in our system equals 99. one one in binary equals 3.\n00000000 = 0\n00000001 = 1\nnow that column is full, so to count to 2 you have to use the next column.\n00000010 = 2\n00000011 = 3\nnow that column is full, so to count to 4 you have to use the next column.\n00000100 = 4\nand so on.\n\nInsted of having the "ones" "tens" "hundreds" etc. columns as our regular system does - \nbinary has the "ones" "twos" "fours" "eights" "sixteens" thirty-twos" etc colmns\n\nA byte is 8 columns.
940062
Binary number system is just like any number system - it has a base which is equal to number of unique digits in that number system. For example:\n\nDecimal Number System has a base of 10 and it has digits from 0-9  (total no of digits = 10 = base). The number system starts with 0 and goes up to the max digit in the number system. Then the digit at that position becomes zero and the number at higher position becomes one. Let's see if you can follow this by looking at decimal no system first\n01, 02, 03, 04, 05, 06, 07, 08, 09 (now the no at least posn is 9 = max possible in a decimal no system, so this posn becomes zero and the no at the next higher level becomes 1)... therefore, after 09 (which is same as 9), you get 10.\n\nNow use the same logic for creating binary numbers (or any no system - octal, hex), just remember binary no. system has only two digits (binary = two) -> 0 and 1... therefore the binary nos go like this (taking 3 leading zeros and keeping in mind that leading zero do not alter the value of a number, eg 09 is same as 9 and 090 is same as 90, similarly for binary 0001 = 1, 0010 =10)\n0000, 0001 - least posn = 1 = max digit in binary no system, so next time we change this to zero and increment the next higher posn with 1 - so we get --- 0010\nNow increment the least posn by 1, we get ---- 0011\nSince both, least n next higher posn is max =1 each, so we make them both zero and the third higher posn = 1 we get -- 0100\n\nand so on we continue - 0100, 0101, 0110, 0111, 1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111. \n\nFinally - the binary nos and their decimal values:\n0000 = 0\n0001 = 1\n0010 = 2\n0011 = 3\n0100 = 4\n0101 = 5\n0110 = 6\n0111 = 7\n1000 = 8\n1001 = 9\n1010 = 10\n\nHope this would help you understand how numbers are generated in binary or any number system :)
909474
You can do this by hand.\nCall your number x.\nWrite the binary digits of x as b_i :\nx = sum_n=0_n=N b_n.2^n\n\nThe algorithm is simple: compare your number x to the largest power of two (2^N), subtract if necessary, and decrease n until finished (n=0).\n\nStart with n=N , the largest integer such that X >=2^n\nIf x >= 2^n, then {\n b_n = 1;\n x = x-2^n;\n n = n-1; \n} else {\n b_n 0;\n}\n\nThe above is pseudocode, you could do it in C, PERL, Python, Java etc.\n\nSo: x= 121 = 64+57=(64+32)+25=(64+32+16)+9\n=(64+32+16+8)+1 =64+32+16+8+0+0+1\n\nThus 121 (base 10) = 1111001 (base 2)
864328
binary 11 is 3 decimal
615376
there are several methods,\n\nalmost all mathematical calculations like add, subtraction...... have many ways to calculate......\n\ncheck the website below to find alternate ways to calculate
605852
multiply n-m by 0.737562
865627
Binary\n101=5\n1010=10\n111=15\n10100=20\n11001=25\n\nAlternate add 7 and 5\n5+7\n12+5\n17+7\n24+5\n29\n\nDouble first and second alternatly\n1*2=2\n16*2=32\n2*2=4\n32*2=64\n4\n64\n\nLast one ????
524465
you must be aware of the logic of half adders. here we will be dealing with a 2 bit number. the addition of the 2 bits is actually a xor function.\n\nA B S C\n0 0 0 0\n1 0 1 0\n0 1 1 0\n1 1 0 1\n\nhere A is the most significant bit and B the least.. (example. if your number is 10 then A is 1 and B is 0).\n\nS stands for the sum \nC stands for carry\n\nthe basic logic is that S is 1 for an odd combination of 1\nand C is 1 only when A=B.\n\nso divide the given number( number input) by 10 (N/10).\nfind a and b and make the program accordingly. it is very simple if a is not equal to b the sum is 1.. and carry is 0. if A=B=0 then S=0 C=0.\nif A=B=1 then S=0 C=1.  \n(all this is binary addition)