Please convert this binary number 100011101 to a decimal number?
95089
1(1)+0(2)+1(4)+1(8)+1(16)+0(32)+0(64)+0(128)+1(256)=\n1+0+4+8+16+0+0+0+256=\n285
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.
16829
Binary numbers are just normal (base 10) numbers written in a different way.\n\nTo convert a binary number to a number you're used to seeing, start at the right end. Each place in the number represents a power of 2, starting with 2^0, or 1. Each place to the left increases by a power of 2. To determine what number you have, simply add the powers of 2 which correspond to the 1's.\n\nFor example, take the binary number 110101.\n\nStarting from the left, we have\n\n1 * 2^0 = 1*1  = 1\n0 * 2^1 = 0*2  = 0\n1 * 2^2 = 1*4  = 4\n0 * 2^3 = 0*8  = 0\n1 * 2^4 = 1*16 = 16\n1 * 2^5 = 1*32 = 32\n\nSo we add 1 + 4 + 16 + 32 = 53\n\nSo 53 expressed in binary is 110101\n\nHope this helps.
341598
If you want to convert a hex string to decimal, you\nneed to do two operations:\n\n- convert the hex string to a number\n- generate a decimal string from that number\n\nIf your system supported a full C implementation,\nit would be as simple as using scanf()/printf().\nI presume you're being asked to implement the\nprimitives yourself :-)\n\nTo convert a hex string to a number, you might\nimplement something like:\n\nint number = 0;\nwhile ((c = getchar()) != '&#92;n') {\n  if (c >= '0' && c <= '9')\n    number = (number<<4) + (c-'0')\n  else if (c>='a' && c<='f')\n    number = (number<<4) + (c-'a')\n  else if (c>='A' && c<='F')\n    number = (number<<4) + (c-'A')\n}\n\nThis could be cleaned up if you have access to the\n<ctype.h> macros [ie. isdigit() and tolower()].\n\nTo convert the number to a decimal string, you need\nto perform successive divisions by 10.  An 8051\ndoesn't have a hardware divide, but that's OK\nbecause you should have a software divide/remainder\nsubroutine.\n\nThere are two tricks to keep in mind.  First to\ngenerate the appropriate character for each digit,\nyou can treat a literal character string as an\narray.  For example:\n\nchar c = "0123456789ABCDEF"[i];\n\nwill assign the correct character to variable "c"\nbased on the value of "i" (if it is between 0 and\n15).\n\nThe other detail is that as you call the divide/\nremainder subroutine, you will generate the digits\nworking from least significant to most significant.\nTherefore, you'll want to fill an array from right\nto left.  Assuming you have a divide routine with\nthe prototype:\n\nint\ndivide(int dividend, int divisor, int *remainder)\n\nand you can use the "puts()" function to display the\ngenerated string, you might use something like:\n\nvoid\ntodecimal(int value)\n{\n  char buffer[20];\n  char*p = &buffer[sizeof(buffer)-1];\n  int remainder;\n  while (value != 0) {\n    value = divide(value, 10, &remainder);\n    *p-- = "0123456789ABCDEF"[remainder];\n  }\n  puts(p+1);\n}
614610
Just subtract 27624824 from both sides, leaving the answer as 7497211600
438874
Binary works in base two so it goes 1 2 4 8 16 32 etc only you work from right to left instead of left to right.\n\nFor instance if you wanted to convert the number 52 you would use the following\n\n52 / 2 = 26 r 0\n26 / 2 = 13 r 0\n13 / 2 = 6 r 1\n6 / 2 = 3 r 0\n3 / 2 = 1 r 1\n1 / 2 = 0 r 1\n\nYou then read from top to bottom so 52 in binary would be 110100\n\nYou can see it better if you put these values under the binary table so\n\n32  16  8  4  2  1 \n1    1   0  1  0  0   \n\n32 + 16 + 4 = 52\n\nHope that makes sense
938068
it is 2 to the power 1. I am assuming b10 means binary 10 which is 2 in decimal. Therfor b10 = 2^1.
580667
Type ' the the number\nlike this\n'12345678901234567890\n\nand it will entered as it is..\n\nXLMan
357910
hello!!!!!!!!!! \n1+1 in binary is 10! it's not ten but one zero!!!
139403
in decimal its 2\nin binary its  0\nin other like octal hexa etc its 2\nin logical or its 1\nnow wat do u want????
366105
I always have to use a table. In the column header put your decimal equivalents of binary. 128 64 32 16 8 4 2 1 etc. \n\nTo convert the 107 to binary start at the left of the table. Find the biggest decimal equivalent that can be subtracted from your decimal number. That column gets a 1. Do the subtraction and then work with the result. If the next binary equivalent is larger than the previous result then that column gets a 0 and you move on to the next column. \n\nThese examples look lousy because of text formatting. If you write them out in a table you should understand them better.\n\n\nSo for 107\n\n64      32      16      8      4      2      1\n107-64  43-32   -       11-8   -     3-2     1-1\n1        1      0       1      0       1       1\n     \n\nFor converting binary to decimal do a table with the decimal equivalents at the top.\nPut your binary number in the next row. Then in the next row carry down the decimal equivelent in each column if there is a 1. Add up the row \n\n128  64  32  16  8   4   2   1\n     \n1    0    1   0  0   1    1  0\n\n128+ 0+32+ 0+  0+  4+  2+  0   =   You work this out
32238
If you write the codes for 16 digits in binary out -\n\n0000 - 0\n0001 - 1\n0010 - 2\n0011 - 3\n0100 - 4\n0101 - 5\n0110 - 6\n0111 - 7\n1000 - 8\n1001 - 9\n1010 - A\n1011 - B\n1100 - C\n1101 - D\n1110 - E\n1111 - F\n\nThen take 3E7C, match the four digits on the chart:\n\n0011 1110 0111 1100 =\n\n11111001111100\n\nThats hex to binary conversion, to decimal, you just add your exponents of two, reading binary number from right to left:\n\n2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^9 + 2^10 + 2^11 + 2^12\n\n= \n\n15996\n\nI admit there's a more direct path to the number, but rememebring the way above allows you do any decimal/binary/hex conversion neededed, so its all I've remembered.
128262
The answer is 110111010.  In decimal, it's the same as 253 + 189 = 442.
428515
Binary numbers use only one (1) and zero (0). \n\n10 in binary is 2 in decimal.\n11 in binary is 3 in decimal.\n110 in binary is 6 in decimal.\n\n(Decimal numbers are normal numbers.)
810525
Move the decimal two places to the left.\n.3455
163688
If you just want an integar value, you can use the 'parseInt()' method of the 'Integer' class which takes a string (the number) and a integer radix (or base).  The radix, or base, can be 2 for binary, 8 for octal, 10 for decimal and 16 if you wish to parse hexadecimal.\n\nfor example:\n\nparseInt("1100100", 2) would return the integer value, '100'.  You can use then use toString() on the resulting integer value in order to get a string representation of the integer back.
840338
In windows, open up the calculator.\nGo to scientific mode.\nSelect "Bin".\nType in the binary number.\nSelect "Dec"\nThe number is now displayed in decimal.\n\n4\n5\n12\n21\n15\n185\n154