Placing only binary operations (+-x/) between them, get the numbers 1,5,6, and 7 to equal 21?
353275
1x5x6x7=21\n210=21\n2+1+0=2+1\n3=3
978426
for example:\n\n-1 < 2\nthen\n1 > -2\n\nanother:\n x < -5\nthen\n-x > 5\n\n-2 x > 9\nx< -9/2\n\ns
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.
315839
keep a table of binary 4 bit equivalents 0 1 2 ... F of hexadecimal degits.\nconvert each 4 bits of your binary no. to equivalent hex. no looking the table.\n\nyou got the answer
152234
1, 3, 5, 7, 11, 17....\n\n\nNumbers that can only be divided by that number or 1 with the result of a whole number.
902614
2/7 *1/6 = 2/42 = 1/21
286367
Take the last number off, double it, and subtract it from the rest of the number.  If that is divisible by 7, then the original number is.\n\nFor example 2457\n\n2457 is 245 & 7; 7*2 = 14\n245-14=231\n\nNot sure?  do it again\n\n231 is 23 & 1; 1*2 = 2\n23 - 2 = 21; this is divisible by 7, so 2457 is also divisible by 7.
352166
8\ntake away o from top and you're left with o on bottom - half - nothing\nor i guess 10 could work, too (or any double digit ending in 0) take away the first number and you're left with 0
703752
simply subtract 7 from both sides. This leave you with\n\nx<_3
284585
It depends on the definition that you use for Gray code since there "is no unique definition for Gray code" [Peter Jackel, p. 84]. The most commonly used (at least in a Sobol sequence context) is:\n \nG(n) = n bitxor [n/2],\nwhere [x] is the integer part of x,\nbitxor is the bitwise operation on the binary representation of to integers.\n\nSo you have to express 17 and 8 in binary and perform bitxor.\n\n17 = ___________10001\n8  = ____________01000\nbitxor(17, 8) =____ 11001 = 25\n\n(I did my best to align the 3 numbers above but am not sure how the'll show up in other pcs)\n\nObservation: the bitxor operation can be performed as a "sum without carry on the binary representation".
560969
23\nif intead of '=' there is +
1007006
add or subtract 1\n\nor else subtract the S at the begining of Seven
277190
1 + 49; 2 + 48; etc. Just match the next highest with the next lowest.
545599
subract 32 then divide by 1.8 (or mulitply by 5/9 --- same thing)
518698
use the inputs i0 - i3 as input and of the 3 selector lines make the last one (s2) 0.\n\nThe possible value will be \n\ns2 s1 s0 out\n0  0  0 - i0\n0  0  1 - i1\n0  1  0 - i2\n0  1  1 - i3\n\nenjoy :)
916538
7+13-4=16\n\ngotta subtract 4 since it is counted twice - both 7 and 13 include it.
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)