0 . . . . . 0 0 0 0 1 . . . . . 0 0 0 1 2 . . . . . 0 0 1 0 3 . . . . . 0 0 1 1 4 . . . . . 0 1 0 0 5 . . . . . 0 1 0 1 6 . . . . . 0 1 1 0 7 . . . . . 0 1 1 1 8 . . . . . 1 0 0 0 9 . . . . . 1 0 0 1 10 . . . . 1 0 1 0
| x | y | x' | y' | x⊕y | x'⊕y' | ---------------------------------- | 0 | 0 | 1 | 1 | 0 | 0 | | 0 | 1 | 1 | 0 | 1 | 1 | | 1 | 0 | 0 | 1 | 1 | 1 | | 1 | 1 | 0 | 0 | 0 | 0 |
Here they are for 1 to 16, at no extra cost:Dec . . . Bin1 . . . . . 12 . . . . . 1 03 . . . . . 1 14 . . . . . 1 0 05 . . . . . 1 0 16 . . . . . 1 1 07 . . . . . 1 1 18 . . . . . 1 0 0 09 . . . . . 1 0 0 110 . . . . 1 0 1 011 . . . . 1 0 1 112 . . . . 1 1 0 013 . . . . 1 1 0 114 . . . . 1 1 1 015 . . . . 1 1 1 116 . . . . 1 0 0 0 0
AnswerAnswer: ( 0! + 0! + 0! + 0! + 0! ) ! = 120 Explanation: Here we have used operator called " factorial ". As you know that 0! = 1 so, = ( 0! + 0! + 0! + 0! + 0! ) ! = ( 1 + 1 + 1 + 1 + 1 ) ! = (5 )! = 120 : ( 0! + 0! + 0! + 0! + 0! ) ! = 120 Explanation: Here we have used operator called " factorial ". As you know that 0! = 1 so, = ( 0! + 0! + 0! + 0! + 0! ) ! = ( 1 + 1 + 1 + 1 + 1 ) ! = (5 )! = 120
0-(-1) is the same as 0+1, so 0-(-1)=1
It depends on the writer of course. But Hulk's powerset simply does not give him the tools to beat The Juggernaut.
Excess-3 BCD a B c d w x y z 0 0 1 1 0 0 0 0 0 1 0 0 0 0 0 1 0 1 0 1 0 0 1 0 0 1 1 0 0 0 1 1 0 1 1 1 0 1 0 0 1 0 0 0 0 1 0 1 1 0 0 1 0 1 1 0 1 0 1 0 0 1 1 1 1 0 1 0 1 0 0 0 1 0 0 0 1 0 0 1 i'm not sure. but it should be the ans
a b c y 0 0 0 0 0 0 0 1 0 0 1 0 0 0 1 1 0 1 0 0 0 1 0 1 0 1 1 0 0 1 1 1 1 0 0 0 1 0 0 1 1 0 1 0 1 0 1 1 1 1 0 0 1 1 0 1 1 1 1 0 1 1 1 1 FORMULA FOR possibilities = 2 ^(no of variables). Here its 4 so, 2n=24=16 Hence we have 16 possibilities.
w x y z 0 0 0 0 0 0 0 1 0 0 1 0 0 0 1 1 0 1 0 0 0 1 0 1 0 1 1 0 0 1 1 1 1 0 0 0 1 0 0 1 1 0 1 0 1 0 1 1 1 1 0 0 1 1 0 1 1 1 1 0 1 1 1 1 FORMULA FOR possibilities = 2 ^(no of variables). Here its 4 so, 2n=24=16 Hence we have 16 possibilities.
0 . . . 0 0 0 0 1 . . . 0 0 0 1 2 . . . 0 0 1 0 3 . . . 0 0 1 1 4 . . . 0 1 0 0 5 . . . 0 1 0 1 6 . . . 0 1 1 0 7 . . . 0 1 1 1 8 . . . 1 0 0 0 9 . . . 1 0 0 1 10 . . 1 0 1 0 11 . . 1 0 1 1 12 . . 1 1 0 0 13 . . 1 1 0 1 14 . . 1 1 1 0 15 . . 1 1 1 1 16. 1 0 0 0 0 . . etc.
11
0 . . . . . 0 0 0 0 1 . . . . . 0 0 0 1 2 . . . . . 0 0 1 0 3 . . . . . 0 0 1 1 4 . . . . . 0 1 0 0 5 . . . . . 0 1 0 1 6 . . . . . 0 1 1 0 7 . . . . . 0 1 1 1 8 . . . . . 1 0 0 0 9 . . . . . 1 0 0 1 10 . . . . 1 0 1 0
1 0 1 0 0 1 0 1 0 1 1 0 1 1 0 1 0 1 1 1 1 1 0 0 0 1 1 0 0 1
AND A B Q 0 0 0 0 1 0 1 0 0 1 1 1 OR A B Q 0 0 0 0 1 1 1 0 1 1 1 1 NOT A Q 0 1 1 0 (I might as well carry on) XOR A B Q 0 0 0 0 1 1 1 0 1 1 1 0 NAND A B Q 0 0 1 0 1 1 1 0 1 1 1 0 NOR A B Q 0 0 1 0 1 0 1 0 0 1 1 1 BUF (NNOT) A Q 0 0 1 1 XNOR A B Q 0 0 1 0 1 0 1 0 0 1 1 1
Strobe A B C D Output (Y) 0 0 0 0 0 D0 0 0 0 0 1 D1 0 0 0 1 0 D2 0 0 0 1 1 D3 0 0 1 0 0 D4 0 0 1 0 1 D5 0 0 1 1 0 D6 0 0 1 1 1 D7 0 1 0 0 0 D8 0 1 0 0 1 D9 0 1 0 1 0 D10 0 1 0 1 1 D11 0 1 1 0 0 D12 0 1 1 0 1 D13 0 1 1 1 0 D14 0 1 1 1 1 D15 1 X X X X 1 where A,B,C,D are the control input or control nibble and the Boolean expression for Y is given as:- Y = A'B'C'D'D0 + A'B'C'DD1 + A'B'CD' D2 + A'B'CDD3 + A'BC'D'D4 + A'BC'DD5+ A'BCD'D6 + A'BCDD7 + AB'C'D'D8 + AB'C'DD9 + AB'CD'D10 + AB'CDD11 + ABC'D'D12 + ABC'DD13 + ABCD'D14 + ABCDD15
1 - 0 - 0 - 1 = 0
| x | y | x' | y' | x⊕y | x'⊕y' | ---------------------------------- | 0 | 0 | 1 | 1 | 0 | 0 | | 0 | 1 | 1 | 0 | 1 | 1 | | 1 | 0 | 0 | 1 | 1 | 1 | | 1 | 1 | 0 | 0 | 0 | 0 |