A) \[{{(111101)}_{2}}\]
B) \[{{(111111)}_{2}}\]
C) \[{{(101111)}_{2}}\]
D) \[{{(111001)}_{2}}\]
Correct Answer: A
Solution :
\[{{(100010)}_{2}}={{2}^{5}}\times 1+{{2}^{4}}\times 0+{{2}^{3}}\times 0+{{2}^{2}}\times 0+\] \[{{2}^{1}}\times 1+{{2}^{0}}\times 0\]\[=32+0+0+0+2+0={{(34)}_{10}}\] and \[{{(11011)}_{2}}={{2}^{4}}\times 1+{{2}^{3}}\times 1+{{2}^{2}}\times 0+{{2}^{1}}\times 1+{{2}^{0}}\times 1\] \[=16+8+0+2+1={{(27)}_{10}}\] \ Sum \[{{(100010)}_{2}}+{{(11011)}_{2}}={{(34)}_{10}}+{{(27)}_{10}}={{(61)}_{10}}\] Now| 2 | 61 | Remainder |
| 2 | 30 | 1 LSD |
| 2 | 15 | 0 |
| 2 | 7 | 1 |
| 2 | 3 | 1 |
| 2 | 1 | 1 |
| 0 | 1 MSD |
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