A) \[0\]
B) \[1\]
C) \[2\]
D) \[{{2}^{16}}\]
Correct Answer: B
Solution :
[b] \[{{2}^{32}}-(2+1)({{2}^{2}}+1)({{2}^{4}}+1)({{2}^{8}}+1)({{2}^{16}}+1)\] \[={{2}^{32}}-(2-1)(2+1)({{2}^{2}}+1)({{2}^{4}}+1)({{2}^{8}}+1)({{2}^{16}}+1)\] \[={{2}^{32}}-({{2}^{2}}-1)({{2}^{2}}+1)({{2}^{4}}+1)({{2}^{8}}+1)({{2}^{16}}+1)\] \[={{2}^{32}}-({{2}^{4}}-1)({{2}^{4}}+1)({{2}^{8}}+1)({{2}^{16}}+1)\] \[={{2}^{32}}-({{2}^{8}}-1)({{2}^{8}}+1)({{2}^{16}}+1)\] \[={{2}^{32}}-({{2}^{16}}-1)({{2}^{16}}+1)\] \[={{2}^{32}}-({{2}^{32}}-1)=1\] |
You need to login to perform this action.
You will be redirected in
3 sec