A) 1
B) 3
C) 5
D) 7
Correct Answer: B
Solution :
We have, \[{{\left( {{\sum\limits_{k=1}^{5}{{}^{20}C}}_{2k-1}} \right)}^{6}}\]\[={{({}^{20}{{C}_{1}}+{}^{20}{{C}_{3}}+{}^{20}{{C}_{5}}+{}^{20}{{C}_{7}}+{}^{20}{{C}_{9}})}^{6}}\] |
\[={{\left( {{2}^{18}} \right)}^{6}}\] |
\[=8\times {{\left( 32 \right)}^{21}}=8{{\left( 33-1 \right)}^{21}}\] |
\[=8\times \left( 11\lambda -1 \right)=8\times 11\lambda -11+3\] |
\[=11\left( 8\lambda -1 \right)+3\] |
Remainder is 3 when divided by 11. |
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