A) 1
B) 3
C) 7
D) 9
Correct Answer: A
Solution :
Unit's digit in \[({{7}^{4}})=1.\]. Therefore, unit's digit in \[{{({{7}^{4}})}^{8}}\]i.e. \[{{7}^{32}}\]will be 1. Hence, unit's digit in \[{{(7)}^{35}}=1\times 7\times 7\times 7=3\] Again, unit's digit in \[{{(3)}^{4}}=1\] Therefore, unit's digit in the expansion of \[{{({{3}^{4}})}^{17}}={{(3)}^{68}}=1\] \[\Rightarrow \] Unit's digit in the expansion of \[({{3}^{71}})=1\times 3\times 3\times 3=7\] and unit's digit in the expanison of \[{{(11)}^{35}}=1\] Hence, unit's digit in the expansion of \[{{7}^{35}}\times {{3}^{71}}\times {{11}^{55}}=3\times 7\times 1=1\]
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