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Home  >>  CBSE XI  >>  Math  >>  Binomial Theorem
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The digit at units place in the number $17^{1995}+11^{1995}-7^{1995}$ is

$\begin{array}{1 1}(A)\;0\\(B)\;1\\(C)\;2\\(D)\;3\end{array} $

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1 Answer

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Toolbox:
  • $(a+b)^n=nC_0a^nb^0+nC_1a^{n-1}b^1.......nC_n a^0b^n$
$17^{1995}+11^{1995}-7^{1995}$
$\Rightarrow (7+10)^{1995}+(1+10)^{1995}-7^{1995}$
$\Rightarrow 7^{1995}+1995C_17^{1994}.10+.....1995C_{1995}10^{1995})+(1+1995C_110+.......1995C_{1995}10^{1995})-7^{1995}$
$\Rightarrow 1995C_17^{1994}.10+......10^{1995})+(1995C_110+.....+10^{1995})$
$\Rightarrow $=(some multiple of 10)+1
$\therefore$ Digits at the unit's lace =1
Hence (B) is the correct answer.
answered Jun 26, 2014 by sreemathi.v
 

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