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Home  >>  CBSE XI  >>  Math  >>  Binomial Theorem

The least (positive) remainder when $17^{30}$ is divided by $5$ is

$\begin{array}{1 1}(A)\;2\\(B)\;1\\(C)\;4\\(D)\;3\end{array} $

1 Answer

Toolbox:
  • $(a+b)^n=nC_0a^n b^0+nC_1 a^{n-1} b^1+......nC_n a^0b^n$
$17^{30}=((17^2))^{15}$
$\Rightarrow (289)^{15}$
$\Rightarrow (290-1)^{15}$
$\Rightarrow 15C_0290^{15}-15C_1(290)^{4}+.......15C_{14} 290(-1)^4+15C_{15}(-1)^{15}$
$\Rightarrow $=(a multiple of 5)-1
$\Rightarrow 5\lambda -1$ (say)
$\Rightarrow 5(\lambda-1)+4$
$\therefore$ Least positive remainder =4
Hence (C) is the correct answer.
answered Jun 26, 2014 by sreemathi.v
 
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