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The digit at unit place in the number $17^{1995}+11^{1995}-7^{1995}$ is ?

$\begin{array}{1 1} 0 \\ 1\\ 2\\ 3 \end{array} $

1 Answer

Toolbox:
  • $^nC_0=^nC_n=1$
  • $^nC_1=n$
$17^{1995}+11^{1995}-7^{1995}=$
$(7+10)^{1995}+(1+10)^{1995}-7^{1995}$
=$\big(7^{1995}+^{1995}C_1.7^{1994}.10+.........^{1995}C_{1995}.10^{1995}\big)+$
$\big(^{1995}C_0+^{1995}C_1.10+......^{1995}C_{1995}.10^{1995}\big)-7^{1995}$
$=\bigg(^{1995}C_1.7^{1994}.10+........10^{1995}\bigg)+\bigg(^{1995}C_1.10+......10^{1995}\bigg)+1$
$\Rightarrow\:$ The term at the unit place is 1
answered Oct 14, 2013 by rvidyagovindarajan_1
 
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