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$53^{53} - 33^{33}$ is divisible by

$\begin{array}{1 1} 53 \\ 33 \\ \text{By both 53 and 33 } \\ 10 \end{array} $

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$53^{53}-33^{33}=(33+20)^{53}-33^{33}$
$=\big[^{53}C_0 .33^{53}+^{53}C_1. 33^{52}.20+^{53}C_2 33^{51}.20^2+........^{53}C_{53}.20^{53}\big]-33^{33}$
$=\big(^{53}C_1. 33^{52}.20+^{53}C_2 33^{51}.20^2+........^{53}C_{53}.20^{53}\big)+33^{33}(33^{20}-1)$
$= A+33^{33}((33^4)^5-1) $ (Where A is multiple of $20$)
But $33^{20}$ has $1$ in its unit's place.
$\therefore\: 33^{33}-1 $ has $0$ in its unit's place.
$\Rightarrow\:33^{20}-1$ is a multiple of $10$
$\therefore\:53^{53}-33^{33}$ is divisible by 10.
answered Nov 22, 2013 by rvidyagovindarajan_1
 

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