Question Number 5947 by Yozzii last updated on 07/Jun/16
$$ \\ $$$${Show}\:{that}\:\mathrm{3}!^{\mathrm{5}!^{\mathrm{7}!^{\mathrm{9}!^{...\mathrm{2013}!} } } } \:\equiv\mathrm{1}\:\left({mod}\:\mathrm{11}\right). \\ $$
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Question Number 5947 by Yozzii last updated on 07/Jun/16 | ||
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$$ \\ $$$${Show}\:{that}\:\mathrm{3}!^{\mathrm{5}!^{\mathrm{7}!^{\mathrm{9}!^{...\mathrm{2013}!} } } } \:\equiv\mathrm{1}\:\left({mod}\:\mathrm{11}\right). \\ $$ | ||