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4 votes

178 mod 47 =4

how do we calculate such huge values

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$17^8\; mod \; 47 = 289^4\; mod\; 47 = (6*47 + 7)^4\;mod\;47 = 7^4\;mod\;47 $
$ 7^4 = 2401$

$\color{red}{2401\;mod\;47 = 4}$
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.==>74mod47= 492mod47 = 2 = 4

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1 Answer

3 votes
3 votes
Best answer

We can calculate this value using property of exponentiation in modular arithmetic

$If \ a \equiv b(modN), then\ a^{k} \equiv b^{k}(modN) \ for \ any \ positive \ integer$ k

$17^{8}(mod47) \\ \equiv (17^{2})^{4}(mod47) \\ \equiv 7^{4}(mod)47 \ \because 289mod47=7 \\ \equiv (7^{2})^{2}(mod47) \\ \equiv 2^{2}(mod47) \ \because \ 49mod47=2 \\ \equiv 4(mod47) \\ \equiv 4$

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