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What is the last digit of:


I am not sure how to approach this problem as 24! is extremely large. Normally I would notice the cyclic nature of powers of 2 i.e. $2^1 = 2, 2^2 = 4, 2^3 = 8, 2^4 = 16, 2^5 = 32$ and so the last digit cycles so all we need to do is determine $n$ mod $4$.

But not sure how to do it with such a large number, thanks.

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up vote 4 down vote accepted

$$24!=1\times2\times\ldots\times10\times\ldots\times20\times \ldots\times 24=100\times(??)\equiv4 \mod{4}$$ Clearly the last digits are $00$ for $24!$, so the last digit of $2^{24!}$ is same as $2^4$ which is $6$.

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...which is $6$. – Lord_Farin Feb 16 '13 at 10:03
Haha, sorry for that. Edit is made. – Michael Li Feb 16 '13 at 10:04
so would $2^{120!}$ for example also be 6? – fosho Feb 16 '13 at 10:06
$2^{n!}$ has last digit $6$ for all $n \ge 4$. – Lord_Farin Feb 16 '13 at 10:07
Yes. In general $4|n!$ for all $n\geq4$, as @Lord_Farin says. – Michael Li Feb 16 '13 at 10:13

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