Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

What is the last digit of:

$2^{24!}$

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.

share|improve this question

1 Answer 1

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 $6$.

share|improve this answer
1  
...which is $6$. –  Lord_Farin Feb 16 '13 at 10:03
1  
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  
$2^{n!}$ 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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.