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.

Let A be:$${2013}^{{2012}^{{2011}^{2010.....}}}$$

This, goes, on, till, $3^{{2}^1}$

Let B be:$${2012}^{{2011}^{{2010}^{2009.....}}}$$

This, goes, on, till, $3^{{2}^1}$,

For A , I did this,

Let f(x) denote , unit digit of x,

$$f(A)=f(2013)^{4s}=f(3)^{4s}=1$$

For B i did this,

$$f(B)=f(2012)^{{(4s-1)}^{4t}}=f(2012)^{4u+1}=2$$

Is my approach right?

share|improve this question
    
I made sure to show all my working this time. –  Tomarinator Mar 18 '12 at 18:45
    
It needs a bit more explanation, and the second term in the last line should be $f(2012)^{(4s-1)^{4t}}$, with parentheses around the lower exponent, but yes, the implied reasoning is correct. –  Brian M. Scott Mar 18 '12 at 18:54
    
I edited that parenthesis thing, apart from that what more explanation does this needs @BrianM.Scott –  Tomarinator Mar 18 '12 at 18:58
    
@5T0M, your approach is right. –  quartz Mar 18 '12 at 19:07
    
@5T0M: I’ll write up an answer the way that I’d like to see a student explain it. –  Brian M. Scott Mar 18 '12 at 19:21

2 Answers 2

up vote 2 down vote accepted

For any integer $n$ let $f(n)$ denote the unit digit of $n$. Since $2012$ is a multiple of $4$, so is any power of $2012$, and therefore $A$ is of the form $2013^{4s}$ for some positive integer $s$. Thus, $$f(A)=f(2013^{4s})=f(3^{4s})=f(81^s)=f(1^s)=1\;.$$

For $B$, note that $2011=4s-1$ for some integer $s$. Moreover, $2010$ is even, so $B$ is of the form $2012^{(4s-1)^{2t}}$ for some positive integers $s$ and $t$. Now $(4s-1)^2=4(4s^2-2s)+1$ is of the form $4u+1$ for some positive integer $u$, and any positive power of $4u+1$ is of the same form, so without loss of generality $$f(B)=f(2012^{(4u+1)^t})=f(2^{4u+1})=f(2\cdot 16^u)\;.$$ It’s easy to see that $f(16^u)=6$ for all $u\in\Bbb Z^+$, so $f(B)=f(2\cdot 6)=2$.

share|improve this answer

Yes, your approach is right. You can also solve it "recursively" with Euler's theorem, which says that for any $a,x,y,k,$

$$x \equiv y \bmod \phi(k) \ \Rightarrow \ a^x \equiv a^y \bmod k.$$

In this case you want to find $A = 2013^{2012^{\ldots}} \bmod 10$ and $B = 2012^{2011^{\ldots}} \bmod 10$. For $A$ we apply the above with $k = 10$ and $\phi(k) = 4$ to get:

$$2012^{2011^{\ldots}} \equiv 0 \bmod 4 \ \Rightarrow \ 2013^{2012^{\ldots}} \equiv 2013^0 \equiv 1 \bmod 10.$$

Applying the above twice, starting with $k = 10$ (so that $\phi(k) = 4$ and $\phi(\phi(k)) = 2$), for $B$ we get

$$2010^{2009^{\ldots}} \equiv 0 \bmod 2 \ \Rightarrow \ 2011^{2010^{\ldots}} \equiv 2011^0 \equiv 1 \bmod 4 \ \Rightarrow \ 2012^{2011^{\ldots}} \equiv 2012^1 \equiv 2 \bmod 10.$$


In other words:

$$\displaystyle a^{\displaystyle b^{\displaystyle c^{\displaystyle d^{\ldots}}}} \equiv (a \bmod k)^{\displaystyle (b \bmod \phi(k))^{\displaystyle (c \bmod \phi^2(k))^{\displaystyle (d \bmod \phi^3(k))^{\ldots}}}} \bmod k,$$

where e.g. $\phi^3(k) = \phi(\phi(\phi(k)))$. Applying this to $A$ we get

$$2013^{\displaystyle 2012^{\ldots}} \equiv (2013 \bmod 10)^{\displaystyle (2012 \bmod 4)^{\ldots}} \equiv 4^{\displaystyle 0^{\ldots}} \equiv 1 \bmod 10,$$

and for $B$ we get

$$2012^{\displaystyle 2011^{\displaystyle 2010^{\ldots}}} \equiv (2012 \bmod 10)^{\displaystyle (2011 \bmod 4)^{\displaystyle (2010 \bmod 2)^{\ldots}}} \equiv 2^{\displaystyle 3^{\displaystyle 0^{\ldots}}} \equiv 2 \bmod 10.$$

share|improve this answer

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.