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.

I'm trying to compute the most right digit of ${{27^{27}}^{27}}^{27}$.

I need to compute ${{27^{27}}^{27}}^{27}(\bmod 10)$.

I now that ${{(27)^{27}}^{27}}^{27}(\bmod 10) \equiv{{(7)^{27}}^{27}}^{27} (\bmod 10)$, so now I need to to compute ${({7^{27})}^{27}}^{27} (\bmod 10)$, since $\gcd(7,10)=1$ and $\phi(10)=4$, $7^{27}=7^{24}\cdot 7^3(\bmod 10)=1 \cdot 7^3 (\bmod 10)=3 (\bmod 10)$ - (Fermat theorem), so I am left with computing ${(3^{27}})^{27} (\bmod 10)$, again $\gcd(3,10)=1$, so $3^{27}= 3^{24}\cdot3^3 \equiv 7(\bmod 10)$, so as I see it the final cut should be again $7^{27}$ which I saw it is already $3 (\bmod 10)$.

Is it correct? what is the correct way to do that?

Thanks

share|improve this question
1  
$(7)^{27^{27^{27}}}$ isn't $(7^{27})^{27^{27}}$ –  Cocopuffs Jun 24 '12 at 20:13
    
Oh no :( What a fatal mistake. What should I do then? –  Jozef Jun 24 '12 at 20:18
    
Except that I confused notation, so scratch that. (Thanks Cocopuffs). –  MGN Jun 24 '12 at 20:18
1  
@Jozef Your method still works fine. Just figure out what $27^{27^{27}}$ is mod $4$. –  Cocopuffs Jun 24 '12 at 20:19
1  
The $\TeX$ code {{27^{27}}^{27}}^{27} looks as if it meant $((27^{27})^{27})^{27}$, but it comes out looking like $\displaystyle{{27^{27}}^{27}}^{27}$. If you write it as 27^{27^{27^{27}}}, so that it looks right to the naked eye reading the $\TeX$ code, then it looks like this: $\displaystyle 27^{27^{27^{27}}}$, which for some purposes might be considered inferior under the circumstances. –  Michael Hardy Jun 24 '12 at 20:23

4 Answers 4

up vote 5 down vote accepted

On the base level, for all $q$ you get $$\forall \ k: \quad a \equiv a - k \cdot q \mod q.$$ Going one level up, as you may know you're not calculating modulo $q$ anymore, but modulo $\phi(q)$: $$\forall \ k: \quad a^{\displaystyle b} \equiv a^{\displaystyle b - k \cdot \phi(q)} \mod q.$$ You can repeat this, and each time you add a $\phi$: $$\forall \ k: \quad a^{\displaystyle b^{\displaystyle c}} \equiv a^{\displaystyle b^{\displaystyle c - k \cdot \phi(\phi(q))}} \mod q, \\ \forall \ k: \quad a^{\displaystyle b^{\displaystyle c^{\displaystyle d}}} \equiv a^{\displaystyle b^{\displaystyle c^{\displaystyle d - k \cdot \phi(\phi(\phi(q)))}}} \mod q.$$

For $q = 10$ we then have \begin{align} q &= 10, \\ \phi(q) &= 4, \\ \phi(\phi(q)) &= 2, \\\phi(\phi(\phi(q))) &= 1. \end{align} So reducing the numbers on each level separately we get \begin{align} 27^{\displaystyle 27^{\displaystyle 27^{\displaystyle 27}}} &\equiv (27 \bmod 10)^{\displaystyle (27 \bmod 4)^{\displaystyle (27 \bmod 2)^{\displaystyle (27 \bmod 1)}}} \mod 10 \\ &\equiv 7^{\displaystyle 3^{\displaystyle 1^{\displaystyle 0}}} \mod 10 \\ &\equiv 7^{\displaystyle 3^{\displaystyle 1}} \mod 10 \\ &\equiv 7^{\displaystyle 3} \mod 10 \\ &\equiv 3 \mod 10 \end{align}

share|improve this answer

No such thing as "the" correct way. First we need to know the meaning of the exponential tower, since parentheses have not been inserted. The usual convention is that $a^{b^c}$ means $a^{(b^c)}$. That convention was violated in your argument, so in principle the argument is not correct. A similar error with different numbers can lead to a wrong answer. In this case it didn't. Apart from that, everything was fine.

For $10$, we might as well let our knowledge of the multiplication table guide us. The powers of $7$, starting with $7^1$, end in $7,9,3,1,7,9,\dots$. We are looking at an odd power of $27$, so the answer must be $7$ or $3$.

To decide which, we have to decide whether the exponent $27^{27^{27}}$ is congruent to $1$ or $3$ modulo $4$. Our exponent is $27$ to an odd power, and $27\equiv -1\pmod{4}$. Thus the exponent $27^{27^{27}}$ is also congruent to $-1$ modulo $4$. So the last digit must be $3$.

Remark: One can mention Euler's Theorem. Since $\varphi(10)=4$, the powers of $27$ must, modulo $10$, cycle with period that divides $4$. However, in this case, the cycling is clear.

Note the almost automatic translation of $x\equiv 3\pmod{4}$ to $x\equiv -1\pmod{4}$. This makes computation much more transparent.

share|improve this answer

Hint $\rm\,\ n\,$ odd $\rm\:\Rightarrow n^{n^{n^{\cdot^{\cdot^{\cdot^n}}}}}\!\!\!\equiv n^3\!\pmod{10}\ $ by $\rm\ \phi(10) = 4,\ \, n^n\equiv n\!\pmod 4\,\ $ $[$if tower has $> 1$ term$]$

share|improve this answer

$\varphi(10)=4$ and $27$ is of the form $(4n+3)\; $. Now $ {(4n+3)^{(4n+3)}} \equiv 3(\bmod 4)=4c+3\; $. Clearly, ${(4n+3)}^{(4n+3)^{(4n+3)}}=(4n+3)^{(4c+3)}\equiv 3(\bmod 4)\; $. This will be held true for any length of such powers of the form $4n+3$.

So ${27}^{27^{27^{27}}}\equiv 7^3(\bmod\ 10)\equiv 3(\bmod\ 10)\; \; $ as $7^4\equiv 1(\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.