2
$\begingroup$

How to calculate the last two digits of ${14}^{{14}^{14}}$? With the help of any method. I have tried and have got the last digit to be $6$. But not sure.

$\endgroup$
  • $\begingroup$ you have to find x $\equiv$ 14^14^14 (mod 100). And you are correct the last digit is 6, but whats the second last digit? $\endgroup$ – stackErr May 6 '14 at 11:55
4
$\begingroup$

Clearly, $14^{14^{14}}$ is a multiple of $4$. To compute $14^n\pmod{25}$ we should know $n\pmod {\phi(25)}$, i.e. $14^{14}\pmod{20}$. Again, $14^{14}$ is a multiple of $4$, and it is $\equiv (-1)^{14}\equiv 1\pmod 5$. Hence $14^{14}\equiv 16\pmod {20}$. Thus $14^{14^{14}}\equiv 14^{16}\pmod {25}$. This can me computed by repeated squareing: $$ 14^{16}=(14^2)^8=196^8\equiv (-4)^8=16^4=256^2\equiv 6^2=36\pmod{25}.$$ Since $36$ is already a multiple of $4$, we have immediately that $14^{14^{14}}\equiv 36\pmod{100}$.

$\endgroup$
3
$\begingroup$

Hints:

  1. Work modulo $100$.
  2. Split $14$ into $7\times 2$.
  3. Note that $7^2\times2=98=-2\pmod{100}$.
$\endgroup$
2
$\begingroup$

Using twice the mod Distributive Law $\ ab\bmod ac = a(b\bmod c)\ $ we have

$\!\bmod 25\!:\,\ 14^{\large\color{#0a0}{10}}\! \equiv\!\!\overbrace{(8^{\large 2})^{\large 10}\! \equiv 1}^{\large\ \ \ 2\cdot 10\ =\ \phi(25)}\!\!,\ $ & $\, \ \color{#c00}{14^{\large 14}}\, \bmod\,\color{#0a0}{10}\ =\ 2\!\!\!\!\!\!\overbrace{(14^{\large 14}/2 \bmod 5)}^{\ \ \ \ \ \large (-1)^{\LARGE 14}/2\ \equiv\ 1/2\ \equiv\ 6/2}\!\!\!\!\!\!\! =\color{#c00}6$

therefore $\ \,14^{\large 14^{\LARGE 14}}\!\!\bmod 100 = 4\, (14^{\large\color{#c00}{14^{\LARGE 14}}}\!\!/4 \bmod 25) =\! \underbrace{4\,(14^{\large\color{#c00}6}/4\ \bmod\, 25)\, =\, 36}_{\large\ (-11)^{\LARGE \color{#c00}6}\equiv\ (121)^{\LARGE 3} \equiv\ (-4)^{\LARGE 3} \equiv\ 36 \ } $


Or $\bmod 25\!:\ (-1\!+\!15)^{\large 14^{\LARGE 14}}\overset{\rm\color{#c00}{BT}}\equiv 1- 15\,(\!\!\underbrace{14^{\large 14}}_{\large (-1)^{\LARGE 14}\ \ }\!\!\!\bmod 5)\equiv 36\ $ via $\,\rm\color{#c00}{BT} = $ Binomial Theoem

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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