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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.

Martin Sleziak
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user143384
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  • 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? – stackErr May 06 '14 at 11:55

4 Answers4

4

Hints:

  1. Work modulo $100$.
  2. Split $14$ into $7\times 2$.
  3. Note that $7^2\times2=98=-2\pmod{100}$.
user1729
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4

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 $14^{14^{14}}\equiv 36\pmod{100}$.

Hagen von Eitzen
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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 Theorem

J. W. Tanner
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Bill Dubuque
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  • Note: Euler's theorem $,a^{\phi(m)}\equiv 1\pmod{!m},$ has hypothesis $,\gcd(a,m)= 1.,$ As above, when this fails we can pull out the gcd of $,a^N$ and $,m,$ via mod Distributive Law, reducing to coprime case where Euler applies – Bill Dubuque Jul 13 '23 at 14:13
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Review the theory and techniques found here.

We write as true (no calculator is necessary),

$\; 14^{14} \equiv (14^2)^7 \equiv 16 \times 16^2 \times 16^2 \times 16^2 \equiv 16 \times 16 \times 16 \times 16 \equiv 16 \pmod{20}$

So (again using mental calculations),

$\; \displaystyle{ {14}^{{14}^{14}} \equiv 76 \times {14}^{16} \equiv 76 \times ({14}^{2})^8 \equiv 76 \times ({96}^{2})^4 \equiv ({16}^{2})^2 \equiv 36 \pmod{100}}$

CopyPasteIt
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