If the last digit of $9^{9^9}$ is $z$ then find the last digit of $2^{z^{100}}$. My try:- As unit digit of $9^{\text{odd}}$ is $9$, then $z=9$. Then we are asked to find the last digit of $2^{9^{100}}$. I'm unable to find it. Please help me.
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$2^{1} = \color{#f00}{2},\quad 2^{2} = 4\quad 2^{3} = 8\quad 2^{4} = 16\quad 2^{5} = 3\color{#f00}{2}$. – Felix Marin Jul 08 '16 at 05:24
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How is this not a duplicate of this? – Jyrki Lahtonen Jul 10 '16 at 13:03
3 Answers
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Note that the units digit of $2^n$ goes through a cycle $2,4,8,6$, so you just need $n \bmod 4$ to figure out the units digit.
Ross Millikan
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$9 \equiv 1 \pmod{4}$ so for any $k$, $9^k \equiv 1 \pmod{4}$.
Also, $2^{4s+1}$ ends in a $2$.
So the answer is the last digit of $$2^{9^{100}}$$ is $2$.
Mark Fischler
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At this level, you should probably say more about why $, 2^{4N+1}!\equiv 2\pmod{10}\ $ for all $,N,\ $ since that is the crux of the matter. – Bill Dubuque Jul 07 '16 at 19:21
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Another approach:
$$\text{Euler's formula: }\quad a^{\phi(n)}\equiv 1 \pmod{n} \text{ when} \gcd(a,n)=1$$
$$\phi(10)=\phi(2)\phi(5)=4$$
$$a^4\equiv 1 \pmod {10}$$
Therefore:
$$\color{green}{9^{100}}\equiv (9^{25})^4\equiv \color{red}1\pmod{10}$$
$$2^{\color{green}{9^{100}}}\equiv 2^\color{red}1\equiv 2\pmod {10}$$
3SAT
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