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Find the last three digits of $2^{2017}$

My approach:

As $125 \times 8=1000$ we have the congruence modulo $$x \equiv 2^{2017}(mod \: 1000)$$ is equivalent to the equations $$x \equiv 2^{2017}(mod \:125) \tag{1}$$ and $$x \equiv 2^{2017}(mod\:8) \tag{2}$$

Clearly from $(2)$ $x=8m$

Now we need to find remainder when $2^{2017}$ is divided by $125$

We have:

$$2^7 \equiv 3(mod \:125)$$

so $$2^{49} \equiv 3^7(mod \:125)\equiv -63(mod \:125)$$

Hence $$2^{50}\equiv -1(mod \:125) \tag{3}$$

From $(3)$ how to find remainder when $2^{2017}$ is divided by $125$?

Umesh shankar
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    Well, since $2^{50}\equiv -1\pmod {125}$ we have $2^{100}\equiv 1 \pmod {125}$ which implies $2^{2017}\equiv 2^{17}\pmod {125}$. Can you finish from there? – lulu May 19 '20 at 23:41
  • note: $2^{17}=131072$ – J. W. Tanner May 19 '20 at 23:45
  • It seems what you are missing is the notion of modular order reduction on exponents. Once you know that $2^{100}\equiv 1$ then you can reduce $\bmod 100$ all exponents on powers of $2$ - see the comment on my answer (this is essentially what is being done implicitly in Don's answer, but it is best to make the idea more explicit in order to best master it). – Bill Dubuque May 20 '20 at 01:00

4 Answers4

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Using $\ ab\bmod ac = a(b\bmod c) = $ mod Distributive Law to factor out $\,a = 8\,$ yields

$\ \ \ 2^{\large 2017}\!\bmod 1000 = 8\left[\dfrac{2^{2017}}8\bmod 125\right] = 8\left[\dfrac{2^{\large\color{#c00}{17}}}8\bmod 125\right] = 8\left[\dfrac{\color{#0a0}{72}}8\bmod 125\right] = 72$

by Euler & $\,\color{#c00}{17} = 2017\bmod 100\!=\!\phi(125),\,$ & $\!\bmod 125\!:\ 2^{\large\color{#c00}{17}}\!\equiv 2(\color{#90f}{2^{\large 8}})^{\large 2}\!\equiv 2(\color{#90f}6)^{\large 2}\!\equiv\color{#0a0}{72}$

Bill Dubuque
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As @lulu noted in the comments, $2^{50}\equiv-1\bmod{125}$, as well as $2^7\equiv3\bmod{125}$ imply that $$2^{2017}=(2^{50})^{40}\cdot(2^7)^2\cdot2^3\equiv(-1)^{40}\cdot3^2\cdot8\bmod{125}$$$$2^{2017}\equiv72\bmod{125}$$

Rushabh Mehta
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$\varphi(125)=100\stackrel{\text{Euler's theorem}}\implies 2^{100}\cong1\bmod {125}$. Thus we get $2^{2017}\cong2^{17}\bmod{125}\implies 2^{2017}\cong2^7\cdot2^{10}\cong3\cdot24\cong72\bmod{125}$.

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Note: I wrote this answer to show how relations / tricks (c.f. wiki article Presentation of a group) can be found to solve these questions; I like working on these questions as puzzles.

We take it from the OP's

Now we need to find remainder when $2^{2017}$ is divided by $125$

So calculating

$\quad x \equiv 2^{2017} \pmod {125}$ and $0 \le x \lt 125$

Since $2^7 \equiv 3 \pmod {125}$,

$\quad x \equiv 2 \times 3^{288} \pmod {125}$

Since $3^4 \equiv -1 \times 4 \times 11 \pmod {125}$,

$\quad x \equiv 2 \times 2^{144} \times 11^{72} \pmod {125}$

Since $11^2 \equiv -4 \pmod {125}$,

$\quad x \equiv 2^{217} \pmod {125}$

Plowing along using our developed relations,

$\quad x \equiv {(2^{7})}^{31} \equiv 3^{31} \equiv 3^3 \times {(3^4)}^7 \equiv (-1) (3^3) (4^7) (11^7) \equiv (3^3)(2^{20}) (11) \equiv$
$\quad \quad (3^5)(2^{6}) (11) \equiv (-7) (-61) (11) \equiv (-61) (-77) \equiv 72 \pmod {125}$

CopyPasteIt
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