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I encounter a modular arithmetic problem. which says:

Find the last Digit of $17^{16}$, by intuition the last digit of a number is the remainder of the number divided by $10$.

so the statement is: $17^{16} \pmod {10}$

According to my knowledge, I figure out that the solution to this problem can be "solved" this way:

** $17^{16}\pmod {10}$ = $(17^8\pmod{10} * 17^8\pmod{10})\pmod{10}$**

then we get $(7 * 7)\pmod{10}$, which is equal to $49\pmod{10}$, and we get a result of $ 9$.

The problem is that when I go to modular calculator around the internet I get a result of 1. and I certainly don't know why.. check for yourself: https://www.mtholyoke.edu/courses/quenell/s2003/ma139/js/powermod.html

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    $\begingroup$ I wish you had verified $17^{8} \equiv 1 \pmod{10}$ from the calculator you mentioned in the post. $\endgroup$ – Math Lover Apr 22 at 22:51
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    $\begingroup$ I don't understand the step $$17^8\cdot 17^8\equiv 7\cdot 7\pmod {10}$$ Could you please elaborate? $\endgroup$ – Dr. Mathva Apr 22 at 22:53
  • $\begingroup$ according to the exponential property : (a^b)mod c = (a mod c)^b (mod c) $\endgroup$ – Teuddy R Apr 22 at 23:03
  • $\begingroup$ You are missing the expontens: $17^8\cdot 17^8\equiv 7^8\cdot 7^8\mod 10$ $\endgroup$ – Cornman Apr 22 at 23:18
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    $\begingroup$ How did you get $\bmod 10\!:\ 17^{\large 8}\!\equiv 7$? ($\equiv 1$ is correct) $\endgroup$ – Bill Dubuque Apr 22 at 23:28
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The exponent $16$ is small enough to calculate by hand:

$$\begin{align*} 17^1 &\equiv 7 \pmod{10}\\ 17^2 &\equiv 7^2 \equiv 9\\ 17^4 &\equiv \left(17^2\right)^2\equiv 9^2 \equiv 1\\ 17^8 &\equiv \left(17^4\right)^2\equiv 1^2 \equiv 1\\ 17^{16} &\equiv \left(17^8\right)^2\equiv 1^2 \equiv 1 \end{align*}$$

Which by the way also shows that $17^8\equiv 1\pmod{10}$.

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This is an elementary approach:

It is $17^{16}\equiv (17^2)^8\equiv (289)^8\equiv 9^8\mod 10$.

And $9^8\equiv (81)^4\equiv 1^4\mod 10$.

So it is $17^{16}\equiv 1\mod 10$.

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It is actually more simple than what you've done: By Euler's Theorem

$$\rm a^{\varphi(10)}\equiv a^4\equiv 1\pmod {10}$$ as long as $10$ and $\rm a$ are relatively prime, and $\rm a=17$ certainly satifies the condition.

Thus

$$\rm \big(17^4\big)\;^4\equiv 1^4\equiv 1\pmod {10}$$


Alternatively Observe that $$17^{16}\equiv (10+7)^{16}\equiv 7^{16}\equiv 49^{8}\equiv 9^8\color{red}{\equiv (-1)^8}\equiv 1\pmod {10}$$

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$$17=2\cdot10-3\\-3^2=10-1\\-1^{2n}=1\\16=2\cdot(2\cdot 4)$$ Therefore:$$17^{16}\equiv-3^{16}\equiv(-1)^{2\cdot4}\equiv 1\pmod{10}$$

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