What is the answer to $17^{16} \pmod {10}$? Is it equal to $ 9 $ or $1$

Question

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 figured 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

Answer 1

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

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

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

Answer 2

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 $17^8\equiv 1\pmod{10}$.

Answer 3

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

Answer 4

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

Answer 5

We can see if there is a pattern.

$17\equiv{10} = 7$

$17^2=289\equiv{10} = 9$

$17^3=4913\equiv{10} = 3$

$17^4=83521\equiv{10} = 1$

This $(7, 9, 3, 1)$ pattern continues. Since 16 is a multiple of 4, the last digit of $17^{16}$ is 1.

I know that the $(7, 9, 3, 1)$ pattern continues from simple multiplication of ones digits:

$7\cdot{7}=49$

$7\cdot{9}=63$

$7\cdot{3}=21$

$7\cdot{1}=7$

$7\cdot{7}=49\dots$

This pattern continues with the ones digits . That's my proof of $17^{16}\equiv{10}=1$.

Answer 6

$17^2 =289$

$\Rightarrow$ $17^2 =-1[5]$ and $ 17^2 =1[2]$

$(17^{2}) ^8=1[5] $$\Rightarrow$$ 17^{16} =1[5]$

So $17^{16} =1[5]$ and $17^{16}=1[2] $

$gcd(5,2)=1$

$\Rightarrow$ $17^{16} =1[10]$

Note:$a=b[n] $ and $a=b[m] $ and $gcd(m, n) =1$$\Rightarrow$ $a=b[mn]$