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# Find the Units digit, challenge!

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I'm the challenger, I post questions every week!

Week 1:

Find the units digit of 3^17*7^23?

May 20, 2018

#1
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As we cycle through the powers of 3 and 7, we see that our answer is $$3*3=\boxed{9}$$  .
May 20, 2018
#2
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That is correct tertre, but we need a solution.

Like what tertre said, let's cycle through the powers of $$3$$ and $$7$$, and see if there's a pattern.

We have, for the powers of $$3$$$$3^1=3, 3^2=9, 3^3=27, 3^4=81$$ . Woah, we found a pattern! The units digit ($$3,9,7,1$$), will repeat forever with a power of $$3.$$ Since we have to find the units digit of $$3^{17}$$ , we can simply do: $$\frac{17}{4}=4 R1$$ . A remainder of $$1$$ , means the first number in the pattern, which is $$3$$ . Next, on to the powers of $$7.$$

We have, for the powers of 7: $$7^1=7, 7^2=49, 7^3=343, 7^4=2401$$ . We found a pattern here, again! The units digit

($$7,9,3,1$$), will repeat forever with a power of $$7.$$ Since we have to find the units digit of $$7^{23}$$ , we simply do $$\frac{23}{4}=5 R3$$ . A remainder of $$3$$ means the third number in the pattern, which is $$3.$$

We then have a units digit of $$3$$ for $$3^{17}$$ , and a units digit of $$3$$ for $$7^{23}$$.

Thus, we have, $$3*3=\boxed{9}$$

.
May 20, 2018
#3
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Good work Azsun. Melody  May 20, 2018
#4
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I'm the challenger, I post questions every week!

Week 1:

Find the units digit of 3^17*7^23?

$$3^{17} * 7 ^{23} \pmod{10} = \ ?$$

$$\begin{array}{|rclcl|} \hline && 3^{17} * 7 ^{23} \pmod{10} \\ &\equiv& (3^2)^{8}*3^1*(7^2)*7^1 \pmod{10} \\\\ &&&& 3^2 \pmod{10} \\ &&&\equiv& 9 \pmod{10} \\ &&&\equiv& 9-10 \pmod{10} \\ &&&\equiv& {\color{red}-1} \pmod{10} \\\\ &&&& 7^2 \pmod{10} \\ &&&\equiv& 49 \pmod{10} \\ &&&\equiv& 9 \pmod{10} \\ &&&\equiv& 9-10 \pmod{10} \\ &&&\equiv& {\color{red}-1} \pmod{10} \\\\ &\equiv& ({\color{red}-1})^{8}*3^1*({\color{red}-1})*7^1 \pmod{10} \\ &\equiv& 3 *(-7) \pmod{10} \\ &\equiv& -21 \pmod{10} \\ &\equiv& -1 \pmod{10} \\ &\equiv& -1+10 \pmod{10} \\ &\equiv& {\color{red}9} \pmod{10} \\ \hline \end{array}$$ May 22, 2018