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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
avatar+4569 
+1

As we cycle through the powers of 3 and 7, we see that our answer is \(3*3=\boxed{9}\)

smileysmiley

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 May 20, 2018
 #2
avatar+191 
+3

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, \(3*3=\boxed{9}\)

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 May 20, 2018
 #3
avatar+109520 
+1

Good work Azsun.   laugh

Melody  May 20, 2018
 #4
avatar+24979 
+2

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

Week 1:

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

 

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

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

 

laugh

 May 22, 2018

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