+50 I'm the challenger, I post questions every week!
Week 1:
Find the units digit of 3^17*7^23?
+4622 As we cycle through the powers of 3 and 7, we see that our answer is \(3*3=\boxed{9}\)
+199 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, \(3*3=\boxed{9}\)
+26393 I'm the challenger, I post questions every week!
Week 1:
Find the units digit of 3^17*7^23?
\(3^{17} * 7 ^{23} \pmod{10} = \ ? \)
\(\begin{array}{|rclcl|} \hline && 3^{17} * 7 ^{23} \pmod{10} \\ &\equiv& (3^2)^{8}*3^1*(7^2)*7^1 \pmod{10} \\\\ &&&& 3^2 \pmod{10} \\ &&&\equiv& 9 \pmod{10} \\ &&&\equiv& 9-10 \pmod{10} \\ &&&\equiv& {\color{red}-1} \pmod{10} \\\\ &&&& 7^2 \pmod{10} \\ &&&\equiv& 49 \pmod{10} \\ &&&\equiv& 9 \pmod{10} \\ &&&\equiv& 9-10 \pmod{10} \\ &&&\equiv& {\color{red}-1} \pmod{10} \\\\ &\equiv& ({\color{red}-1})^{8}*3^1*({\color{red}-1})*7^1 \pmod{10} \\ &\equiv& 3 *(-7) \pmod{10} \\ &\equiv& -21 \pmod{10} \\ &\equiv& -1 \pmod{10} \\ &\equiv& -1+10 \pmod{10} \\ &\equiv& {\color{red}9} \pmod{10} \\ \hline \end{array}\)
