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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