Hi everyone,

Two comments on the Exam 3 Review sheet:

- In class, I proposed a strategy for solving problem #2, which was to break into cases based on the remainder when n was divided by 4. While I believe this strategy will work, it is simpler to simply look at the cases “n is odd” and “n is even”. Despite my comments in class, this method really will work (can anyone explain why?)
- It looks like the answer for problem #7 in the answer key was all wrong – I’m not sure what happened. In any case, I’ve updated it, so it should now be correct.

Both of these corrections were due to a diligent student who will remain nameless (but, in fact, it was Irania – nice work!).

Let me know if you find anything else,

Prof. Reitz

Yes, Prof. Reitz, for question #2, I found that we did this problem for homework. We used the even and odd two cases. Since 4k and 4k + 2 can be written as 2n as even, and 4k +1 and 4k +3 can be written as 2n + 1 as odd; therefore, it is better to use even and odd two cases to analyze this problem.

Great – thanks Mei!