### Instruction: If $m$ and $n$ are odd integers, how many terms in the expansion of $(m+n)^6$ are odd?### Response: I know that the expansion of $(m+n)^6$ will have seven terms, each of the form $\binom{6}{k}m^{6-k}n^k$, where $k$ ranges from $0$ to $6$.
I also know that $\binom{6}{k}$ is the number of ways to choose $k$ items from a set of $6$ items, and that this number is even or odd depending on the parity of $k$.
For example, $\binom{6}{0}=\binom{6}{6}=1$, which is odd, $\binom{6}{1}=\binom{6}{5}=6$, which is even, $\binom{6}{2}=\binom{6}{4}=15$, which is odd, and $\binom{6}{3}=20$, which is even.
So, in order to have an odd term in the expansion, I need both $\binom{6}{k}$ and $m^{6-k}n^k$ to be odd.
Since $m$ and $n$ are odd, their powers are also odd, so $m^{6-k}n^k$ is always odd.
Therefore, the only thing that matters is the parity of $\binom{6}{k}$.
As I mentioned before, $\binom{6}{k}$ is odd when $k=0,2,4,6$ and even when $k=1,3,5$.
This means that there are four odd terms in the expansion, corresponding to $k=0,2,4,6$.
The other three terms are even.