Question:
Find the number of solutions of $x_1+x_2+x_3 = 51$ for $x_1,x_2,x_3$ being odd numbers
Not sure how I would even begin this question. It would be simple except for the condition given of being odd numbers. Please help by giving a hint to get started.
The substitution $x_i=2y_i+1$ would help; then you need to find the number of solutions to this equation for nonnegative integers.
If the requirement of $x_1,x_2,$ and $x_3$ is just that they are odd integers, then there are infinitely many solutions (e.g. $x_1=51, x_2=2n+1, x_3=-2n-1$). If, as is more likely, the requirement is that they are positive odd numbers, then there are a finite number of solutions.
There are $25$ possibilities for $x_1$: the odd numbers between $1$ and $49$, inclusive.
Given the value of $x_1$, $x_2$ can be any of the odd numbers between $1$ and $50-x_1$, inclusive. For each of the $25$ choices for $x_1$, the number of choices for $x_2$ is thus $$\frac{51-x_1}{2}$$
Given $x_1$ and $x_2$, $x_3$ is fixed. Thus we want to evaluate the sum
$$\displaystyle\sum_{\substack{x_1 \text{odd} \\ 1 \leq x_1 \leq 49}} \frac{51-x_1}{2}$$
This is equivalent to
\begin{align} \sum_{k=1}^{25} \frac{51-(2k-1)}{2} &=\sum_{k=1}^{25} \frac{52-2k}{2}\\ &=25 \cdot 26 - \sum_{k=1}^{25} k\\\\ &=25(26-13)\\ \\ &=\boxed{325} \end{align}
