# Number Theory: Problem 6

### Solution to (b)

Since all the students get the same number of muffins, and there were $$n + 11$$ students in total, then $$n + 11$$ must divide $$n^2 + 9n - 2$$.

Notice that $$n^2 + 9n - 2 = (n +11) (n -2 ) + 20$$. So $$n + 11$$ divides $$(n +11) (n -2 ) + 20$$. This means that $$(n + 11) \mid 20$$.

Since n is a positive integer and $$(n + 11) \mid 20$$, $$( n + 11) \leq 20$$. So $$n\leq 9$$. Hence there can be no solution when $$n > 11$$.

In the second paragraph, we're using the fact that if $$d \mid (a+b)$$ and $$d\mid a$$, then $$d \mid b$$.
We also rewrote $$n^2 + 9n - 2$$ as $$(n +11) (n -2 ) + 20$$. There are many ways to derive this, e.g. long division of polynomials, attempting to factor it. It's not necessary to explain how you derived this, since the reader can easily verify that it is correct.