# 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 \(n > 11\).

### Comments

In fact, the only solution is n=9. But the problem didn't ask you to prove that.

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.

### Self check

Did your solution mention muffins and students?
It's important that your answer speak to the original word problem, not just the mathematics
that underlie it.