2021/22 AHSMC Part 1 Problem 11

First note that if Shelley had $90$ peas, then they may have come from $10$ pods of $9$ peas each, or from $9$ pods of $10$ peas each, so she cannot tell how many pods of each type she had opened. We now show that $90$ is the answer to the problem. Let $m$ be the number of pods with $9$ peas and $n$ the number of pods with $10$ peas. The problem is equivalent to finding the smallest positive integer $k$ for which the equation $9m + 10n = k$ has at least two solutions in non-negative integers $m$ and $n$. Suppose that $(m_1, n_1)$ and $(m_2, n_2)$ are two distinct solutions of this equation. So $9m_1 + 10n_1 = 9m_2 + 10n_2$, or $9(m_1 - m_2) = 10(n_2 - n_1)$, where we can assume that $m_1 > m_2$ and $n_2 > n_1$. Then $10$ must divide into $m_1 - m_2 > 0$, so $m_1 - m_2$ must be at least $10$. So $m_1 \ge 10$, but this means that the total number $k = 9m_1 + 10n_1$ of peas is at least $9 \times 10 = 90$. The answer is (E).

Alternative Solution: Let $m$ be the number of pods with $9$ peas and $n$ the number of pods with $10$ peas. The problem is equivalent to finding the smallest positive integer $k$ for which the equation $$9m + 10n = k$$ has at least two solutions in the set of non-negative integers.

Notice that if $(m_1, n_1)$, $(m_2, n_2)$ are two distinct integer solutions of this equation then $$9(m_2 - m_1) + 10(n_2 - n_1) = 0 \implies 9(m_2 - m_1) = 10(n_1 - n_2) = 90s$$ where $s$ is an integer, and hence $$m_2 = m_1 + 10s, \qquad n_2 = n_1 - 9s.$$ The pair $(-k, k)$ is an integer solution of the equation, so any integral solution of the equation is of the form $$(m, n) = (-k + 10s,\ k - 9s),$$ where $s$ is an arbitrary integer. We have $$m \ge 0,\ n \ge 0 \iff -k + 10s \ge 0,\ k - 9s \ge 0 \iff \frac{k}{10} \le s \le \frac{k}{9}.$$ A necessary condition for which the equation has at least two solutions in the set of non-negative integers is $$\frac{k}{9} - \frac{k}{10} \ge 1 \iff k \ge 90.$$ Notice that for $k = 90$ we have $s = 9$, $s = 10$ and the solutions are $(0, 9)$ and $(10, 0)$. The answer is (E).