Disclaimer: these are my personal solutions; they have not been reviewed, so be careful relying on them as there might be errors. Some competitions may have official solutions from the contest publishers which you may wish to refer to instead if you’re looking for verified solutions.
Problem 2
A box contains marbles, each of which is red, white or blue. The number of blue marbles is a least half the number of white marbles and at most one-third the number of red marbles. The number which are white or blue is at least $55$. Find the minimum possible number of red marbles.
Hints
Hint #1
Consider all the given constraints; how can you put them to use together?
Solution
Open to see the solution; try solving it first or see the hints above!
Let $r$ represent the number of red marbles, $w$ represent the number of white marbles, and $b$ represent the number of blue marbles.
From the problem statement, we know that:
The number of blue marbles is a least half the number of white marbles and at most one-third the number of red marbles.
thus:
$$ \frac{w}{2} \le b \le \frac{r}{3} $$
We also know that:
The number which are white or blue is at least $55$.
so we also have:
$$ w + b \ge 55 $$
Our goal is to find $x$ such that $r \ge x$.
$$ \begin{align*} \frac{r}{3} & \ge b & \text{above} \\ r& \ge 3b & \\ \end{align*} $$We also have that:
$$ \begin{align*} b & \ge \frac{w}{2} \\ 2b & \ge w \\ 2b + b & \ge w + b \\ 3b & \ge w + b \ge 55 \\ 3b & \ge 55 \\ b & \ge \frac{55}{3} = 18.\bar{3} \\ \end{align*} $$The smallest integer to satisfy $b \ge 18.\bar{3}$ is $b = 19$. We need the smallest $b$ because $r$ is constrained by $b$; since $r \ge 3b = 57$, we can take $r = 57$ as the minimal value of $r$. Let’s see if we can come up with an assignment of $w$ to satisfy the remaining conditions of the problem.
Since $w \le 2b = 38$ and $w + b \ge 55$; taking $b = 19$, we know that $w \ge 36$; thus, $36 \le w \le 38$ satisfy the remaining constraints.
Thus, the minimal value of $r$ is $\boxed{r = 57} \quad \blacksquare$
Verification
What if we consider whether $r = 56$ is a feasible solution to this problem?
In that case, $b \le \dfrac{r}{3} = 18.\bar{6}$, so $b \le 18$. Given that $b \ge \dfrac{w}{2}$, we need to consider taking the largest $b$ since we also need to ensure that $w + b \ge 55$.
If $b = 18$, then we need $w \ge 37$ to satisfy $w + b \ge 55$; however, even the minimal $w$ cannot satisfy $b \ge \dfrac{w}{2}$ since $18 \not\ge \dfrac{37}{2} = 18.5$.
As above, the issue comes down to ensuring that $b \ge \dfrac{55}{3} = 18.\bar{3}$, which means $b \le 18$ cannot be a solution to the problem, and we arrive back to $b \ge 19$ and hence, $\boxed{ r \ge 57 }$.