2009 AMC 10B Problem 14

P11-15~5 minPrint

Sequences & Series Ratios, Percents & Averages Compute small cases, spot the pattern, generalize Define states/recurrence and iterate

On Monday, Millie puts a quart of seeds, $25\%$ of which are millet, into a bird feeder. On each successive day she adds another quart of the same mix of seeds without removing any seeds that are left. Each day the birds eat only $25\%$ of the millet in the feeder, but they eat all of the other seeds. On which day, just after Millie has placed the seeds, will the birds find that more than half the seeds in the feeder are millet?

Show solution without answering

Key insight

After each refill there is always exactly 3/4 quart of other seeds, since the birds eat them all. Track only the millet until it passes 3/4 quart.

Solution

Split the seeds into millet and "other" seeds, because the birds treat them differently.

1. Each quart of mix has $\frac14$ quart of millet and $\frac34$ quart of other seeds.
2. Other seeds: the birds eat all of them every day. So right after each refill, the feeder holds exactly the $\frac34$ quart that was just added. This never changes.
3. Millet: the birds eat $25\%$ of it, so $\frac34$ of it is left. Then Millie adds $\frac14$ quart more. So each day:
$\text{new millet} = \tfrac34 \times (\text{old millet}) + \tfrac14.$
4. More than half the seeds are millet exactly when the millet is more than the other seeds, that is, more than $\frac34$ quart.
5. Make a table of the millet just after each refill:

Day	Millet (quarts)	More than $\frac34$ ?
Monday	$\frac14$	no
Tuesday	$\frac34 \cdot \frac14 + \frac14 = \frac{7}{16}$	no
Wednesday	$\frac34 \cdot \frac{7}{16} + \frac14 = \frac{37}{64}$	no
Thursday	$\frac34 \cdot \frac{37}{64} + \frac14 = \frac{175}{256}$	no
Friday	$\frac34 \cdot \frac{175}{256} + \frac14 = \frac{781}{1024}$	yes

To compare with $\frac34$ , write $\frac34$ with the same bottom: $\frac34 = \frac{192}{256}$ , which is more than $\frac{175}{256}$ ; and $\frac34 = \frac{768}{1024}$ , which is less than $\frac{781}{1024}$ . So Thursday falls just short and Friday is the first day that works.

The answer is $\boxed{\textbf{(D)}\ \text{Friday}}$ .

Why this works

Splitting the mix into two kinds of seeds shows that one amount stays fixed and only the other one grows. "More than half" is easiest to test as "this part is bigger than the rest," so you never need the total. A short day-by-day table is all it takes.

Alternative approach

Track the millet that is missing from a full quart. On Monday the millet is $\frac14$ , so it is $\frac34$ short of $1$ . Each day the shortfall is multiplied by $\frac34$ (you can check: $1 - \frac{7}{16} = \frac{9}{16} = \frac34 \cdot \frac34$ ). So on day $n$ the shortfall is $\left(\frac34\right)^n$ . We need the millet to be more than $\frac34$ , so the shortfall must be less than $\frac14$ . The powers of $\frac34$ are about $0.75, 0.56, 0.42, 0.32, 0.24$ , and $0.24$ is the first one below $0.25$ . That is day $5$ , Friday.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Measure in quarts. Each fresh quart contains $\frac14$ millet and $\frac34$ other seeds. The birds eat all the other seeds every day, so just after a refill the non-millet amount is always exactly $\frac34$ . Millet, on the other hand, survives at $75\%$ and is topped up by $\frac14$ .

Let $m_n$ be the millet just after the refill on day $n$ (Monday is day $1$ ). Then $m_1 = \frac14$ and $m_{n} = \frac34 m_{n-1} + \frac14$ :
$m_2 = \tfrac{7}{16}, \quad m_3 = \tfrac{37}{64}, \quad m_4 = \tfrac{175}{256}, \quad m_5 = \tfrac{781}{1024}.$

Millet exceeds half the feeder exactly when $m_n > \frac34$ (the amount of other seeds). Now $\frac{175}{256} < \frac{192}{256} = \frac34$ , while $\frac{781}{1024} > \frac{768}{1024} = \frac34$ . So day $5$ is the first such day: Friday.

The answer is $\boxed{\textbf{(D)}\ \text{Friday}}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 21: Sequences & Series (§21.1, §21.2, §21.3, §21.4)
Introduction to Algebra, Chapter 6: Ratios and Percents
Intermediate Counting & Probability, Chapter 10: Recursion
Prealgebra, Chapter 15: Problem-Solving Strategies (§15.1)