2009 AMC 10B Problems/Problem 14

The following problem is from both the 2009 AMC 10B #14 and 2009 AMC 12B #11, so both problems redirect to this page.

Problem

On Monday, Millie puts a quart of seeds, 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 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?

Solution 1

On Monday, day 1, the birds find quart of millet in the feeder. On Tuesday they find quarts of millet. On Wednesday, day 3, they find quarts of millet. The number of quarts of millet they find on day is The birds always find quart of other seeds, so more than half the seeds are millet if , that is, when . Because and , this will first occur on day which is . The answer is .

Solution 2 (Brute Force)

Notice that, every day, there is always quarts of other seeds in the feeder, so we simply need to calculate when the quantity of millet seeds is greater than quarts. Every day that elapses, a fourth of the millet is taken away (eaten by birds), and then a fourth of a quart is added (by Millie). Thus, if was the quantity of millet seeds the previous day, then on the next day, the quantity is
Monday:

Tuesday:

Wednesday:

Thursday:

Friday:

Brute forcing, we see that it takes days. Thus, the answer is , or . ~ aidan0626, lucaswujc, & Technodoggo
Note: Yes there is a pattern to it, where is the number of days, as shown in Solution 1. This solution is only if you didn't notice the pattern. -aidan0626

Solution 3 (Overkill)

Let denote the the amount of millet that is in the bird feeder after Millie puts the seeds into the feeder on day . Here and corresponds to Monday. Similarly, let denote the amount of other seeds in the feeder after Millie puts the seeds in on day . Notice that and satisfy the following: \begin{align} \begin{cases} a_1=\frac{1}{4}\\ a_n=\frac{3}{4}a_{n-1}+\frac{1}{4}\quad (n>1) \end{cases} \end{align}

\begin{align} b_n=\frac{3}{4} \end{align}

We can solve for using difference equations to get . We want to find the least for which . In other words, . Substituting in what we know about and we get:

\begin{gather} -\left(\frac{3}{4}\right)^n+1>\frac{3}{4}\\

\left(\frac{3}{4}\right)^n<\frac{1}{4}\\

3^n<4^{n-1}\\

n\ge5 \end{gather}

Hence, the answer is , or .

Video Solution

https://www.youtube.com/watch?v=jj3eCwD7Bms

~savannahsolver

See also

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.