AMC 10 · 2009 · #14

Grade 6 algebra
Archetype Geometric Series Share └ Sub-archetype Cyclic Participant Share
sequences-geometricfraction-arithmeticpattern-recognition pattern-recognition ↑ Prerequisites: sequences-geometric
📏 Long solution 💡 3 insights
Problem
Each day, starting Monday, Millie adds one quart of a seed mix that is 1/4 millet and 3/4 other seeds, without removing what is already in the feeder. Before each new day's seeds go in, the birds eat 25% of the millet already there and 100% of the other seeds. Find the first day, measured right after Millie adds the seeds, on which more than half of everything in the feeder is millet.

Pick an answer.

(A)
Tuesday
(B)
Wednesday
(C)
Thursday
(D)
Friday
(E)
Saturday

AMC 10 2009 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Look for a Pattern

The other seeds are the easy half: birds eat all of them every day, so right after adding, there is always exactly 3/4 quart of other seeds. That turns the whole question into one moving quantity — the millet. Tool #4 (Introduce a Variable) names the millet on day n as m_n and captures the daily rule m_n+1=3/4 m_n+1/4. Tool #6 (Guess and Check) runs the rule day by day, and Tool #5 (Look for a Pattern) spots the clean closed form m_n = 1-(3/4)ⁿ, which makes the winning condition transparent.

1STEP 1

Fix the other seeds at 3/4

Yesterday's other seeds are all eaten, so right after the pour only the fresh 3/4 quart remains — a fixed finish line the millet must beat.

other seeds = 3/4 (always) → need millet ＞ 3/4
2STEP 2

Write the daily millet rule

Let m_n be the millet just after day n's pour. Day 1 gives m₁=1/4; each day 3/4 survives and a fresh 1/4 arrives, so m_n+1=3/4 m_n+1/4.

m₁=1/4, m_n+1=3/4 m_n+1/4
3STEP 3

Run it day by day

Run the rule against the 0.75 line: m₂=7/16, m₃=37/64, m₄=175/256≈0.684 still short, m₅=781/1024≈0.763 clears it.

m₁=1/4, m₂=7/16, m₃=37/64, m₄=175/256 ＜ 3/4, m₅=781/1024 ＞ 3/4
4STEP 4

See the pattern behind the numbers

Write each as 1 minus something: 1/4=1-3/4, 7/16=1-9/16, 37/64=1-27/64. So m_n=1-(3/4)ⁿ, and millet ＞ 3/4 becomes (3/4)ⁿ ＜ 1/4.

m_n=1-(3/4)ⁿ → m_n ＞ 3/4 ⇔ (3/4)ⁿ ＜ 1/4
5STEP 5

Find the first winning day

(3/4)⁴≈0.316 still tops 1/4, but (3/4)⁵≈0.237 does not, so n=5; counting Monday as day 1, day 5 is Friday — (D).

(3/4)⁴≈0.316 ＞ 1/4, (3/4)⁵≈0.237 ＜ 1/4 → n=5=Friday → (D)
Answer
Friday
On day 4 the millet is 175/256≈0.684 quart against 0.75 quart of other seeds — still losing, so Thursday is too early, which rules out (C) and everything before it. On day 5 the millet is 781/1024≈0.763 against 0.75, a slim but real win, so Friday is exactly the first day it crosses over. The margin being small and positive is a good sign we did not overshoot to Saturday (E). Everything is consistent with (D).
💡Key takeaway

When one part gets wiped out and refilled to the same size each day, it becomes a fixed target the other part just has to grow past.

  • Fix the other seeds at 3/4
  • Write the daily millet rule
  • Run it day by day
  • See the pattern behind the numbers
  • Find the first winning day

More like this

Same archetype — closest grade level first.