AMC 10 · 2015 · #15

Grade 6 arithmetic
Archetype Small-Domain Constrained Search └ Sub-archetype Answer-Choice Plug-In
ratio-proportionlinear-diophantine casework ↑ Prerequisites: ratio-proportion
📏 Medium solution 💡 3 insights
Problem
Hamlet's animal counts are locked by three rules: there are 3 people per horse, 4 sheep per cow, and 3 ducks per person. Find which of the listed totals for people, horses, sheep, cows, and ducks combined can never happen.

Pick an answer.

(A)
41
(B)
47
(C)
59
(D)
61
(E)
66

AMC 10 2015 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Convert to Algebra

The ratios tie every count back to just two free choices: how many horses and how many cows. Tool #4 (Introduce a Variable) names those two, and Tool #13 (Convert to Algebra) turns the whole word problem into one expression for the total. Once the total has a fixed form, Tool #3 (Eliminate Possibilities) tests each choice to see which one can never be built from whole numbers — exactly what the question asks.

1STEP 1

Name the two free counts

Every count chains back to horses and cows, so let h be the number of horses and c the number of cows — whole numbers you choose freely.

h = horses, c = cows, h, c ≥ 0 integers
2STEP 2

Write every count in h and c

Chain the ratios: people = 3h, ducks = 9h, sheep = 4c, plus horses h and cows c — all five counts now live in h and c.

people=3h, horses=h, ducks=9h, cows=c, sheep=4c
3STEP 3

Add and combine like terms

Sum all five and group: horse terms 3h + h + 9h = 13h, cow terms 4c + c = 5c, so every possible total has the form 13h + 5c.

total = 3h + h + 9h + 4c + c = 13h + 5c
4STEP 4

Test each choice for whole-number blocks

A total works only if 13h + 5c hits it; peel off 13s and check for a multiple of 5 — only 47 fails every time, so (B) can never happen.

47-13h ∈ {47,34,21,8}, none divisible by 5 → (B)
Answer
47
Four of the five totals split cleanly into 13h+5c with small whole numbers, which is what should happen for a reachable count. Only 47 resists every horse-group from 0 to 3 (and 13·4=52 already overshoots), leaving no whole number of cow-groups to finish it. So 47 standing alone as impossible is consistent.
💡Key takeaway

Chase the ratios back to a couple of free counts, collect like terms, and the total has to be built from those fixed-size blocks — if the leftovers never line up, that total can't happen.

  • Name the two free counts
  • Write every count in h and c
  • Add and combine like terms
  • Test each choice for whole-number blocks

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