AMC 10 · 2008 · #8

Grade 6 counting
Archetype Small-Domain Constrained Search └ Sub-archetype Sum-Target Search
linear-diophantineparitydivisibility-rules bound-inequality-then-enumerate ↑ Prerequisites: linear-equations-two-var
📏 Medium solution 💡 2 insights
Problem
A class has exactly 50 dollars to spend on a bouquet for a classmate. Roses cost 3 dollars each, carnations cost 2 dollars each, and no other flowers may be used. The whole 50 dollars must be spent. How many different bouquets of roses and carnations cost exactly 50 dollars?

Pick an answer.

(A)
$\ 1$
(B)
$\ 7$
(C)
$\ 9$
(D)
$\ 16$
(E)
$\ 17$

AMC 10 2008 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Introduce a Variable

Tool #4 (Introduce a Variable) turns the money story into one clean equation, 3r+2c=50, which pins down exactly what must be true. From there the choices are still infinite-looking, so Tool #3 (Eliminate Possibilities) uses a parity fact to throw away every odd number of roses at once. Tool #14 (Extreme Principle) finds the biggest number of roses that still leaves room, capping the range. Then Tool #2 (Make a Systematic List) walks the survivors in order so none are missed and none are double-counted.

1STEP 1

Turn the money into an equation

Let r be the roses and c the carnations; every valid bouquet is a whole-number solution of 3r+2c=50 with r and c at least 0.

3r+2c=50, r ≥ 0, c ≥ 0
2STEP 2

Only an even number of roses works

2c is always even and 50 is even, so 3r must be even; since 3 is odd, that happens only when r is even — every odd rose count is out.

50-2c=3r→ 3r even→ r even
3STEP 3

Find the largest number of roses that fits

Roses alone cannot cost more than 50, so 3r is at most 50 and the largest even r is 16: 3×16=48 fits, but 3×18=54 does not.

3r ≤ 50→ r ≤ 16 (even); 3×16=48, 3×18=54
4STEP 4

List the survivors and count them

Each even r from 0 to 16 gives a whole c=50−3r2c=\frac{50-3r}{2}: 25, 22, 19, 16, 13, 10, 7, 4, 1 — that is 9 bouquets, choice (C).

r=0,2,4,6,8,10,12,14,16→ 9 bouquets→(C)
Answer
9
The even rose counts from 0 to 16 are 0,2,4,6,8,10,12,14,16; that is (16-0)/2+1=9 values, matching the direct list. Each carnation total 25,22,19,16,13,10,7,4,1 steps down by 3 every time r climbs by 2 (because two more roses cost 6, which is three fewer carnations), and all stay whole and non-negative, so no combination was wrongly kept or dropped. The answer 9 is choice (C). It sensibly sits between 1 (far too few) and 17 (which would wrongly count odd rose amounts too).
💡Key takeaway

Turn the spending into one equation, use even-and-odd to knock out half the cases, then list the rest in order and count.

  • Turn the money into an equation
  • Only an even number of roses works
  • Find the largest number of roses that fits
  • List the survivors and count them

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