AMC 10 · 2009 · #22
Grade 10 countingnumber-theoryPick an answer.
The picture has far less freedom than it looks. A is nailed to the origin, B is dragged onto the line y = x, and D is dragged onto a line through the origin of integer slope, so three whole numbers — the slope k and the two x-coordinates s and t — already fix the whole figure. Write the area in those three letters and the geometry collapses to the single equation (k-1)st = 1,000,000, which is a factoring question, not a shape question. The trap is to stop there. That equation is only a necessary condition: it says a legal parallelogram forces a factorization. Counting factorizations answers the question only if the traffic runs both ways, so two extra checks carry the argument. First, build the parallelogram back from an arbitrary factorization and confirm it really lands in the first quadrant with lattice corners and the right area — otherwise some triples would be counted that no parallelogram matches. Second, confirm no single parallelogram is described by two different triples, in particular that the roles of B and D can never be swapped. Only after both checks is the count of ordered triples the count of parallelograms, and that count is then done one prime at a time.
Name the two corners beside A
Two parameters name the corners beside the origin.
A point stuck on a line through the origin costs only one number, so the whole picture rides on three integers.
9.A-CED.A.2Introduce A VariableTurn the area into one equation
The area becomes a single product of three whole numbers.
Area measures how far D has been tilted away from the line AB, and that tilt is exactly the gap between slope k and slope 1.
10.G-GPE.B.7Convert To AlgebraCheck every triple builds a real figure
Every triple builds a real figure.
An equation that must hold is not the same as an equation that can be met, so draw the figure back out of the numbers.
10.G-GPE.B.4Identify SubproblemsCheck no figure is counted twice
No figure is counted twice.
The two corners next to A can never trade places, because only one of them is allowed to sit on the 45° line.
10.G-GPE.B.4Eliminate PossibilitiesSplit the count prime by prime
The count splits prime by prime.
The 2s and the 5s never interfere with each other, so one hard count becomes two easy identical counts.
4.OA.B.4Solve An Easier Related ProblemSplit 6 three ways: 28 ways
Each prime contributes 28 ways.
Once the first two exponents are chosen the third has no freedom left, so the list is just a triangle of pairs.
Once the first two exponents are chosen the third has no freedom left, so the list is a triangle of pairs.
▸ Why?
The three exponents always add to the same fixed total, so naming two names the third.
▸ Why?
The first two are chosen independently within their range, so the count is a product read off the grid.
Multiply the two primes' counts
Multiplying gives 784, choice (D).
Two independent choices multiply, so one count for the 2s times one count for the 5s finishes the job.
7.SP.C.8Organize Information In More WaysWhen a shape is pinned down by a few whole numbers, count the numbers instead of the shapes — but only after checking that every set of numbers really draws a legal shape, and that no shape gets drawn twice.
- Name the two corners beside A
- Turn the area into one equation
- Check every triple builds a real figure
- Check no figure is counted twice
- Split the count prime by prime
- Split 6 three ways: 28 ways
- Multiply the two primes' counts