AMC 10 · 2010 · #20
Grade 6 number-theoryalgebraPick an answer.
Tool #4 (Introduce a Variable): each sequence starts at 1, so naming the two step sizes x and y captures every bit of freedom the problem has, and turns a_n and b_n into 1+(n-1)x and 1+(n-1)y. That form says something sharp: subtract 1 from either nth term and you get a multiple of n-1. Tool #2 (Make a Systematic List): 2010 only splits into a factor pair in a handful of ways, so I can list every split, subtract 1 from both halves, and read off the largest n-1 that could possibly divide both. Tool #14 (Extreme Principle): the question asks for a maximum, so I need two separate things — a ceiling nobody can beat, and one actual pair of sequences that reaches it. Listing gives the ceiling; building the sequences proves the ceiling is reached. Tool #15 (Organize Information in More Ways): the whole problem cracks open by looking at a_n-1 and b_n-1 instead of a_n and b_n. Tool #3 (Eliminate Possibilities): with the seven splits written out, ruling each one out in turn is what makes the ceiling airtight rather than a guess.
Name the two step sizes
Two step sizes describe both sequences.
Two step sizes are the only choices you get to make, so every fact about the sequences has to be squeezed out of x and y.
6.EE.A.2Introduce A VariableSubtract 1 to expose multiples
Dropping by one exposes a shared divisor.
Both sequences pass through 1, so shifting the target numbers down by 1 lines them up with the step sizes.
4.OA.B.4Organize Information In More WaysRule out the 1 times 2010 split
The lopsided split is ruled out at once.
A sequence forced to move at least one unit per step cannot still be sitting at 1 by its nth term.
6.EE.B.5Extreme PrincipleList every factor pair of 2010
Only seven factor pairs need checking.
Four distinct primes make only sixteen divisors, so the full search space is small enough to write out.
4.OA.B.4Make A Systematic ListTake the gcd of each shifted pair
The biggest shared divisor caps the index at 8.
The two shifted halves have to share the step count n-1 as a common factor, so their gcd is a hard ceiling on n-1.
The two shifted halves must both hold the step count as a factor, so their greatest common factor caps it.
▸ Why?
Each number has one prime recipe, so what two numbers can share is settled before anything else.
▸ Why?
Splitting a value into equal whole steps works only when the division leaves nothing over.
Build sequences that reach n = 8
A real pair of sequences reaches that index.
The gcd of 14 and 133 being 7 is precisely the statement that both halves can be split into 7 equal steps.
4.OA.C.5Introduce A VariableCheck that no case was skipped
No case was skipped, so 8 stands, choice (C).
Because the seven factor pairs are the complete list, checking all of them leaves no gap for a bigger n to hide in.
6.NS.B.4Eliminate PossibilitiesBoth sequences start at 1, so subtract 1 from each half of a factor pair of 2010 — the biggest common factor you can find among those shifted pairs is exactly how many steps the sequences are allowed to take.
- Name the two step sizes
- Subtract 1 to expose multiples
- Rule out the 1 times 2010 split
- List every factor pair of 2010
- Take the gcd of each shifted pair
- Build sequences that reach n = 8
- Check that no case was skipped