AMC 10 · 2006 · #24

Grade 8 countingalgebra
Archetype Overlap / Complement Counting └ Sub-archetype Parity Partition: Complement via Odd-Even Split
binomial-theoremparitystars-and-bars easier-related-problemsystematic-enumeration ↑ Prerequisites: binomial-theoremparity
📏 Long solution 💡 3 insights
Problem
Two large powers of a three-variable expression are expanded and added, and like terms are combined. Count how many terms are left standing.

Pick an answer.

(A)
6018
(B)
671,676
(C)
1,007,514
(D)
1,008,016
(E)
2,015,028
How to solve
Strategy Introduce a Variable

Nobody can write out a 2006th power, so stop thinking about the expansion as a list of terms and start thinking about one general term. Name the exponents a, b, c (Tool #4): every possible term of either expansion is x^a y^b z^c with a+b+c=2006, so 'how many terms' becomes 'how many whole-number triples (a,b,c) survive'. First shrink the exponent from 2006 to 2 (Tool #9) to see with your own eyes what the second expansion does to a term — that only tells you what to prove, not that it is true. Then line the two expansions up term against term (Tool #15) to get an exact rule for which coefficients double and which hit 0. The rule turns into a plain parity condition on a, and counting the triples is then an orderly sweep (Tool #2) that ends in a sum of odd numbers with a familiar shape (Tool #5).

1STEP 1

Shrink the exponent to see the mechanism

A tiny exponent shows the mechanism in four terms.

(x+y+z)²+(x-y-z)² = 2x²+2y²+2z²+4yz — 4 terms
2STEP 2

Name a general term with a, b, c

Naming a general term by its three exponents makes the count concrete.

(x+y+z)²⁰⁰⁶=Σ_a+b+c=2006 N(a,b,c) x^a y^b z^c, N(a,b,c)=2006!/a! b! c! ≥ 1
3STEP 3

Compare the two expansions term by term

Comparing the expansions doubles or kills each term.

coefficient in the sum = N(a,b,c)(1+(-1)^b+c) = 2N(a,b,c), & b+c even ; 0, & b+c odd
4STEP 4

Settle exactly which terms survive

So survival depends only on a parity condition.

#terms = #{(a,b,c)inZ_ ≥ 0³ : a+b+c=2006, b+c even}
5STEP 5

Trade the condition on b+c for one on a

That condition transfers onto a single exponent.

a+(b+c)=2006 even ⟹ b+c even ⇔ a even
6STEP 6

Count the triples with a even

Counting the triples leaves a sum of odd numbers.

#terms=Σ_k=0¹⁰⁰³(2007-2k) = 2007+2005+2003+…+3+1
7STEP 7

Add the odd numbers and answer

That sum is a perfect square, giving 1,008,016, choice (D).

1+3+…+2007 = 502 · 2008 = 1004² = 1,008,016 → (D)
Answer
1,008,016
The same method on small exponents can be checked by hand. For exponent 2 it predicts (2/2+1)² = 4 terms, matching the 4 terms found in the first step. For exponent 4 it predicts 3²=9: the even values a=0,2,4 contribute 5+3+1=9 terms, and expanding by hand confirms it. The big count also fits its surroundings. Ignoring all cancellation, degree-2006 monomials in three letters number C(2008, 2)=2,015,028, which is choice (E) — the count before any cancelling, so the true answer must be smaller, and it is. The dead terms are the ones with a odd: 2006+2004+…+2 = 1003 · 1004 = 1,007,012, and 1,008,016+1,007,012 = 2,015,028 exactly, so the survivors and the casualties account for every monomial with nothing left over. Slightly more than half survive, as expected, since the surviving slice includes the largest block (a=0, with 2007 terms). The distractors line up with the usual slips: (E) forgets cancellation, (C) =1,007,514 is half of (E) and comes from assuming exactly half the terms die, and (A) =3 · 2006 counts nothing meaningful. One popular shortcut says '1004 of the 2007 possible values of a are even, so 1004/2007 of the 2,015,028 terms survive', and it does land on 1,008,016 — but not for the reason given, since the terms are spread very unevenly over a (2007 terms at a=0 against 1 at a=2006). It works only because 1+3+…+2007=1004² and 1004²/(1004 · 2007)=1004/2007, so it is a coincidence worth verifying, not a reason to trust.
💡Key takeaway

Flipping the signs of y and z multiplies each term by -1 once per y and once per z, so a term doubles when it holds an even number of them and vanishes when odd; counting the survivors gives 1+3+5+…+2007=1004²=1,008,016 terms.

  • Shrink the exponent to see the mechanism
  • Name a general term with a, b, c
  • Compare the two expansions term by term
  • Settle exactly which terms survive
  • Trade the condition on b+c for one on a
  • Count the triples with a even
  • Add the odd numbers and answer

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