AMC 10 · 2021 · #12

Grade 12 counting
Archetype Modular Cycle Frequency Count └ Sub-archetype Floor Count Across Residue Bins
binomial-theoremexponentslcmconsecutive-integer-count convert-to-algebraidentify-subproblems ↑ Prerequisites: binomial-theorem
📏 Medium solution 💡 3 insights
Problem
A two-term sum raised to the thousandth power is expanded and collected, giving 1001 terms. Each term's coefficient is built from a binomial coefficient, a power of the cube root of two, and a power of the square root of three. Count how many of those coefficients are rational.

Pick an answer.

(A)
0
(B)
166
(C)
167
(D)
500
(E)
501
How to solve
Strategy Introduce a Variable

Writing out 1001 terms is impossible, so name a term instead of listing them. The Binomial Theorem gives a single formula for all 1001 terms once an index k is introduced for how many of the 1000 factors hand over x∛(2). Rewriting the roots as fractional exponents turns the coefficient into C(1000, k) times 2 raised to k/3 times 3 raised to (1000-k)/2, so "is the coefficient rational" becomes "are those two exponents whole numbers". That is a divisibility condition on k, and counting the k that satisfy it is elementary.

1STEP 1

Name one term with an index

Name one term with an index.

(x∛(2)+y√(3))¹⁰⁰⁰=Σ_k=0¹⁰⁰⁰C(1000, k)(x∛(2))^k(y√(3))¹000-k
2STEP 2

Peel the numbers off the variables

Peel the numeric part off the variables.

C(1000, k)(x∛(2))^k(y√(3))¹000-k=C(1000, k) 2^k/3 3^(1000-k)/2_coefficient x^ky¹000-k
3STEP 3

Both exponents must be whole

Both exponents must be whole.

2^k/3 · 3^(1000-k)/2 is rational⇔ k/3 and (1000-k)/2 are both integers
4STEP 4

Two conditions collapse into one

They collapse into multiples of six.

3 ∣ k and 2 ∣ k ⇔ 6 ∣ k, lcm(2,3)=6
5STEP 5

Count the multiples of six

Counting gives 167.

k ∈ {0, 6, 12, …, 996}, 996/6 + 1 = 166 + 1 = 167
Answer
167
Multiples of 6 occupy about 1/6 of the whole numbers, and 1001/6 is close to 167, so the size is right. The real risk is the fencepost, and the choices are built around it: 166 is exactly what you get if you forget that k = 0 gives a genuine term, namely 3⁵⁰⁰y¹⁰⁰⁰, whose coefficient 3⁵⁰⁰ is rational. Counting 0 through 996 in steps of 6 gives 166 + 1 = 167. The choices 500 and 501 are what you get from the even condition alone, ignoring the cube root, and 0 would require no term at all to be rational, which is already contradicted by the k = 0 term.
💡Key takeaway

When an expansion is far too big to write out, name one term with an index and ask what the index must satisfy: here the coefficient is rational only when the index is a multiple of both 3 and 2, that is, a multiple of 6.

  • Name one term with an index
  • Peel the numbers off the variables
  • Both exponents must be whole
  • Two conditions collapse into one
  • Count the multiples of six

More like this

Same archetype — closest grade level first.