AMC 10 · 2025 · #21

Grade 11 algebranumber-theory

Archetype Small-Domain Constrained Search └ Sub-archetype Factor / Divisibility Constraint

sequences-geometricdifference-of-squaresprime-factorizationexponentsparity identify-subproblems ↑ Prerequisites: sequences-geometricprime-factorization

📏 Medium solution 💡 3 insights

Problem

The top and bottom of a big fraction are each a finite geometric-style sum: the exponents run a, a+k, a+2k, …, a+mk, with base 4 on top and base 2 on the bottom. The whole fraction equals 964. Exactly one triple of nonnegative integers (a,k,m) makes this work; find a+k+m.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

How to solve

Strategy Identify Subproblems

A three-unknown equation with sums-of-sums looks hopeless head-on, so Tool #7 (Identify Subproblems) sets the whole route: first crush the giant fraction into something small, then peel the unknowns off one at a time. Tool #5 (Look for a Pattern) powers the crushing step — spotting that the top sum is the bottom sum with 2^k replaced by 2²k makes a difference-of-squares cancellation appear, turning the monster into 2^a·(2^k(m+1)+1)/(2^k+1). Tool #3 (Eliminate Possibilities) then does the number-theory finish: because 964=2²·241 splits cleanly into an even part and an odd prime, matching even-with-even and odd-with-odd forces a, then k, then m with no guessing left over.

1STEP 1

Write both sums in closed form

Each line is a geometric series; the finite-sum formula 1+r+…+r^m= $\frac{r^(m+1)-1}{r-1}$ collapses both into one closed ratio.

2²a·(4^k(m+1)-1)/(4^k-1){2^a·(2^k(m+1)-1)/(2^k-1)} = 2^a·(4^k(m+1)-1)/(4^k-1)·(2^k-1)/(2^k(m+1)-1)

2STEP 2

Collapse with difference of squares

Base 4 is base 2 squared, so each top piece is x²-1=(x-1)(x+1); the minus factors cancel, leaving 2^a· $\frac{2^(k(m+1))+1}{2^k+1}$ =964.

2^a·(2^k(m+1)-1)(2^k(m+1)+1)/(2^k-1)(2^k+1)·(2^k-1)/(2^k(m+1)-1)=2^a·(2^k(m+1)+1)/(2^k+1)=964

3STEP 3

Split off the power of two to fix a

For k≥1 the fraction $\frac{2^(k(m+1))+1}{2^k+1}$ is odd, and 964=2²·241 (241 prime), so all the 2's must come from 2^a: a=2.

2^a=2² → a=2, (2^k(m+1)+1)/(2^k+1)=241

4STEP 4

Peel off k, then m

Clearing gives 2^k(2^(km)-241)=240=2⁴·15; the odd part forces 2^k=16 so k=4, then 2⁴m=256 gives m=2, and a+k+m=2+4+2.

2^k (2^km-241)=240=2⁴ · 15 → 2^k=16, k=4; 2⁴m=256, m=2 → a+k+m=8 (A)

Answer

Plug (a,k,m)=(2,4,2) straight into the original expression, no formulas: the top is 4²+4⁶+4¹⁰=16+4096+1,048,576=1,052,688 and the bottom is 2²+2⁶+2¹⁰=4+64+1024=1092. Dividing gives

\frac{1,052,688}{1092}

=964 exactly, so the triple is confirmed and a+k+m=8. Every alternative letter would need a different triple, but the parity split (964's single odd prime 241, and 240=16 · 15's single odd part) leaves no other integer solution — matching the problem's promise that the triple is unique.

💡Key takeaway

When a base-4 sum sits over a base-2 sum, remember 4 is 2 squared: difference of squares cancels the mess, and then even-matches-even, odd-matches-odd locks in every unknown.

Write both sums in closed form
Collapse with difference of squares
Split off the power of two to fix a
Peel off k, then m

Write both sums in closed form

Collapse with difference of squares

Split off the power of two to fix a

Peel off k, then m

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