AMC 10 · 2009 · #9

Grade 6 number-theory
Archetype Small-Domain Constrained Search └ Sub-archetype Factor / Divisibility Constraint
sequences-geometricprime-factorizationperfect-squares convert-to-algebracasework ↑ Prerequisites: sequences-geometric
📏 Medium solution 💡 2 insights
Problem
Three positive integers a, b, and 2009 form a geometric sequence, in that order, with a ＜ b ＜ 2009. The common ratio is a whole number. Find the value of a.

Pick an answer.

(A)
$\ 7$
(B)
$\ 41$
(C)
$\ 49$
(D)
$\ 289$
(E)
$\ 2009$

AMC 10 2009 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Introduce a Variable

The sequence hides one number, the ratio, so Tool #4 (Introduce a Variable) names it r and turns the three terms into a, ar, ar². That collapses the whole problem into one equation, a · r²=2009. From there Tool #7 (Identify Subproblems) splits the work: first factor 2009, then decide which integer ratios can fit. Tool #3 (Eliminate Possibilities) uses the factorization to rule out every ratio except one, and Tool #6 (Guess and Check) plugs that ratio back to confirm the sequence really works.

1STEP 1

Name the ratio

Call the common ratio r. Each term is the previous one times r, so the three terms are a, a·r, a·r².

b=a · r, 2009=b · r=a · r²
2STEP 2

One equation for a and r

The third term is 2009, so a·r²=2009. For a to be a positive integer, r² must divide 2009 exactly.

a · r²=2009 → a=2009/r²
3STEP 3

Factor 2009

2009 has no factor of 2, 3, or 5, but 2009=7×287 and 287=7×41 with 41 prime, so 2009=7²×41.

2009=7×287=7×7×41=7²×41
4STEP 4

Only one ratio fits

A square factor needs primes in pairs, and only the two 7s pair up, so r²=49 is the sole square above 1 and r=7.

r² ∣ 7²×41, r ＞ 1 → r²=49 → r=7
5STEP 5

Find a and check

With r=7, a=2009÷49=41, and 41, 287, 2009 do satisfy 41 ＜ 287 ＜ 2009 — choice (B).

a=2009/49=41, 41, 287, 2009 (41 ＜ 287 ＜ 2009) → (B)
Answer
41
The sequence 41, 287, 2009 multiplies by 7 each step: 41×7=287 and 287×7=2009, and 41 ＜ 287 ＜ 2009 holds. Testing the other choices as a fails: a=7 needs r²=287, a=49 needs r²=41, a=289 needs r²=2009/289, and a=2009 needs r²=1 (which breaks a ＜ 2009) — none give an integer ratio bigger than 1. Only a=41 works, confirming (B).
💡Key takeaway

In a geometric jump-by-r sequence, the last term is a × r², so r² must be a square that divides 2009=7²×41 — only 49 fits, making a=41.

  • Name the ratio
  • One equation for a and r
  • Factor 2009
  • Only one ratio fits
  • Find a and check

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