AMC 10 · 2009 · #9
Grade 6 number-theoryPositive integers a, b, and 2009, with a<b<2009, form a geometric sequence with an integer ratio. What is a?
Pick an answer.
AMC 10 2009 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: 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 $a$.
Givens: $a$, $b$, $2009$ are positive integers forming a geometric sequence in this order; The common ratio between consecutive terms is an integer; $a<b<2009$; Answer choices: (A) $7$, (B) $41$, (C) $49$, (D) $289$, (E) $2009$
Unknowns: The value of the first term $a$
Understand
Restated: 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 $a$.
Givens: $a$, $b$, $2009$ are positive integers forming a geometric sequence in this order; The common ratio between consecutive terms is an integer; $a<b<2009$; Answer choices: (A) $7$, (B) $41$, (C) $49$, (D) $289$, (E) $2009$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #7 Identify Subproblems, #3 Eliminate Possibilities, #6 Guess and Check
The sequence hides one number, the ratio, so Tool #4 (Introduce a Variable) names it $r$ and turns the three terms into $a$, $ar$, $ar^2$. That collapses the whole problem into one equation, $a\cdot r^2=2009$. 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.
Execute — Answer: B
6.RP.A.1 Step 1 Name the ratio
- Let the common ratio be $r$.
- In a geometric sequence each term is the one before it times $r$.
- So starting from $a$, the next term is $b=a\cdot r$, and the term after that is $2009=b\cdot r$.
- Writing everything from $a$, the three terms are $a$, $a\cdot r$, and $a\cdot r^2$.
💡 A geometric sequence multiplies by the same number each step, so naming that number captures the whole pattern.
6.EE.A.1 Step 2 One equation for a and r
- The third term equals $2009$, so $a\cdot r^2=2009$.
- Solving for $a$ gives $a=\dfrac{2009}{r^2}$.
- Because $a$ has to be a positive integer, $r^2$ must divide $2009$ exactly.
- The whole puzzle is now: which whole-number ratio $r$ makes $2009$ split evenly by $r^2$?
💡 If $a$ must come out a whole number, then $r^2$ is forced to be a factor of $2009$.
4.OA.B.4 Step 3 Factor 2009
- Break $2009$ into primes.
- It is not even and not a multiple of $3$ or $5$.
- Dividing by $7$ gives $2009=7\times287$, and $287=7\times41$, where $41$ is prime.
- So $2009=7\times7\times41=7^2\times41$.
💡 Prime factors show every square that could possibly hide inside $2009$.
6.NS.B.4 Step 4 Only one ratio fits
- We need $r^2$ to be a factor of $2009=7^2\times41$, with $r$ a whole number bigger than $1$.
- A perfect square built from these primes can only use pairs of the same prime.
- The pair $7^2=49$ is available, but $41$ appears just once so it cannot form a square, and $r=2009$ or other choices give $r^2$ far too big.
- The one perfect square that divides $2009$ and is greater than $1$ is $49$, so $r^2=49$ and $r=7$.
💡 A perfect-square factor needs its primes in matched pairs, and only the two $7$s pair up.
6.RP.A.3 Step 5 Find a and check
- With $r=7$, $a=\dfrac{2009}{49}=41$.
- Then $b=a\cdot r=41\times7=287$, and the third term is $287\times7=2009$.
- The three terms $41$, $287$, $2009$ are positive integers with $41<287<2009$, exactly as required.
- So $a=41$, which is (B).
💡 Plugging the ratio back rebuilds the sequence, proving the answer is consistent.
6.RP.A.1 Let the common ratio be $r$. In a geometric sequence each term is the one before 6.EE.A.1 The third term equals $2009$, so $a\cdot r^2=2009$. Solving for $a$ gives $a=\df 4.OA.B.4 Break $2009$ into primes. It is not even and not a multiple of $3$ or $5$. Divid 6.NS.B.4 We need $r^2$ to be a factor of $2009=7^2\times41$, with $r$ a whole number bigg 6.RP.A.3 With $r=7$, $a=\dfrac{2009}{49}=41$. Then $b=a\cdot r=41\times7=287$, and the th Review
Reasonableness: The sequence $41,\ 287,\ 2009$ multiplies by $7$ each step: $41\times7=287$ and $287\times7=2009$, and $41<287<2009$ holds. Testing the other choices as $a$ fails: $a=7$ needs $r^2=287$, $a=49$ needs $r^2=41$, $a=289$ needs $r^2=2009/289$, and $a=2009$ needs $r^2=1$ (which breaks $a<2009$) — none give an integer ratio bigger than $1$. Only $a=41$ works, confirming (B).
Alternative: Skip the algebra and use the choices directly (Tool #3). For each listed $a$, compute $2009\div a$ and ask whether it is a perfect square (that value must equal $r^2$). $2009/41=49=7^2$ is the only clean perfect square among the options, so $a=41$.
CCSS standards used (min grade 6)
6.RP.A.1Understand the concept of a ratio and use ratio language (Reading the geometric sequence as a fixed common ratio $r$ so that $b=ar$ and $2009=ar^2$.)6.EE.A.1Write and evaluate numerical expressions involving whole-number exponents (Writing the third term as $a\cdot r^2$ and concluding $a=2009/r^2$ must be a whole number.)4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Factoring $2009$ into primes as $7^2\times41$ to expose its square factors.)6.NS.B.4Find greatest common factor and least common multiple of two numbers (Using the prime factorization to see that $49$ is the only perfect-square factor of $2009$ greater than $1$, forcing $r=7$.)6.RP.A.3Use ratio and rate reasoning to solve real-world and mathematical problems (Computing $a=2009/49=41$ and rebuilding the sequence $41,287,2009$ to verify the ordering.)
⭐ In a geometric jump-by-$r$ sequence, the last term is $a\times r^2$, so $r^2$ must be a square that divides $2009=7^2\times41$ — only $49$ fits, making $a=41$.
⭐ In a geometric jump-by-$r$ sequence, the last term is $a\times r^2$, so $r^2$ must be a square that divides $2009=7^2\times41$ — only $49$ fits, making $a=41$.
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