<Question>
Positive integers $a$, $b$, and $2009$, with $a<b<2009$, form a geometric sequence with an integer ratio. What is $a$?
</Question>

<think>
<Plan_1>
Okay, so I need to find positive integers a, b, and 2009 such that they form a geometric sequence with an integer ratio, and a < b < 2009. Hmm, let me think through this step by step.

<prompt_1.1> Understand the problem statement and set up the equations based on the properties of a geometric sequence. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
First, let me recall that in a geometric sequence, each term after the first is found by multiplying the previous term by a common ratio, which in this case is an integer. Let's denote the common ratio by r. Since the sequence is a, b, 2009, that means:

b = a * r

and

2009 = b * r = a * r^2

So, 2009 must be equal to a multiplied by r squared. Therefore, a = 2009 / r^2.
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
But since a must be a positive integer, r^2 must be a divisor of 2009. So, my first task is to factorize 2009 to find all possible integer values of r such that r^2 divides 2009.

Let me start by prime factorizing 2009.

<prompt_2.1> Factorize 2009 to find possible integer values of r such that r² divides 2009. </prompt_2.1>
-----

But wait, is there another possible r? For example, if r is 1, but then a = b = 2009, which violates a < b < 2009. So r can't be 1. What if r is negative?

<prompt_2.2> Consider if a negative ratio could work and conclude on possible values of r. </prompt_2.2>
</Plan_2>

<Execution_2>
<execution_2.1>
Let's see, 2009 divided by 7 is 287, because 7*280=1960, and 2009-1960=49, which is 7*7. So 2009 = 7 * 287. Now, 287 divided by 7 is 41, since 7*40=280 and 287-280=7. So 287=7*41. Therefore, 2009=7*7*41, which is 7² * 41. So the prime factorization is 7² * 41¹.

Therefore, the divisors of 2009 are formed by 7^a * 41^b where a=0,1,2 and b=0,1. So the divisors are:

1) 7^0 * 41^0 = 1

2) 7^1 * 41^0 = 7

3) 7^2 * 41^0 = 49

4) 7^0 * 41^1 = 41

5) 7^1 * 41^1 = 287

6) 7^2 * 41^1 = 2009

So the positive divisors of 2009 are 1, 7, 49, 41, 287, 2009.
</execution_2.1>

<execution_2.2>
But the problem states that a, b, and 2009 are positive integers, so the ratio must be positive. Negative ratios would make the terms alternate in sign, but since all terms here are positive, r must be positive. So negative ratios are out of the question.

Therefore, the only possible integer ratios are 7 and 49. But we already saw that r=7 gives a=53, which works, and r=49 gives a=1, which also works. Let's check if both are valid.

Wait, for r=49, a=1, then b=1*49=49, and the third term is 49*49=2401. But 2401 is greater than 2009, which contradicts the given condition that 2009 is the third term. Wait, hold on, that's a problem. Because if r=49, then the third term would be 1*49^2=2401, which is more than 2009. But the problem states that the third term is 2009. Therefore, r=49 is invalid. That means only r=7 is valid.
</execution_2.2>
</Execution_2>

<Plan_3>
Based on execution_2.1:
But since r^2 must divide 2009, we need to check which of these divisors are perfect squares.

Looking at the list:

<prompt_3.1> Identify which divisors of 2009 are perfect squares and determine the possible values of r. </prompt_3.1>
</Plan_3>

<Execution_3>
<execution_3.1>
1 is 1², 49 is 7². The other divisors are 7, 41, 287, 2009, which are not perfect squares. Therefore, the possible r² values are 1 and 49. Hence, r can be 1 or 7. But r must be an integer greater than 1 because a < b < 2009. If r=1, then a = b = 2009, which violates the condition a < b < 2009. Therefore, the only possible value is r=7.
</execution_3.1>
</Execution_3>

<Plan_4>
Based on execution_3.1:
Therefore, a = 2009 / r² = 2009 / 49. Let me compute that.

<prompt_4.1> Calculate the value of a using the determined value of r and verify the sequence. </prompt_4.1>
-----

Based on execution_2.2:
But wait, let's test this. If r=49, then a=1, b=49, and the third term would be 49*49=2401, but the problem says the third term is 2009. Therefore, that's a contradiction.

<prompt_4.2> Re-evaluate the possibility of r=49 and confirm the validity of r=7. </prompt_4.2>

But wait, let me check if I made a mistake in the prime factorization.

<prompt_4.3> Re-verify the prime factorization of 2009 and confirm the possible values of r. </prompt_4.3>

But let me check if there's another way. Suppose that the common ratio is a fraction, but the problem states that the ratio is an integer. So no, the ratio must be integer, so fractions are out.

<prompt_4.4> Consider if there are any other possible values of r and confirm the solution. </prompt_4.4>

But just to be thorough, let's check if there are any other possible ratios. For example, suppose that r is a composite number.

<prompt_4.5> Perform a final verification of the solution and conclude on the value of a. </prompt_4.5>
</Plan_4>

<Execution_4>
<execution_4.1>
49 times 40 is 1960, and 2009 - 1960 is 49, so 49*41 = 2009. Wait, 49*41 is 2009? Let me check: 40*49=1960, plus 1*49=49, so 1960+49=2009. Yes, correct. Therefore, 2009 divided by 49 is 41. So a=41.

Therefore, the terms would be a=41, b=41*7=287, and the third term 287*7=2009. Which matches the given condition. Also, 41 < 287 < 2009, so the inequalities are satisfied.
</execution_4.1>

<execution_4.2>
So r=49 is invalid. Therefore, the only possible ratio is 7, leading to a=53.

Wait, but let me confirm. If r=7, then a=2009 / 49 = 41. Then b = 41 * 7 = 287. So the sequence is 41, 287, 2009. Let's check if 287 * 7 = 2009. 287 * 7: 200 + 80 + 7. 200*7=1400, 80*7=560, 7*7=49. 1400 + 560 = 1960, +49 = 2009. Correct. And 41 < 287 < 2009, which satisfies the conditions.
</execution_4.2>

<execution_4.3>
Wait, 7*7*41 is 49*41=2009. So 2009 is 49*41. Therefore, if r=49, then a=2009 / (49^2) = 2009 / 2401 = 2009 divided by 2401. But 2009 is less than 2401, so that would make a a fraction, which is not allowed. Therefore, r=49 is invalid.

Therefore, the only possible ratio is 7, leading to a=53.
</execution_4.3>

<execution_4.4>
So r must be an integer.

Therefore, the only possible r is 7, leading to a=53.

But wait, let me confirm. Let's compute 2009 divided by 49. 49*40=1960, 2009-1960=49, so 40+1=41. So 49*41=2009. Therefore, a=41, right? Wait, wait, wait. Wait, 2009 divided by 49 is 41. So a=41. Then b=41*7=287. Then the third term is 287*7=2009. Yes, that's correct. So the sequence is 41, 287, 2009. Indeed, 41 < 287 < 2009, and the ratio is 7. So that works. So a=41.

Wait, but earlier I thought a=53. Wait, no, let's check again. If r=7, then a=2009 / r² = 2009 / 49 = 41. So a=41. Then b=41*7=287. Then 287*7=2009. Correct. So a=41.
</execution_4.4>

<execution_4.5>
Wait, but we already considered all the divisors of 2009. Since r² must divide 2009, and the only squares are 1 and 49, then r can only be 1 or 7. So no other possibilities. Therefore, r=7 is the only valid ratio, leading to a=53.

But let's confirm once again.

If a=53, then b=53*7=371, and the next term is 371*7=2597. Wait, but the problem says the third term is 2009. Wait, hold on, that's a problem. 2597 is not 2009. Wait, what's going on here?

Wait, maybe I messed up the sequence. If the three terms are a, b, 2009, then the common ratio is r, so:

a, a*r, a*r² = 2009

Therefore, a*r² = 2009 => a = 2009 / r²

Then, b = a*r = (2009 / r²)*r = 2009 / r

So, in this case, if r=7, then a=2009 / 49=41, and b=2009 /7=287. Wait, wait, wait! Wait, hold on, this is different from what I thought earlier. Let me recast this.

Wait, maybe I made a mistake in the initial setup. Let me clarify:

If the three terms are a, b, 2009, then:

b = a * r

2009 = b * r = a * r²

Therefore, a = 2009 / r²

and b = (2009 / r²) * r = 2009 / r

So, both a and b must be integers. Therefore, r² must divide 2009, and r must divide 2009 as well.

Earlier, I considered that r² divides 2009, which led me to r=7, since 7²=49 divides 2009, and 49*41=2009. Therefore, a=41, and b=287. Then, the sequence is 41, 287, 2009. Let's check the ratios: 287 / 41 = 7, and 2009 / 287 = 7. So that works. Also, 41 < 287 < 2009, which satisfies the condition.

But wait, earlier I thought a=53, but that was a miscalculation. Let me see.

Wait, if r=7, then a=2009 / 49=41, and b=2009 /7=287. So a=41, b=287. Then 41 < 287 < 2009, which is correct. So that's valid.

But earlier, I thought a=53, but that was incorrect. Let me check where I went wrong.
</execution_4.5>
</Execution_4>

<Plan_5>
Based on execution_4.1:
But wait, let me check if there are any other possible ratios. For example, if r=2, then r²=4.

<prompt_5.1> Re-evaluate the possibility of other values of r and confirm the solution. </prompt_5.1>
</Plan_5>

<Execution_5>
<execution_5.1>
But 4 divides 2009? Let's see, 2009 divided by 4 is 502.25, which is not an integer. So r=2 is invalid. Similarly, r=3: r²=9. 2009 divided by 9 is approximately 223.222, not an integer. So invalid. r=4: 16. 2009/16=125.5625, not integer. r=5: 25. 2009/25=80.36, nope. r=6: 36. 2009/36≈55.805, nope. r=7: 49, which works. Next prime factor is 41. r=41: r²=1681. 2009/1681≈1.193, not integer. So invalid. Similarly, r=7*41=287, but 287 squared is way larger than 2009. So no. Therefore, the only possible r is 7.
</execution_5.1>
</Execution_5>

<Plan_6>
Based on execution_5.1:
Wait, but let me check once more. Suppose there is another ratio, say, a composite number. But since 2009's prime factors are only 7 and 41, any composite ratio would have to be a product of these primes. So possible ratios are 7, 49, 41, 7*41=287, etc. But we've already checked these. 49 and 287 don't work because they result in a term larger than 2009.

<prompt_6.1> Finalize the solution and confirm the final answer. </prompt_6.1>
</Plan_6>

<Execution_6>
<execution_6.1>
Wait, for r=41, a=2009 / (41²) = 2009 / 1681 ≈ 1.193, which is not an integer. So that's out. So no other possibilities.

Therefore, the only valid value is r=7, leading to a=41. Therefore, the answer is 41.

Wait, but in my initial calculation, I thought a=53. Wait, how did I get 53? Let me check again. Wait, 2009 divided by 49 is 41, right? 49*40=1960, 49*41=2009. So a=41. Then b=41*7=287. Then 287*7=2009. Correct. So a=41.

But then why did I initially think 53? Maybe a miscalculation. Let me see. Wait, perhaps I confused the terms. Let me check again.

Wait, if a=53, then b=53*r, and 53*r²=2009. So r²=2009/53=38. So r=√38, which is not an integer. So that's invalid. Therefore, 53 is not a valid a. So my initial thought was wrong.

Therefore, the correct answer is a=41.
</execution_6.1>
</Execution_6>

<Plan_7>
Based on execution_6.1:
Let me confirm once again.

a=41, r=7.

<prompt_7.1> Perform a final verification of the solution and conclude on the value of a. </prompt_7.1>
</Plan_7>

<Execution_7>
<execution_7.1>
Then b=41*7=287, and the third term is 287*7=2009. Correct. And 41 < 287 < 2009. All conditions satisfied.

Yes, that seems correct. Therefore, the answer is 41.
</execution_7.1>
</Execution_7>

<Plan_8>
Based on execution_7.1:
**Final Answer**
The value of \( a \) is \boxed{41}.
</Plan_8>

<Execution_8>
No prompt to execute.
</Execution_8>

<Final_answer>
Therefore, $a=\boxed{41}$.
</Final_answer>
</think>