<Question>
If $\frac{a}{b}$ is the probability that the reciprocal of a randomly selected positive odd integer less than 2010 gives a terminating decimal, with $a$ and $b$ being relatively prime positive integers, what is $a+b$?
</Question>

<think>
<Plan_1>
Okay, let's try to figure out this problem. So, we need to find the probability that the reciprocal of a randomly selected positive odd integer less than 2010 gives a terminating decimal. Then express that probability as a reduced fraction a/b and find a + b. Hmm, okay. Let me break this down step by step.

First, I remember that a fraction in its simplest form will have a terminating decimal if and only if the denominator (after simplifying the fraction) has no prime factors other than 2 or 5. So, for example, 1/2, 1/4, 1/5, 1/8, 1/10, etc. , all terminate because their denominators are products of 2s and/or 5s. But fractions like 1/3, 1/6, 1/7, etc.

<prompt_1.1> Recall the condition for a fraction to have a terminating decimal expansion. </prompt_1.1>

The problem is about selecting a positive odd integer less than 2010. So first, let's clarify the sample space here. We need to count all positive odd integers less than 2010. Then, among those, how many have reciprocals that are terminating decimals. The probability will be the number of favorable outcomes (odd integers whose reciprocals terminate) divided by the total number of possible outcomes (all odd integers less than 2010).

So, first step: Determine the total number of positive odd integers less than 2010.
Let's compute that.

<prompt_1.2> Calculate the total number of positive odd integers less than 2010. </prompt_1.2>
-----

Now, the next part is to find how many odd integers less than 2010 have denominators with only 2 and 5 when written in simplest form. But wait, since the integers are odd, their only possible prime factors are 5 and others, but since the integer is odd, it can't have 2 as a factor. Therefore, for the reciprocal 1/n to terminate, n must only have 5 as a prime factor because 2 is excluded (since n is odd). So, n must be of the form 5^k, where k is a non-negative integer.

Wait, let me confirm that. The condition for a fraction 1/n to terminate is that in the prime factorization of n, the only primes are 2 and 5. But since n is odd, it can't have 2 as a factor. Therefore, n must be a power of 5. So, n = 5^k for some integer k ≥ 0. Therefore, the numbers we are looking for are the powers of 5 that are odd and less than 2010.

So, let's list the powers of 5:

<prompt_1.3> Identify the powers of 5 that are odd and less than 2010. </prompt_1.3>
</Plan_1>

<Execution_1>
<execution_1.1>
don't. So, the key here is that after simplifying the fraction, the denominator should only have 2s and/or 5s. Since we're dealing with reciprocals of integers, the fraction is already in its simplest form (because the numerator is 1, so gcd(1, n) = 1). Therefore, for 1/n to terminate, n itself must only have prime factors 2 and/or 5. But wait, in this problem, n is a positive odd integer. So, odd integers can't have 2 as a prime factor. Therefore, for the reciprocal 1/n to terminate, n must be a product of only 5s. So, n must be a power of 5.
</execution_1.1>

<execution_1.2>
The positive odd integers start at 1 and go up to 2009 (since 2010 is even). So, how many odd numbers are there between 1 and 2009 inclusive? Well, since every other number is odd, we can calculate this as (2009 - 1)/2 + 1 = 2008/2 + 1 = 1004 + 1 = 1005. Wait, let me check that again. If you have numbers from 1 to n inclusive, the number of odd numbers is (n + 1)/2 if n is odd. Here, 2009 is the last number, which is odd. So, the number of odd numbers is (2009 + 1)/2 = 2010/2 = 1005. Yes, that's correct. So total numbers: 1005.
</execution_1.2>

<execution_1.3>
5^0 = 1

5^1 = 5

5^2 = 25

5^3 = 125

5^4 = 625

5^5 = 3125

Wait, 5^5 is 3125, which is greater than 2010. So the powers of 5 less than 2010 are up to 5^4 = 625. Therefore, the numbers are 1, 5, 25, 125, 625. That's 5 numbers.
</execution_1.3>
</Execution_1>

<Plan_2>
Based on execution_1.2:
So, the total number of possible outcomes is 1005.

Now, the favorable outcomes are the odd integers less than 2010 whose reciprocals are terminating decimals. As we discussed earlier, since the integer is odd, it can't have 2 as a prime factor.

<prompt_2.1> Determine the condition for an odd integer to have a terminating decimal reciprocal. </prompt_2.1>
-----

Based on execution_1.3:
Wait, but hold on. Let me check each of these to make sure they are indeed odd and less than 2010.

<prompt_2.2> Verify the list of powers of 5 that are odd and less than 2010. </prompt_2.2>

Based on execution_1.1, execution_1.3:
But wait, is that all? Let me think again. The original problem says "a randomly selected positive odd integer less than 2010". So, the integers considered are all positive odd integers less than 2010. Then, the favorable ones are those integers which are powers of 5 (since they are odd and their only prime factor is 5). So, yes, the count is 5. Therefore, the probability is 5/1005.

<prompt_2.3> Calculate the probability and reduce the fraction to find a/b. </prompt_2.3>
</Plan_2>

<Execution_2>
<execution_2.1>
Therefore, the denominator (which is the integer n) can only have 5 as a prime factor for the reciprocal to terminate. So, n must be of the form 5^k where k is a non-negative integer. But n also has to be less than 2010. So, we need to find all numbers of the form 5^k that are less than 2010 and are odd (which they all are, since 5 is odd and any power of an odd number is odd).

So, let's list all such numbers. Starting from k=0:

5^0 = 1

5^1 = 5

5^2 = 25

5^3 = 125

5^4 = 625

5^5 = 3125

Wait, 5^5 is 3125, which is greater than 2010, so we can't include that. Therefore, the numbers are 1, 5, 25, 125, 625. Let's check if 625 is less than 2010. Yes, 625 is 625, which is much less than 2010. So, those are the numbers.
</execution_2.1>

<execution_2.2>
1 is 5^0, which is 1. 5^1 is 5, 5^2 is 25, 5^3 is 125, 5^4 is 625. All of these are odd and less than 2010. Then 5^5 is 3125, which is over 2010, so that's excluded. So, there are 5 such numbers.
</execution_2.2>

<execution_2.3>
Simplify that.

First, let's simplify 5/1005. Both numerator and denominator are divisible by 5. 5 ÷ 5 = 1, 1005 ÷ 5 = 201. So, 1/201. Now, check if 1 and 201 are coprime. The gcd of 1 and any number is 1, so yes, 1 and 201 are relatively prime. Therefore, a = 1, b = 201. Therefore, a + b = 1 + 201 = 202.
</execution_2.3>
</Execution_2>

<Plan_3>
Based on execution_2.3:
Wait, but hold on. Let me verify once more. The total number of favorable cases is 5 (the powers of 5: 1, 5, 25, 125, 625).

<prompt_3.1> Verify the probability calculation and ensure the fraction is in simplest terms. </prompt_3.1>
</Plan_3>

<Execution_3>
<execution_3.1>
All of these are less than 2010 and odd. The total number of possible cases is the number of positive odd integers less than 2010, which we calculated as 1005. So 5/1005 reduces to 1/201. Therefore, a = 1, b = 201, so a + b = 202.
</execution_3.1>
</Execution_3>

<Plan_4>
Based on execution_3.1:
But let me double-check the total number of odd integers less than 2010. The first odd integer is 1, then 3, 5, ...., up to 2009.

<prompt_4.1> Recalculate the total number of odd integers less than 2010 to verify the previous result. </prompt_4.1>

But wait, another thought: Is 1 considered a valid odd integer here? Yes, 1 is odd. And 1 is allowed because 1 is a positive odd integer less than 2010. So, 1 is included in both the total count and the favorable outcomes.

Based on execution_3.1, execution_2.2:
But let me check if there are any other numbers that are not powers of 5 but still have reciprocals that terminate. For example, 5 × 2 = 10, but 10 is even, so it's not in our sample space. Since we're only considering odd integers, any number that has 2 as a factor is excluded. Therefore, the only possible prime factor is 5.

<prompt_4.2> Confirm that only powers of 5 are favorable and that the probability is 1/201. </prompt_4.2>
</Plan_4>

<Execution_4>
<execution_4.1>
The number of terms in this sequence can be calculated as follows: The nth term of the sequence of odd numbers is 2n - 1. So, setting 2n - 1 = 2009, we get 2n = 2010, so n = 1005. Therefore, there are 1005 odd numbers less than 2010. That's correct.
</execution_4.1>

<execution_4.2>
Therefore, the only numbers in the sample space (odd integers less than 2010) that have terminating reciprocals are the powers of 5. So, yes, 1, 5, 25, 125, 625. That's 5 numbers.

Wait, but let me check if, for example, 25 is indeed less than 2010. Yes, 25 is much less. Similarly, 625 is 625, which is less than 2010. So, all those are included.
</execution_4.2>
</Execution_4>

<Plan_5>
Based on execution_4.2:
So, 5 numbers. Therefore, 5 favorable outcomes. So, probability is 5/1005 = 1/201. So, reduced fraction is 1/201, so a + b = 202.

<prompt_5.1> Finalize the probability calculation, reduce the fraction, and find a + b. </prompt_5.1>
</Plan_5>

<Execution_5>
<execution_5.1>
That seems correct.

But just to be thorough, let me check if there's any other number that's not a power of 5 but still has a reciprocal that terminates. For example, 5 × 5 × 5 = 125, which is a power of 5. 5^4 is 625, 5^5 is 3125. So, those are all the powers. Then 1 is 5^0. So, that's all.

Alternatively, maybe 1 is considered a special case here. But 1 is 5^0, so it's included in the powers of 5. So, that's okay.

Therefore, the answer is 202.
</execution_5.1>
</Execution_5>

<Plan_6>
Based on execution_5.1:
**Final Answer**
\boxed{202}
</Plan_6>

<Execution_6>
No prompt to execute.
</Execution_6>

<Final_answer>
Given that $a = 1$ and $b = 201$, we find $a + b = 1 + 201 = \boxed{202}$.
</Final_answer>
</think>