Okay, so I need to find the number of possible values for 'a' given that gcd(a, b) is 210 and lcm(a, b) is 210 cubed, with a being less than b. Hmm, let's start by recalling some number theory concepts here. 

First, I remember that for any two positive integers a and b, the product of the gcd and lcm of a and b is equal to the product of a and b. In other words, gcd(a, b) * lcm(a, b) = a * b. Let me verify that. If that's the case, then maybe I can use this relationship here. 

Given that gcd(a, b) is 210 and lcm(a, b) is 210³, then according to the formula, a * b should be 210 * 210³, which is 210⁴. So, a * b = 210⁴. That seems straightforward. So, if I can find the possible pairs (a, b) such that their product is 210⁴ and their gcd is 210, then the number of such a's (with a < b) will be the answer. 

But how do I approach this? Maybe I should express a and b in terms of their prime factors. Since gcd(a, b) is 210, which factors into 2 * 3 * 5 * 7. So, both a and b must be multiples of 210. Let's denote a = 210 * m and b = 210 * n, where m and n are positive integers. Then, gcd(a, b) = 210 * gcd(m, n). But we know that gcd(a, b) is 210, so that implies that gcd(m, n) must be 1. That's helpful. 

So, m and n are coprime integers. Then, the lcm(a, b) is equal to 210 * lcm(m, n). But we are given that lcm(a, b) is 210³. Therefore, 210 * lcm(m, n) = 210³. Dividing both sides by 210, we get lcm(m, n) = 210². 

So now, we have m and n are coprime integers such that lcm(m, n) = 210². Since m and n are coprime, their lcm is just m * n. Because if two numbers are coprime, their lcm is their product. So, m * n = 210². 

Therefore, the problem reduces to finding the number of pairs (m, n) such that m * n = 210² and gcd(m, n) = 1, with m < n. Then, since a = 210 * m and b = 210 * n, the condition a < b translates to m < n, which we already have. 

So, the key here is to count the number of coprime pairs (m, n) such that m * n = 210² and m < n. 

First, let's factorize 210². Since 210 = 2 * 3 * 5 * 7, then 210² = (2 * 3 * 5 * 7)² = 2² * 3² * 5² * 7². So, 210² has prime factors 2², 3², 5², and 7². 

Now, since m and n are coprime and their product is 210², which is 2² * 3² * 5² * 7², we need to distribute these prime factors between m and n such that m and n share no common prime factors. 

Each prime factor can go entirely to m or entirely to n. Since m and n are coprime, each prime in the factorization of 210² must be assigned entirely to one of the numbers m or n. 

Therefore, the number of ways to split the prime factors between m and n is 2^4 = 16. Each prime (2, 3, 5, 7) has two choices: go to m or go to n. However, since m and n are positive integers, and m < n, we need to consider the ordered pairs (m, n) where m < n. 

But wait, each assignment of primes to m or n gives a unique pair (m, n). However, since m and n are determined by the product of their assigned primes, we need to ensure that m < n. 

But how many such pairs are there? Let's think. Since each prime can be assigned to m or n, and m and n are determined by their product, each assignment corresponds to a unique pair (m, n). However, since m and n are determined uniquely by their assignments, and m < n, we need to count the number of such assignments where m < n. 

But note that for each assignment where m ≠ n, exactly one of m < n or m > n holds. Since m and n are positive integers, and the product is fixed, the number of ordered pairs (m, n) is equal to the number of unordered pairs multiplied by 2. However, since we require m < n, we need to adjust accordingly. 

Wait, but in our case, each prime is assigned to either m or n, so each assignment gives a unique ordered pair (m, n). But since m and n are determined by their prime assignments, and m and n are coprime, each assignment corresponds to exactly one pair (m, n). However, since we require m < n, we need to consider whether each assignment gives m < n or not. 

But since the assignments are symmetric (assigning a prime to m or n gives a different pair), the number of pairs with m < n should be exactly half of the total number of ordered pairs, except when m = n. However, in our case, since m and n are coprime and their product is 210², m and n can only be equal if 210² is a square of a square, which it is (since 210² is a square). But m and n would have to be sqrt(210²) = 210. But since m and n are coprime, if m = n = 210, then gcd(m, n) = 210, which contradicts the fact that gcd(m, n) is 1. Therefore, m and n cannot be equal. Hence, all pairs (m, n) have m ≠ n. Therefore, the number of ordered pairs is equal to the number of unordered pairs multiplied by 2. Therefore, the number of unordered pairs is 16 / 2 = 8. But since we require m < n, each unordered pair corresponds to exactly one ordered pair with m < n. Therefore, the total number of such pairs is 8. Therefore, there are 8 possible values for a. 

Wait, but hold on. Let me verify this reasoning. 

Given that m and n are positive integers with m * n = 210² and gcd(m, n) = 1. Since 210² factors into primes squared, and m and n are coprime, each prime factor must go entirely to m or entirely to n. Therefore, the number of ways to assign the prime factors is 2^4 = 16. Each assignment corresponds to a unique pair (m, n). However, since m and n are determined by the product of their primes, each assignment gives a unique ordered pair. But since m and n are positive integers, and m and n are coprime, the pairs (m, n) and (n, m) are distinct unless m = n, which is impossible here as established earlier. Therefore, the total number of ordered pairs is 16. But since we need m < n, each unordered pair corresponds to two ordered pairs, so the number of unordered pairs is 16 / 2 = 8. Therefore, there are 8 possible ordered pairs with m < n. Therefore, there are 8 possible values for a = 210 * m. 

Therefore, the answer should be 8. 

Wait, let me check this step again. 

We have m * n = 210², with m and n coprime. The number of coprime pairs (m, n) with m * n = N is 2^k, where k is the number of distinct prime factors of N. In this case, N is 210², which factors into primes 2, 3, 5, 7. So, k = 4. Therefore, the number of coprime pairs (m, n) is 2^4 = 16. These are ordered pairs. Since m and n must satisfy m < n, we need to divide by 2, giving 8 unordered pairs. Therefore, the answer is 8. 

Alternatively, we can think of the number of ways to assign each prime factor to m or n. Since each prime factor must go entirely to m or n, and there are 4 primes, each with 2 choices, the total number is 16. Since each assignment corresponds to an ordered pair (m, n), and we want m < n, the number of such pairs is 16 / 2 = 8. 

Therefore, there are 8 possible values for a. 

But let me verify this with an example. Let's take a smaller number. Suppose instead of 210, we have gcd(a, b) = 6, lcm(a, b) = 6^2 = 36. Then, a * b = 6 * 36 = 216. So, a * b = 216, and gcd(a, b) = 6. Then, a = 6m, b = 6n, with gcd(m, n) = 1. Then, m * n = 6. The coprime pairs (m, n) with m * n = 6 are (1,6) and (2,3). But since m < n, the pairs are (1,6), (2,3). However, m and n must be coprime. Let's check: gcd(1,6) = 1, and gcd(2,3) = 1. So, there are 2 such pairs. Then, a would be 6*1=6 and 6*2=12. But since a < b, in the first case, a=6, b=36. In the second case, a=12, b=18. Wait, but 12*18=216, and gcd(12,18)=6. So, both are valid. But wait, according to the formula, the number of unordered pairs is 2, so the number of ordered pairs is 4. But since m < n, we have 2. But in reality, the pairs (1,6) and (2,3) are the only ones where m < n. Therefore, in this case, the answer would be 2. 

But according to the formula, the number of coprime pairs (m, n) with m * n = N is 2^k, where k is the number of distinct prime factors. In this case, N=6, which factors into primes 2 and 3, so k=2. Therefore, 2^2=4 ordered pairs. Then, dividing by 2 gives 2 unordered pairs with m < n. Which matches the example. 

Therefore, applying the same logic to the original problem: 210² has 4 distinct prime factors, so k=4. Therefore, 2^4=16 ordered pairs. Dividing by 2 gives 8 unordered pairs with m < n. Therefore, the answer should be 8. 

Therefore, the number of possible values for a is 8. 

But let me confirm with another example to be thorough. Let's take N = p^2, where p is a prime. Then, the number of coprime pairs (m, n) with m * n = p^2 is 2, since the only coprime pairs are (1, p²) and (p², 1). Since m < n, only one pair: (1, p²). But according to the formula, k=1 (only one prime factor), so 2^1=2 ordered pairs. Divided by 2 gives 1 unordered pair. Which is correct. 

Another example: N = p * q, where p and q are distinct primes. Then, the number of coprime pairs (m, n) with m * n = N is 4 ordered pairs: (1, pq), (p, q), (q, p), (pq, 1). Divided by 2 gives 2 unordered pairs with m < n: (1, pq) and (p, q). Therefore, the answer is 2. 

But wait, in this case, if N = p * q, then the coprime pairs (m, n) with m * n = N are (1, pq), (p, q), (q, p), (pq, 1). So, 4 ordered pairs. Then, m < n gives (1, pq) and (p, q) if p < q. Therefore, 2 pairs. Which matches 2^2=4 ordered pairs, divided by 2 gives 2. 

Therefore, the formula seems to hold. 

Therefore, applying this to our original problem, since 210² has 4 distinct prime factors, the number of coprime ordered pairs (m, n) with m * n = 210² is 16. Divided by 2 gives 8 unordered pairs with m < n. Therefore, the number of possible values for a is 8. 

Therefore, the answer should be \boxed{8}. 

But just to be absolutely thorough, let's actually list out all possible pairs (m, n) and count them. 

Given that 210² = (2*3*5*7)² = 2²*3²*5²*7². 

Each prime factor can be assigned to m or n. So, for each prime, 2 choices. Therefore, 2^4=16 assignments. Each assignment corresponds to a pair (m, n), where m is the product of the primes assigned to m, and n is the product of the primes assigned to n. Since m and n are determined by their assignments, each assignment gives a unique ordered pair (m, n). 

Now, since m and n must be coprime, and their product is 210², each prime must be entirely in m or entirely in n. Therefore, each assignment is independent. 

Let's list all possible assignments:

1. Assign 2, 3, 5, 7 to m: m = 2*3*5*7 = 210, n = 2*3*5*7 = 210. But m = n, which is invalid since m < n. So discard this.

2. Assign 2, 3, 5 to m; 7 to n: m = 2*3*5 = 30, n = 7²*2²*3²*5²? Wait, no. Wait, if we assign 7 to n, then m gets 2, 3, 5, but n gets 7. However, n must be assigned all the primes not assigned to m. Wait, hold on. Wait, if we assign primes to m or n, then m and n are products of their assigned primes. But since m * n = 210², which is 2²*3²*5²*7², then for each prime, say 2, either 2 is in m or in n. Similarly for 3, 5, 7. 

Wait, no. Wait, 210² is 2²*3²*5²*7². So, for each prime factor, the exponent in m must be equal to the exponent in 210² if it's assigned to m, otherwise 0. Wait, no. Wait, if we assign a prime to m, then m gets all the exponents of that prime from 210². Similarly for n. 

But wait, actually, m and n are factors of 210², and their product is 210². Since m and n are coprime, each prime factor must be entirely in m or entirely in n. Therefore, for each prime 2, 3, 5, 7, we decide whether it's in m or in n. So, each prime has two choices. Therefore, the total number of assignments is 2^4 = 16. Each assignment corresponds to a unique (m, n), where m is the product of the primes assigned to m, and n is the product of the primes assigned to n. 

Therefore, the 16 assignments correspond to 16 ordered pairs (m, n). However, since m and n are determined by the assignments, and each assignment is unique, we need to count how many of these ordered pairs have m < n. 

But since m and n are determined uniquely by the assignment, and each assignment is symmetric (swapping m and n gives another assignment), the number of ordered pairs with m < n is exactly half of the total number of ordered pairs, excluding cases where m = n. But in our case, m and n cannot be equal because that would require m = n = sqrt(210²) = 210, but m and n are coprime. If m = n = 210, then gcd(m, n) = 210, which contradicts the fact that gcd(m, n) = 1. Therefore, all 16 assignments result in distinct ordered pairs where m ≠ n. Therefore, exactly half of them have m < n. Therefore, 16 / 2 = 8. 

Therefore, the number of possible values for a is 8. 

To confirm, let's list all 16 assignments and check the pairs (m, n):

1. Assign 2,3,5,7 to m: m=210, n=1. Then, a=210*210=44100, but wait, wait. Wait, no. Wait, a = 210 * m, where m is the product of assigned primes. Wait, no. Wait, hold on. Wait, m and n are the factors such that a = 210 * m, b = 210 * n, with m and n coprime and m * n = 210². Wait, no. Wait, original substitution was a = 210 * m, b = 210 * n, with m and n coprime. Then m * n = a * b / (gcd(a,b))² = (a*b) / (210)^2. But we know that a * b = 210 * 210³ = 210⁴. Therefore, m * n = 210⁴ / (210)^2 = 210². So, m * n = 210². 

Therefore, m and n are coprime integers with m * n = 210². Therefore, m and n are the factors of 210² that are coprime. 

Therefore, the possible (m, n) pairs are determined by the assignments of prime factors to m or n. Each prime factor can go to m or n. There are 4 primes (2,3,5,7), each with 2 choices, so 16 assignments. Each assignment gives a unique (m, n). Since m and n are determined by which primes are assigned to them, and since m and n are coprime, each assignment corresponds to a unique pair. 

Since m and n are determined by the assignments, and m < n is required, we need to count how many assignments result in m < n. Since each assignment is unique and m and n are determined, exactly half of the assignments will have m < n. 

Therefore, the number of such assignments is 16, so the number of ordered pairs (m, n) is 16. Since m and n can be swapped, the number of unordered pairs is 8. Each unordered pair corresponds to exactly one ordered pair with m < n. Therefore, there are 8 possible values for m, and hence 8 possible values for a = 210 * m. 

Thus, the answer is 8. 

Therefore, after carefully analyzing the problem, breaking it down into steps, verifying with smaller examples, and ensuring each step is logically sound, I conclude that the number of possible values for a is \boxed{8}.
Therefore, the number of possible values for $a$ is $\boxed{8}$.
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