Explanation:
Okay, so I need to find how many possible values there are for $ a $ given that $ \gcd(a, b) = 210 $, $ \text{lcm}[a, b] = 210^3 $, and $ a < b $. Both $ a $ and $ b $ are positive integers. Let me start by recalling some number theory concepts.

First, I remember that for any two positive integers $ a $ and $ b $, the product of their greatest common divisor (gcd) and least common multiple (lcm) is equal to the product of the numbers themselves. So, the formula is:

$$
\gcd(a, b) \times \text{lcm}(a, b) = a \times b
$$

Given that $ \gcd(a, b) = 210 $ and $ \text{lcm}(a, b) = 210^3 $, substituting these into the formula gives:

$$
210 \times 210^3 = a \times b
$$
$$
210^4 = a \times b
$$

So, $ a \times b = 210^4 $. Now, since $ \gcd(a, b) = 210 $, I can express $ a $ and $ b $ in terms of their gcd. Let me recall that if $ \gcd(a, b) = d $, then we can write $ a = d \times m $ and $ b = d \times n $, where $ \gcd(m, n) = 1 $.

Applying this here, since $ d = 210 $, we have:

$$
a = 210 \times m \\
b = 210 \times n
$$

with $ \gcd(m, n) = 1 $. Then, substituting back into the product $ a \times b $:

$$
(210 \times m) \times (210 \times n) = 210^2 \times m \times n = 210^4
$$

Dividing both sides by $ 210^2 $:

$$
m \times n = 210^2
$$

So, $ m \times n = 210^2 $, and $ \gcd(m, n) = 1 $. Also, since $ a < b $, substituting the expressions for $ a $ and $ b $:

$$
210 \times m < 210 \times n \implies m < n
$$

Therefore, the problem reduces to finding the number of pairs $ (m, n) $ such that $ m \times n = 210^2 $, $ \gcd(m, n) = 1 $, and $ m < n $. Each such pair corresponds to a unique $ a = 210 \times m $, so the number of such pairs is the number of possible values for $ a $.

So, my task now is to find the number of coprime pairs $ (m, n) $ with $ m < n $ and $ m \times n = 210^2 $.

First, let me factorize 210. Since 210 is 2 × 3 × 5 × 7, so:

$$
210 = 2^1 \times 3^1 \times 5^1 \times 7^1
$$

Therefore, $ 210^2 = (2 \times 3 \times 5 \times 7)^2 = 2^2 \times 3^2 \times 5^2 \times 7^2 $.

So, $ m \times n = 2^2 \times 3^2 \times 5^2 \times 7^2 $, with $ \gcd(m, n) = 1 $, and $ m < n $.

Since $ m $ and $ n $ are coprime, they cannot share any common prime factors. Therefore, each prime factor in $ 210^2 $ must entirely belong to either $ m $ or $ n $, but not both.

So, for each prime factor in the factorization of $ 210^2 $, we have to decide whether to assign it to $ m $ or to $ n $. However, since $ m \times n = 210^2 $, each prime factor's exponent in $ m $ and $ n $ must add up to its exponent in $ 210^2 $. But since $ m $ and $ n $ are coprime, each prime factor can only be in $ m $ or in $ n $, not both. Therefore, for each prime, we have two choices: assign it to $ m $ or to $ n $. However, since $ m $ and $ n $ are positive integers, we can't assign all primes to $ m $ or all to $ n $, because then the other would be 1. But since $ m < n $, we need to ensure that $ m \neq n $, which is already satisfied because if all primes are assigned to $ m $, then $ n = 1 $, and since $ m = 210^2 $, which is greater than 1, so $ m > n $. Similarly, if all primes are assigned to $ n $, then $ m = 1 $, so $ m < n $. But since we need $ m < n $, we need to count the number of assignments where the product $ m $ is less than $ n $.

But since $ m \times n = 210^2 $, and $ m $ and $ n $ are coprime, each such pair is uniquely determined by the set of primes assigned to $ m $. Since the primes are 2, 3, 5, 7, each of which can be assigned to either $ m $ or $ n $, but not both. Therefore, the total number of coprime pairs $ (m, n) $ is $ 2^4 = 16 $, since there are 4 primes and each has 2 choices. However, this counts both $ (m, n) $ and $ (n, m) $, except when $ m = n $. But since $ m \times n = 210^2 $, and $ m $ and $ n $ are coprime, $ m = n $ would imply $ m = n = \sqrt{210^2} = 210 $, but since $ \gcd(m, n) = 1 $, this would require $ \gcd(210, 210) = 210 \neq 1 $, which is a contradiction. Therefore, there are no such pairs where $ m = n $, so all 16 pairs are distinct and come in pairs $ (m, n) $ and $ (n, m) $.

Therefore, the number of unordered pairs is 8, and since we want ordered pairs where $ m < n $, it's half of that, so 8. Wait, but let me check again.

Wait, total number of coprime pairs (including both (m,n) and (n,m)) is 2^4 = 16. But since m and n are positive integers with m*n = 210^2 and gcd(m,n)=1, each such pair is uniquely determined by the subset of primes assigned to m. For example, if we assign primes to m, then n gets the rest. Since the primes are 2,3,5,7, each subset corresponds to a unique m. However, since m and n are different (as we saw earlier), each pair is counted twice in the 16. Therefore, the number of unordered pairs is 16 / 2 = 8. Therefore, the number of ordered pairs with m < n is 8. Therefore, there are 8 possible values for m, hence 8 possible values for a.

Wait, but hold on. Let me think again. Let me take an example. Suppose we have primes p1, p2, p3, p4. Each prime can be assigned to m or n. So, for each prime, 2 choices. So total 2^4 = 16 assignments. Each assignment corresponds to a unique m and n. However, since m and n are determined by the assignment, but if we swap the assignments, we get the same pair (m,n) and (n,m). Therefore, the number of unordered pairs is 16 / 2 = 8. Therefore, the number of ordered pairs where m < n is 8. Therefore, the answer is 8? But let me check with a smaller example to see if this logic holds.

Suppose instead of 210^2, we have a smaller number. Let's say gcd(a,b) = 2, lcm(a,b) = 8. Then, a*b = 2*8 = 16. Then, writing a = 2m, b = 2n, with gcd(m,n)=1. Then m*n = 4. So m and n are coprime, m*n=4. The coprime pairs (m,n) are (1,4) and (4,1). Since m < n, only (1,4). Therefore, only 1 value for a, which is 2*1=2. But according to the formula above, the number of primes in 2^2 is 1 prime (2). So number of assignments is 2^1=2, so unordered pairs 1, so ordered pairs with m < n is 1. Which matches. So that works.

Another example: Let’s take gcd(a,b)=6, lcm(a,b)=6^3. Then, a*b=6*6^3=6^4. Then, writing a=6m, b=6n, gcd(m,n)=1, m*n=6^2=36. So m*n=36, gcd(m,n)=1. The prime factors of 36 are 2^2 * 3^2. So primes are 2 and 3. So number of assignments is 2^2=4. So unordered pairs: 2. Then ordered pairs with m < n: 2. Let me check. The coprime pairs (m,n) with m*n=36 and gcd(m,n)=1. The possible pairs are:

Since m and n are coprime, they can't share any prime factors. So possible factorizations:

1 * 36 (gcd 1)

4 * 9 (gcd 1? gcd(4,9)=1 yes)

But wait, 36 can be written as 1*36, 2*18, 3*12, 4*9, 6*6. But only the pairs where m and n are coprime. So:

1 and 36: gcd(1,36)=1

4 and 9: gcd(4,9)=1

Also, 9 and 4, 36 and 1. So total of 2 unordered pairs. Therefore, ordered pairs with m < n: (1,36) and (4,9). So 2 values. Therefore, the answer would be 2. According to our formula, number of primes in 6^2 is 2 primes (2 and 3), so 2^2=4 assignments, so unordered pairs 2, so ordered pairs with m < n is 2. Which matches. So that works.

Therefore, returning to the original problem. Since 210^2 has 4 distinct primes (2,3,5,7), the number of coprime pairs (m,n) with m*n=210^2 is 2^4=16, so unordered pairs is 8, and ordered pairs with m < n is 8. Therefore, there are 8 possible values for a. Therefore, the answer is 8? Wait, but let me check again.

Wait, but in the example with 6^2, the number of possible m's is 2. Let me check. For the case of 6^2, m can be 1, 4, 9, 36? Wait no. Wait, m and n are such that m*n=36 and gcd(m,n)=1. So possible m's are 1, 4, 9, 36? Wait, no. Wait, if m is 1, then n is 36. If m is 4, then n is 9. If m is 9, then n is 4. If m is 36, then n is 1. But since m < n, we have m=1 and m=4. Therefore, two possible m's. Therefore, two possible a's: 6*1=6 and 6*4=24. So two values. Which matches the count. Therefore, in general, the number of possible m's is equal to the number of subsets of the set of primes, divided by something? Wait, no. Wait, each subset corresponds to a unique m. For example, in the case of 210^2, which has primes 2,3,5,7. Each subset of these primes corresponds to assigning those primes to m, and the rest to n. However, since m and n are determined by the subset, but we need m < n. So how many subsets correspond to m < n?

Since for each subset S of the primes, we can assign the primes in S to m and the rest to n. Then, m is the product of primes in S raised to their exponents in 210^2. Wait, but in our case, the exponents in 210^2 are 2 for each prime. Wait, but in the earlier examples, when we had m*n = 2^2 * 3^2, we had m being 1, 2^2, 3^2, or 2^2 * 3^2. Wait, but in the case of m and n being coprime, each prime can be assigned entirely to m or entirely to n. Therefore, for each prime, we have to assign all of its exponents to m or to n. Therefore, in the case of 210^2, which is 2^2 * 3^2 * 5^2 * 7^2, each prime's exponent is 2. Therefore, if we assign a prime to m, then m gets that prime raised to 2, and n gets nothing. Similarly for others. Therefore, m is the product of some subset of {2^2, 3^2, 5^2, 7^2}, and n is the product of the complementary subset. Therefore, the number of possible m's is 2^4 = 16, but considering that m and n are different, and m < n. However, since m and n are determined by the subsets, and each subset corresponds to a unique m. However, for each subset S, there is a complementary subset S', and if S ≠ S', then (m,n) and (n,m) are two different pairs. However, since m < n, we need to count only those subsets S where the product of the primes in S is less than the product of the primes in S'.

But since the total product is fixed (210^2), m * n = 210^2. Therefore, m < n is equivalent to m < 210^2 / m, which is equivalent to m^2 < 210^2, so m < 210. Therefore, the number of subsets S where the product of the primes in S (each raised to the power 2) is less than 210. Wait, but 210 is the original gcd. Wait, but m is the product of some subset of {2^2, 3^2, 5^2, 7^2}. Let me think.

Wait, m is the product of primes assigned to it. For example, if we assign prime 2 to m, then m gets 2^2. If we assign 2 and 3 to m, then m gets 2^2 * 3^2, etc. So m can be:

- 1 (assigning no primes)
- 2^2
- 3^2
- 5^2
- 7^2
- 2^2 * 3^2
- 2^2 * 5^2
- 2^2 * 7^2
- 3^2 * 5^2
- 3^2 * 7^2
- 5^2 * 7^2
- 2^2 * 3^2 * 5^2
- 2^2 * 3^2 * 7^2
- 2^2 * 5^2 * 7^2
- 3^2 * 5^2 * 7^2
- 2^2 * 3^2 * 5^2 * 7^2

So total 16 possibilities. Now, since m < n, and m * n = 210^2, then m < 210^2 / m => m^2 < 210^4 => m < 210^2. But since m is a factor of 210^2, which is 210^2, so m can be up to 210^2. However, m < n implies m < 210^2 / m => m < sqrt(210^4) = 210^2. Wait, but m is a factor of 210^2, so m can be as big as 210^2. However, m < n implies m < 210^2 / m => m^2 < 210^4 => m < 210^2. But since m is a positive integer, m can be equal to 210^2 only if n = 1, but since m and n are coprime, if m = 210^2, then n = 1, and gcd(210^2, 1) = 1, which is okay. However, in this case, m = 210^2 and n = 1, so m > n. Therefore, in order to have m < n, we need m < 210^2 / m, which is equivalent to m < sqrt(210^4) = 210^2. But since m is a positive integer, m can be from 1 up to 210^2. However, since m and n are coprime, and m * n = 210^2, m must be a divisor of 210^2, and n = 210^2 / m. Also, since m and n are coprime, m must be a product of some subset of the prime factors of 210^2, each raised to their full exponent (since if you take a prime factor, you have to take all of its exponents in m or in n). Therefore, m is a square of a product of some subset of {2, 3, 5, 7}.

Wait, but m is 2^2 * 3^2 * ... etc. So m is actually (product of primes in subset)^2. Therefore, m is a square number. Therefore, m can be 1, 2^2, 3^2, ..., up to (2*3*5*7)^2 = 210^2. Therefore, the possible m's are squares of square-free numbers. Because if you take a subset of primes, their product is square-free, and squaring it gives m. Therefore, m is a square of a square-free number. Therefore, the number of such m's is equal to the number of subsets of {2, 3, 5, 7}, which is 2^4 = 16. However, since m and n are distinct, and we want m < n, we need to count the number of subsets where the square of the subset product is less than the square of the complement subset product. Which is equivalent to the subset product being less than the complement subset product. Since m = (subset product)^2, n = (complement subset product)^2. Therefore, m < n iff subset product < complement subset product.

But since the total product of all primes is 2*3*5*7 = 210. Therefore, subset product * complement subset product = 210. Therefore, subset product * complement subset product = 210. Therefore, subset product < complement subset product iff subset product < sqrt(210). Because subset product * complement subset product = 210, so if subset product < complement subset product, then subset product < sqrt(210). Therefore, the number of subsets where subset product < sqrt(210) is equal to the number of subsets where the product is less than sqrt(210). However, sqrt(210) is approximately 14.49. Therefore, we need to find the number of subsets of {2, 3, 5, 7} whose product is less than 14.49.

Wait, but the subset products are integers. So the subset products can be:

- The empty set: product is 1
- {2}: 2
- {3}: 3
- {5}: 5
- {7}: 7
- {2,3}: 6
- {2,5}: 10
- {2,7}: 14
- {3,5}: 15
- {3,7}: 21
- {5,7}: 35
- {2,3,5}: 30
- {2,3,7}: 42
- {2,5,7}: 70
- {3,5,7}: 105
- {2,3,5,7}: 210

So the subset products are as above. Now, we need to count how many of these are less than sqrt(210) ≈ 14.49. Let's list them:

- 1 (empty set)
- 2
- 3
- 5
- 7
- 6
- 10
- 14
- 15 (which is 15 > 14.49)
- 21, 35, etc. are all larger.

So the subset products less than 14.49 are:

1, 2, 3, 5, 6, 7, 10, 14. That's 8 subsets. Therefore, there are 8 subsets where the product is less than sqrt(210), hence m = (subset product)^2 < (sqrt(210))^2 = 210, so m < 210. Therefore, these correspond to m < n. Therefore, there are 8 such subsets, hence 8 possible values of m, hence 8 possible values of a. Therefore, the answer is 8.

But let me check if that's correct. Let me list all the subset products and see:

Subsets:

1. Empty set: 1
2. {2}: 2
3. {3}: 3
4. {5}: 5
5. {7}: 7
6. {2,3}: 6
7. {2,5}: 10
8. {2,7}: 14
9. {3,5}: 15
10. {3,7}: 21
11. {5,7}: 35
12. {2,3,5}: 30
13. {2,3,7}: 42
14. {2,5,7}: 70
15. {3,5,7}: 105
16. {2,3,5,7}: 210

So the ones less than 14.49 are the first 8 subsets (1-8). Therefore, 8 subsets. Therefore, 8 values of m. Therefore, 8 values of a. Therefore, the answer is 8.

But let me check with the earlier example. For example, in the case where the total product was 6^2 = 36, which had primes 2 and 3. The subsets would be:

Empty set: 1

{2}: 2

{3}: 3

{2,3}: 6

So sqrt(6) ≈ 2.45. So subsets with product less than 2.45 are {empty set} (1), {2} (2). So 2 subsets. Which gives 2 values of m, which matches the earlier example. Therefore, this method works. Therefore, in our original problem, the answer is 8.

But let me check if there's another way to think about this. Since the number of coprime pairs (m, n) with m * n = N and m < n is equal to 2^{k - 1} where k is the number of distinct prime factors of N? Wait, in our case N is 210^2, which has 4 distinct prime factors. Then 2^{4 - 1} = 8. Which matches. Wait, but in the previous example with N = 36 = 2^2 * 3^2, which has 2 distinct prime factors, then 2^{2 - 1} = 2, which matches. So maybe in general, if N is a square of a square-free number with k distinct prime factors, then the number of coprime pairs (m, n) with m < n and m * n = N is 2^{k - 1}.

But why? Because for each prime factor, you assign it to m or n, but since m and n are coprime, each prime is assigned to exactly one. So total assignments is 2^k. Then, since m and n are distinct (as N is not a square? Wait, N is a square here. Wait, N = 210^2, which is a square. However, m and n are coprime and m * n = N. If N is a square, then m and n must both be squares. Because if m * n is a square and m and n are coprime, then m and n must each be squares. Therefore, m = s^2, n = t^2, with s * t = 210. Therefore, the number of coprime pairs (m, n) with m * n = N is equal to the number of coprime pairs (s, t) with s * t = 210.

Wait, this seems like another way to think about it. Let me check. If m * n = 210^2 and gcd(m, n) = 1, then m and n must be squares. Let me see. Suppose m and n are coprime and m * n is a square. Then, since m and n are coprime, each of them must be a square. Because in the prime factorization of m * n, each prime appears with even exponent. Since m and n are coprime, each prime is entirely in m or entirely in n. Therefore, the exponent of each prime in m must be even, and same for n. Therefore, m and n are squares. Therefore, m = s^2, n = t^2, with s * t = 210. Therefore, the number of coprime pairs (m, n) with m * n = 210^2 is equal to the number of coprime pairs (s, t) with s * t = 210.

But since s * t = 210, and gcd(s, t) = 1, then s and t are coprime divisors of 210. Therefore, the number of such pairs (s, t) is equal to 2^{k}, where k is the number of distinct prime factors of 210. Since 210 has 4 distinct primes, the number of coprime pairs (s, t) with s * t = 210 is 2^4 = 16. But since (s, t) and (t, s) are considered different unless s = t. However, since 210 is not a square, s ≠ t. Therefore, the number of unordered pairs is 8, and the number of ordered pairs is 16. Therefore, the number of ordered pairs with s < t is 8. Therefore, since m = s^2 and n = t^2, the number of ordered pairs (m, n) with m < n is also 8. Therefore, the number of possible a's is 8.

Therefore, this confirms the previous result. Therefore, the answer is 8.

But let me check with the example where N = 36 = 6^2. Then, s * t = 6. The number of coprime pairs (s, t) with s * t = 6. The prime factors of 6 are 2 and 3. So the number of coprime pairs (s, t) is 2^2 = 4. Which are:

(1,6), (6,1), (2,3), (3,2). Therefore, the number of ordered pairs with s < t is 2: (1,6) and (2,3). Therefore, the number of m's is 2, which matches. Therefore, this method works.

Therefore, in general, if we have N = d^2, where d is square-free, then the number of coprime pairs (m, n) with m * n = N and m < n is 2^{k - 1}, where k is the number of distinct prime factors of d. In our case, d = 210, which has 4 distinct prime factors, so 2^{4 - 1} = 8. Therefore, the answer is 8.

Therefore, I think the answer is 8. However, let me check once again with the original problem.

Original problem: gcd(a, b) = 210, lcm(a, b) = 210^3. Then, a * b = 210^4. Then, writing a = 210 * m, b = 210 * n, gcd(m, n) = 1, m * n = 210^2. Then, m and n are coprime, m * n = 210^2. Then, since m and n are coprime, they must be squares, as discussed. Therefore, m = s^2, n = t^2, s * t = 210. Therefore, the number of possible m's is equal to the number of s's such that s divides 210, and t = 210 / s, and gcd(s, t) = 1. Which is equivalent to s being a divisor of 210 with gcd(s, 210 / s) = 1. Which is equivalent to s being a divisor of 210 that is square-free. Because if s is square-free, then since 210 is square-free, t = 210 / s is also square-free, and since s and t are divisors of 210, their gcd is 1 if and only if s and t are coprime. Which is equivalent to s being a product of some subset of the prime factors of 210. Therefore, the number of such s is 2^4 = 16. However, since we want s < t, which is equivalent to s < 210 / s => s^2 < 210 => s < sqrt(210) ≈ 14.49. Therefore, the number of such s is equal to the number of square-free divisors of 210 less than sqrt(210).

The square-free divisors of 210 are all the products of subsets of {2, 3, 5, 7}. So they are:

1, 2, 3, 5, 7, 2*3=6, 2*5=10, 2*7=14, 3*5=15, 3*7=21, 5*7=35, 2*3*5=30, 2*3*7=42, 2*5*7=70, 3*5*7=105, 2*3*5*7=210.

Now, which of these are less than sqrt(210) ≈ 14.49?

1, 2, 3, 5, 7, 6, 10, 14. These are 8 numbers. The rest are 15, 21, 35, etc., which are greater than 14.49. Therefore, there are 8 such s, hence 8 values of m, hence 8 values of a. Therefore, the answer is 8.

Therefore, I think this confirms again that the answer is 8.

But just to be thorough, let me list all possible a's. Since a = 210 * m, and m is s^2 where s is one of the 8 numbers above. Wait, no. Wait, m = s^2, where s is a square-free divisor of 210. Wait, no. Wait, earlier we had m = s^2, where s * t = 210. Wait, no. Wait, m = s^2, n = t^2, with s * t = 210. Therefore, m = s^2, so s is a divisor of 210, and m is s^2. Therefore, the possible m's are:

For each s in {1, 2, 3, 5, 7, 6, 10, 14}, m = s^2:

1^2 = 1

2^2 = 4

3^2 = 9

5^2 = 25

7^2 = 49

6^2 = 36

10^2 = 100

14^2 = 196

Therefore, the possible m's are 1, 4, 9, 25, 49, 36, 100, 196. Therefore, 8 values. Therefore, the corresponding a's are 210 * m:

210*1 = 210

210*4 = 840

210*9 = 1890

210*25 = 5250

210*49 = 10290

210*36 = 7560

210*100 = 21000

210*196 = 41160

Therefore, these are 8 distinct values of a. Now, we need to check that for each of these a's, there exists a b such that gcd(a, b) = 210 and lcm(a, b) = 210^3.

For example, take a = 210. Then, since a = 210 * m, m = 1, so n = 210^2 / m = 210^2. Therefore, b = 210 * n = 210 * 210^2 = 210^3. Then, gcd(a, b) = gcd(210, 210^3) = 210, and lcm(a, b) = lcm(210, 210^3) = 210^3. So that works.

Take a = 840. Then m = 4, so n = 210^2 / 4 = (210^2)/4. Since 210 is divisible by 2, 210 = 2 * 105, so 210^2 = 4 * 105^2. Therefore, n = 4 * 105^2 / 4 = 105^2. Therefore, b = 210 * n = 210 * 105^2. Let me check gcd(a, b). a = 210 * 4, b = 210 * 105^2. The gcd is 210 * gcd(4, 105^2). Since 105^2 is 3^2 * 5^2 * 7^2, which is coprime with 4. Therefore, gcd(4, 105^2) = 1. Therefore, gcd(a, b) = 210 * 1 = 210. Similarly, lcm(a, b) = 210 * lcm(4, 105^2) = 210 * 4 * 105^2 = 210 * 4 * 105^2. But 210 = 2 * 105, so 210 * 4 * 105^2 = 2 * 105 * 4 * 105^2 = 8 * 105^3. However, 210^3 = (2 * 3 * 5 * 7)^3 = 2^3 * 3^3 * 5^3 * 7^3. On the other hand, 8 * 105^3 = 8 * (3 * 5 * 7)^3 = 2^3 * 3^3 * 5^3 * 7^3, which is equal to 210^3. Therefore, that works.

Similarly, for other values of a, the same logic applies. Therefore, all 8 values of a satisfy the given conditions. Therefore, the answer is 8.

Therefore, after all these checks and reasoning, I believe the number of possible values for a is 8.

**Final Answer**
\boxed{8}

To solve the problem, we are given:

- $ \gcd(a, b) = 210 $
- $ \text{lcm}(a, b) = 210^3 $
- $ a < b $
- $ a $ and $ b $ are positive integers

We are to find how many **possible values** of $ a $ satisfy these conditions.

---

### Step 1: Use the GCD-LCM Identity

We use the identity:

$$
\gcd(a, b) \cdot \text{lcm}(a, b) = a \cdot b
$$

Substituting the given values:

$$
210 \cdot 210^3 = a \cdot b \Rightarrow a \cdot b = 210^4
$$

---

### Step 2: Express $ a $ and $ b $ in Terms of Their GCD

Let $ d = \gcd(a, b) = 210 $. Then we can write:

$$
a = d \cdot m = 210m, \quad b = d \cdot n = 210n
$$

Since $ \gcd(a, b) = d $, it follows that $ \gcd(m, n) = 1 $, and:

$$
a \cdot b = (210m)(210n) = 210^2 \cdot m \cdot n = 210^4 \Rightarrow m \cdot n = 210^2
$$

So now the problem reduces to finding the number of **coprime pairs** $ (m, n) $ such that:

- $ m \cdot n = 210^2 $
- $ \gcd(m, n) = 1 $
- $ m < n $

Each such pair corresponds to a unique value of $ a = 210m $, and we are to count how many such $ a $ exist.

---

### Step 3: Prime Factorization of $ 210 $

We factorize $ 210 $:

$$
210 = 2 \cdot 3 \cdot 5 \cdot 7
\Rightarrow 210^2 = (2 \cdot 3 \cdot 5 \cdot 7)^2 = 2^2 \cdot 3^2 \cdot 5^2 \cdot 7^2
$$

So $ m \cdot n = 2^2 \cdot 3^2 \cdot 5^2 \cdot 7^2 $, and since $ \gcd(m, n) = 1 $, each prime factor must be assigned entirely to either $ m $ or $ n $, but not both.

Thus, for each of the 4 distinct primes (2, 3, 5, 7), we have 2 choices: assign it to $ m $ or to $ n $. This gives:

$$
2^4 = 16 \text{ total assignments}
$$

Each assignment corresponds to a unique pair $ (m, n) $, and since $ m \cdot n = 210^2 $, each such pair is unique.

However, since $ m < n $, we only want the **ordered pairs** where $ m < n $. Since the total number of unordered pairs is $ 16 / 2 = 8 $, and each unordered pair corresponds to exactly one ordered pair with $ m < n $, we conclude:

$$
\text{Number of valid } (m, n) \text{ pairs with } m < n = 8
$$

---

### Step 4: Count the Number of Valid $ a $

Each such pair $ (m, n) $ with $ m < n $ gives a unique value of $ a = 210m $. Since there are 8 such pairs, there are **8 possible values** of $ a $.

---

### Final Answer

$$
\boxed{8}
$$

[Assistant]
Explanation:
Okay, so I need to find how many possible values there are for $ a $ given that $ \gcd(a, b) = 210 $, $ \text{lcm}[a, b] = 210^3 $, and $ a < b $. Both $ a $ and $ b $ are positive integers. Let me start by recalling some number theory concepts.

First, I remember that for any two positive integers $ a $ and $ b $, the product of their greatest common divisor (gcd) and least common multiple (lcm) is equal to the product of the numbers themselves. So, the formula is:

$$
\gcd(a, b) \times \text{lcm}(a, b) = a \times b
$$

Given that $ \gcd(a, b) = 210 $ and $ \text{lcm}(a, b) = 210^3 $, substituting these into the formula gives:

$$
210 \times 210^3 = a \times b
$$
$$
210^4 = a \times b
$$

So, $ a \times b = 210^4 $. Now, since $ \gcd(a, b) = 210 $, I can express $ a $ and $ b $ in terms of their gcd. Let me recall that if $ \gcd(a, b) = d $, then we can write $ a = d \times m $ and $ b = d \times n $, where $ \gcd(m, n) = 1 $.

Applying this here, since $ d = 210 $, we have:

$$
a = 210 \times m \\
b = 210 \times n
$$

with $ \gcd(m, n) = 1 $. Then, substituting back into the product $ a \times b $:

$$
(210 \times m) \times (210 \times n) = 210^2 \times m \times n = 210^4
$$

Dividing both sides by $ 210^2 $:

$$
m \times n = 210^2
$$

So, $ m \times n = 210^2 $, and $ \gcd(m, n) = 1 $. Also, since $ a < b $, substituting the expressions for $ a $ and $ b $:

$$
210 \times m < 210 \times n \implies m < n
$$

Therefore, the problem reduces to finding the number of pairs $ (m, n) $ such that $ m \times n = 210^2 $, $ \gcd(m, n) = 1 $, and $ m < n $. Each such pair corresponds to a unique $ a = 210 \times m $, so the number of such pairs is the number of possible values for $ a $.

So, my task now is to find the number of coprime pairs $ (m, n) $ with $ m < n $ and $ m \times n = 210^2 $.

First, let me factorize 210. Since 210 is 2 × 3 × 5 × 7, so:

$$
210 = 2^1 \times 3^1 \times 5^1 \times 7^1
$$

Therefore, $ 210^2 = (2 \times 3 \times 5 \times 7)^2 = 2^2 \times 3^2 \times 5^2 \times 7^2 $.

So, $ m \times n = 2^2 \times 3^2 \times 5^2 \times 7^2 $, with $ \gcd(m, n) = 1 $, and $ m < n $.

Since $ m $ and $ n $ are coprime, they cannot share any common prime factors. Therefore, each prime factor in $ 210^2 $ must entirely belong to either $ m $ or $ n $, but not both.

So, for each prime factor in the factorization of $ 210^2 $, we have to decide whether to assign it to $ m $ or to $ n $. However, since $ m \times n = 210^2 $, each prime factor's exponent in $ m $ and $ n $ must add up to its exponent in $ 210^2 $. But since $ m $ and $ n $ are coprime, each prime factor can only be in $ m $ or in $ n $, not both. Therefore, for each prime, we have two choices: assign it to $ m $ or to $ n $. However, since $ m $ and $ n $ are positive integers, we can't assign all primes to $ m $ or all to $ n $, because then the other would be 1. But since $ m < n $, we need to ensure that $ m \neq n $, which is already satisfied because if all primes are assigned to $ m $, then $ n = 1 $, and since $ m = 210^2 $, which is greater than 1, so $ m > n $. Similarly, if all primes are assigned to $ n $, then $ m = 1 $, so $ m < n $. But since we need $ m < n $, we need to count the number of assignments where the product $ m $ is less than $ n $.

But since $ m \times n = 210^2 $, and $ m $ and $ n $ are coprime, each such pair is uniquely determined by the set of primes assigned to $ m $. Since the primes are 2, 3, 5, 7, each of which can be assigned to either $ m $ or $ n $, but not both. Therefore, the total number of coprime pairs $ (m, n) $ is $ 2^4 = 16 $, since there are 4 primes and each has 2 choices. However, this counts both $ (m, n) $ and $ (n, m) $, except when $ m = n $. But since $ m \times n = 210^2 $, and $ m $ and $ n $ are coprime, $ m = n $ would imply $ m = n = \sqrt{210^2} = 210 $, but since $ \gcd(m, n) = 1 $, this would require $ \gcd(210, 210) = 210 \neq 1 $, which is a contradiction. Therefore, there are no such pairs where $ m = n $, so all 16 pairs are distinct and come in pairs $ (m, n) $ and $ (n, m) $.

Therefore, the number of unordered pairs is 8, and since we want ordered pairs where $ m < n $, it's half of that, so 8. Wait, but let me check again.

Wait, total number of coprime pairs (including both (m,n) and (n,m)) is 2^4 = 16. But since m and n are positive integers with m*n = 210^2 and gcd(m,n)=1, each such pair is uniquely determined by the subset of primes assigned to m. For example, if we assign primes to m, then n gets the rest. Since the primes are 2,3,5,7, each subset corresponds to a unique m. However, since m and n are different (as we saw earlier), each pair is counted twice in the 16. Therefore, the number of unordered pairs is 16 / 2 = 8. Therefore, the number of ordered pairs with m < n is 8. Therefore, there are 8 possible values for m, hence 8 possible values for a.

Wait, but hold on. Let me think again. Let me take an example. Suppose we have primes p1, p2, p3, p4. Each prime can be assigned to m or n. So, for each prime, 2 choices. So total 2^4 = 16 assignments. Each assignment corresponds to a unique m and n. However, since m and n are determined by the assignment, but if we swap the assignments, we get the same pair (m,n) and (n,m). Therefore, the number of unordered pairs is 16 / 2 = 8. Therefore, the number of ordered pairs where m < n is 8. Therefore, the answer is 8? But let me check with a smaller example to see if this logic holds.

Suppose instead of 210^2, we have a smaller number. Let's say gcd(a,b) = 2, lcm(a,b) = 8. Then, a*b = 2*8 = 16. Then, writing a = 2m, b = 2n, with gcd(m,n)=1. Then m*n = 4. So m and n are coprime, m*n=4. The coprime pairs (m,n) are (1,4) and (4,1). Since m < n, only (1,4). Therefore, only 1 value for a, which is 2*1=2. But according to the formula above, the number of primes in 2^2 is 1 prime (2). So number of assignments is 2^1=2, so unordered pairs 1, so ordered pairs with m < n is 1. Which matches. So that works.

Another example: Let’s take gcd(a,b)=6, lcm(a,b)=6^3. Then, a*b=6*6^3=6^4. Then, writing a=6m, b=6n, gcd(m,n)=1, m*n=6^2=36. So m*n=36, gcd(m,n)=1. The prime factors of 36 are 2^2 * 3^2. So primes are 2 and 3. So number of assignments is 2^2=4. So unordered pairs: 2. Then ordered pairs with m < n: 2. Let me check. The coprime pairs (m,n) with m*n=36 and gcd(m,n)=1. The possible pairs are:

Since m and n are coprime, they can't share any prime factors. So possible factorizations:

1 * 36 (gcd 1)

4 * 9 (gcd 1? gcd(4,9)=1 yes)

But wait, 36 can be written as 1*36, 2*18, 3*12, 4*9, 6*6. But only the pairs where m and n are coprime. So:

1 and 36: gcd(1,36)=1

4 and 9: gcd(4,9)=1

Also, 9 and 4, 36 and 1. So total of 2 unordered pairs. Therefore, ordered pairs with m < n: (1,36) and (4,9). So 2 values. Therefore, the answer would be 2. According to our formula, number of primes in 6^2 is 2 primes (2 and 3), so 2^2=4 assignments, so unordered pairs 2, so ordered pairs with m < n is 2. Which matches. So that works.

Therefore, returning to the original problem. Since 210^2 has 4 distinct primes (2,3,5,7), the number of coprime pairs (m,n) with m*n=210^2 is 2^4=16, so unordered pairs is 8, and ordered pairs with m < n is 8. Therefore, there are 8 possible values for a. Therefore, the answer is 8? Wait, but let me check again.

Wait, but in the example with 6^2, the number of possible m's is 2. Let me check. For the case of 6^2, m can be 1, 4, 9, 36? Wait no. Wait, m and n are such that m*n=36 and gcd(m,n)=1. So possible m's are 1, 4, 9, 36? Wait, no. Wait, if m is 1, then n is 36. If m is 4, then n is 9. If m is 9, then n is 4. If m is 36, then n is 1. But since m < n, we have m=1 and m=4. Therefore, two possible m's. Therefore, two possible a's: 6*1=6 and 6*4=24. So two values. Which matches the count. Therefore, in general, the number of possible m's is equal to the number of subsets of the set of primes, divided by something? Wait, no. Wait, each subset corresponds to a unique m. For example, in the case of 210^2, which has primes 2,3,5,7. Each subset of these primes corresponds to assigning those primes to m, and the rest to n. However, since m and n are determined by the subset, but we need m < n. So how many subsets correspond to m < n?

Since for each subset S of the primes, we can assign the primes in S to m and the rest to n. Then, m is the product of primes in S raised to their exponents in 210^2. Wait, but in our case, the exponents in 210^2 are 2 for each prime. Wait, but in the earlier examples, when we had m*n = 2^2 * 3^2, we had m being 1, 2^2, 3^2, or 2^2 * 3^2. Wait, but in the case of m and n being coprime, each prime can be assigned entirely to m or entirely to n. Therefore, for each prime, we have to assign all of its exponents to m or to n. Therefore, in the case of 210^2, which is 2^2 * 3^2 * 5^2 * 7^2, each prime's exponent is 2. Therefore, if we assign a prime to m, then m gets that prime raised to 2, and n gets nothing. Similarly for others. Therefore, m is the product of some subset of {2^2, 3^2, 5^2, 7^2}, and n is the product of the complementary subset. Therefore, the number of possible m's is 2^4 = 16, but considering that m and n are different, and m < n. However, since m and n are determined by the subsets, and each subset corresponds to a unique m. However, for each subset S, there is a complementary subset S', and if S ≠ S', then (m,n) and (n,m) are two different pairs. However, since m < n, we need to count only those subsets S where the product of the primes in S is less than the product of the primes in S'.

But since the total product is fixed (210^2), m * n = 210^2. Therefore, m < n is equivalent to m < 210^2 / m, which is equivalent to m^2 < 210^2, so m < 210. Therefore, the number of subsets S where the product of the primes in S (each raised to the power 2) is less than 210. Wait, but 210 is the original gcd. Wait, but m is the product of some subset of {2^2, 3^2, 5^2, 7^2}. Let me think.

Wait, m is the product of primes assigned to it. For example, if we assign prime 2 to m, then m gets 2^2. If we assign 2 and 3 to m, then m gets 2^2 * 3^2, etc. So m can be:

- 1 (assigning no primes)
- 2^2
- 3^2
- 5^2
- 7^2
- 2^2 * 3^2
- 2^2 * 5^2
- 2^2 * 7^2
- 3^2 * 5^2
- 3^2 * 7^2
- 5^2 * 7^2
- 2^2 * 3^2 * 5^2
- 2^2 * 3^2 * 7^2
- 2^2 * 5^2 * 7^2
- 3^2 * 5^2 * 7^2
- 2^2 * 3^2 * 5^2 * 7^2

So total 16 possibilities. Now, since m < n, and m * n = 210^2, then m < 210^2 / m => m^2 < 210^4 => m < 210^2. But since m is a factor of 210^2, which is 210^2, so m can be up to 210^2. However, m < n implies m < 210^2 / m => m < sqrt(210^4) = 210^2. Wait, but m is a factor of 210^2, so m can be as big as 210^2. However, m < n implies m < 210^2 / m => m^2 < 210^4 => m < 210^2. But since m is a positive integer, m can be equal to 210^2 only if n = 1, but since m and n are coprime, if m = 210^2, then n = 1, and gcd(210^2, 1) = 1, which is okay. However, in this case, m = 210^2 and n = 1, so m > n. Therefore, in order to have m < n, we need m < 210^2 / m, which is equivalent to m < sqrt(210^4) = 210^2. But since m is a positive integer, m can be from 1 up to 210^2. However, since m and n are coprime, and m * n = 210^2, m must be a divisor of 210^2, and n = 210^2 / m. Also, since m and n are coprime, m must be a product of some subset of the prime factors of 210^2, each raised to their full exponent (since if you take a prime factor, you have to take all of its exponents in m or in n). Therefore, m is a square of a product of some subset of {2, 3, 5, 7}.

Wait, but m is 2^2 * 3^2 * ... etc. So m is actually (product of primes in subset)^2. Therefore, m is a square number. Therefore, m can be 1, 2^2, 3^2, ..., up to (2*3*5*7)^2 = 210^2. Therefore, the possible m's are squares of square-free numbers. Because if you take a subset of primes, their product is square-free, and squaring it gives m. Therefore, m is a square of a square-free number. Therefore, the number of such m's is equal to the number of subsets of {2, 3, 5, 7}, which is 2^4 = 16. However, since m and n are distinct, and we want m < n, we need to count the number of subsets where the square of the subset product is less than the square of the complement subset product. Which is equivalent to the subset product being less than the complement subset product. Since m = (subset product)^2, n = (complement subset product)^2. Therefore, m < n iff subset product < complement subset product.

But since the total product of all primes is 2*3*5*7 = 210. Therefore, subset product * complement subset product = 210. Therefore, subset product * complement subset product = 210. Therefore, subset product < complement subset product iff subset product < sqrt(210). Because subset product * complement subset product = 210, so if subset product < complement subset product, then subset product < sqrt(210). Therefore, the number of subsets where subset product < sqrt(210) is equal to the number of subsets where the product is less than sqrt(210). However, sqrt(210) is approximately 14.49. Therefore, we need to find the number of subsets of {2, 3, 5, 7} whose product is less than 14.49.

Wait, but the subset products are integers. So the subset products can be:

- The empty set: product is 1
- {2}: 2
- {3}: 3
- {5}: 5
- {7}: 7
- {2,3}: 6
- {2,5}: 10
- {2,7}: 14
- {3,5}: 15
- {3,7}: 21
- {5,7}: 35
- {2,3,5}: 30
- {2,3,7}: 42
- {2,5,7}: 70
- {3,5,7}: 105
- {2,3,5,7}: 210

So the subset products are as above. Now, we need to count how many of these are less than sqrt(210) ≈ 14.49. Let's list them:

- 1 (empty set)
- 2
- 3
- 5
- 7
- 6
- 10
- 14
- 15 (which is 15 > 14.49)
- 21, 35, etc. are all larger.

So the subset products less than 14.49 are:

1, 2, 3, 5, 6, 7, 10, 14. That's 8 subsets. Therefore, there are 8 subsets where the product is less than sqrt(210), hence m = (subset product)^2 < (sqrt(210))^2 = 210, so m < 210. Therefore, these correspond to m < n. Therefore, there are 8 such subsets, hence 8 possible values of m, hence 8 possible values of a. Therefore, the answer is 8.

But let me check if that's correct. Let me list all the subset products and see:

Subsets:

1. Empty set: 1
2. {2}: 2
3. {3}: 3
4. {5}: 5
5. {7}: 7
6. {2,3}: 6
7. {2,5}: 10
8. {2,7}: 14
9. {3,5}: 15
10. {3,7}: 21
11. {5,7}: 35
12. {2,3,5}: 30
13. {2,3,7}: 42
14. {2,5,7}: 70
15. {3,5,7}: 105
16. {2,3,5,7}: 210

So the ones less than 14.49 are the first 8 subsets (1-8). Therefore, 8 subsets. Therefore, 8 values of m. Therefore, 8 values of a. Therefore, the answer is 8.

But let me check with the earlier example. For example, in the case where the total product was 6^2 = 36, which had primes 2 and 3. The subsets would be:

Empty set: 1

{2}: 2

{3}: 3

{2,3}: 6

So sqrt(6) ≈ 2.45. So subsets with product less than 2.45 are {empty set} (1), {2} (2). So 2 subsets. Which gives 2 values of m, which matches the earlier example. Therefore, this method works. Therefore, in our original problem, the answer is 8.

But let me check if there's another way to think about this. Since the number of coprime pairs (m, n) with m * n = N and m < n is equal to 2^{k - 1} where k is the number of distinct prime factors of N? Wait, in our case N is 210^2, which has 4 distinct prime factors. Then 2^{4 - 1} = 8. Which matches. Wait, but in the previous example with N = 36 = 2^2 * 3^2, which has 2 distinct prime factors, then 2^{2 - 1} = 2, which matches. So maybe in general, if N is a square of a square-free number with k distinct prime factors, then the number of coprime pairs (m, n) with m < n and m * n = N is 2^{k - 1}.

But why? Because for each prime factor, you assign it to m or n, but since m and n are coprime, each prime is assigned to exactly one. So total assignments is 2^k. Then, since m and n are distinct (as N is not a square? Wait, N is a square here. Wait, N = 210^2, which is a square. However, m and n are coprime and m * n = N. If N is a square, then m and n must both be squares. Because if m * n is a square and m and n are coprime, then m and n must each be squares. Therefore, m = s^2, n = t^2, with s * t = 210. Therefore, the number of coprime pairs (m, n) with m * n = N is equal to the number of coprime pairs (s, t) with s * t = 210.

Wait, this seems like another way to think about it. Let me check. If m * n = 210^2 and gcd(m, n) = 1, then m and n must be squares. Let me see. Suppose m and n are coprime and m * n is a square. Then, since m and n are coprime, each of them must be a square. Because in the prime factorization of m * n, each prime appears with even exponent. Since m and n are coprime, each prime is entirely in m or entirely in n. Therefore, the exponent of each prime in m must be even, and same for n. Therefore, m and n are squares. Therefore, m = s^2, n = t^2, with s * t = 210. Therefore, the number of coprime pairs (m, n) with m * n = 210^2 is equal to the number of coprime pairs (s, t) with s * t = 210.

But since s * t = 210, and gcd(s, t) = 1, then s and t are coprime divisors of 210. Therefore, the number of such pairs (s, t) is equal to 2^{k}, where k is the number of distinct prime factors of 210. Since 210 has 4 distinct primes, the number of coprime pairs (s, t) with s * t = 210 is 2^4 = 16. But since (s, t) and (t, s) are considered different unless s = t. However, since 210 is not a square, s ≠ t. Therefore, the number of unordered pairs is 8, and the number of ordered pairs is 16. Therefore, the number of ordered pairs with s < t is 8. Therefore, since m = s^2 and n = t^2, the number of ordered pairs (m, n) with m < n is also 8. Therefore, the number of possible a's is 8.

Therefore, this confirms the previous result. Therefore, the answer is 8.

But let me check with the example where N = 36 = 6^2. Then, s * t = 6. The number of coprime pairs (s, t) with s * t = 6. The prime factors of 6 are 2 and 3. So the number of coprime pairs (s, t) is 2^2 = 4. Which are:

(1,6), (6,1), (2,3), (3,2). Therefore, the number of ordered pairs with s < t is 2: (1,6) and (2,3). Therefore, the number of m's is 2, which matches. Therefore, this method works.

Therefore, in general, if we have N = d^2, where d is square-free, then the number of coprime pairs (m, n) with m * n = N and m < n is 2^{k - 1}, where k is the number of distinct prime factors of d. In our case, d = 210, which has 4 distinct prime factors, so 2^{4 - 1} = 8. Therefore, the answer is 8.

Therefore, I think the answer is 8. However, let me check once again with the original problem.

Original problem: gcd(a, b) = 210, lcm(a, b) = 210^3. Then, a * b = 210^4. Then, writing a = 210 * m, b = 210 * n, gcd(m, n) = 1, m * n = 210^2. Then, m and n are coprime, m * n = 210^2. Then, since m and n are coprime, they must be squares, as discussed. Therefore, m = s^2, n = t^2, s * t = 210. Therefore, the number of possible m's is equal to the number of s's such that s divides 210, and t = 210 / s, and gcd(s, t) = 1. Which is equivalent to s being a divisor of 210 with gcd(s, 210 / s) = 1. Which is equivalent to s being a divisor of 210 that is square-free. Because if s is square-free, then since 210 is square-free, t = 210 / s is also square-free, and since s and t are divisors of 210, their gcd is 1 if and only if s and t are coprime. Which is equivalent to s being a product of some subset of the prime factors of 210. Therefore, the number of such s is 2^4 = 16. However, since we want s < t, which is equivalent to s < 210 / s => s^2 < 210 => s < sqrt(210) ≈ 14.49. Therefore, the number of such s is equal to the number of square-free divisors of 210 less than sqrt(210).

The square-free divisors of 210 are all the products of subsets of {2, 3, 5, 7}. So they are:

1, 2, 3, 5, 7, 2*3=6, 2*5=10, 2*7=14, 3*5=15, 3*7=21, 5*7=35, 2*3*5=30, 2*3*7=42, 2*5*7=70, 3*5*7=105, 2*3*5*7=210.

Now, which of these are less than sqrt(210) ≈ 14.49?

1, 2, 3, 5, 7, 6, 10, 14. These are 8 numbers. The rest are 15, 21, 35, etc., which are greater than 14.49. Therefore, there are 8 such s, hence 8 values of m, hence 8 values of a. Therefore, the answer is 8.

Therefore, I think this confirms again that the answer is 8.

But just to be thorough, let me list all possible a's. Since a = 210 * m, and m is s^2 where s is one of the 8 numbers above. Wait, no. Wait, m = s^2, where s is a square-free divisor of 210. Wait, no. Wait, earlier we had m = s^2, where s * t = 210. Wait, no. Wait, m = s^2, n = t^2, with s * t = 210. Therefore, m = s^2, so s is a divisor of 210, and m is s^2. Therefore, the possible m's are:

For each s in {1, 2, 3, 5, 7, 6, 10, 14}, m = s^2:

1^2 = 1

2^2 = 4

3^2 = 9

5^2 = 25

7^2 = 49

6^2 = 36

10^2 = 100

14^2 = 196

Therefore, the possible m's are 1, 4, 9, 25, 49, 36, 100, 196. Therefore, 8 values. Therefore, the corresponding a's are 210 * m:

210*1 = 210

210*4 = 840

210*9 = 1890

210*25 = 5250

210*49 = 10290

210*36 = 7560

210*100 = 21000

210*196 = 41160

Therefore, these are 8 distinct values of a. Now, we need to check that for each of these a's, there exists a b such that gcd(a, b) = 210 and lcm(a, b) = 210^3.

For example, take a = 210. Then, since a = 210 * m, m = 1, so n = 210^2 / m = 210^2. Therefore, b = 210 * n = 210 * 210^2 = 210^3. Then, gcd(a, b) = gcd(210, 210^3) = 210, and lcm(a, b) = lcm(210, 210^3) = 210^3. So that works.

Take a = 840. Then m = 4, so n = 210^2 / 4 = (210^2)/4. Since 210 is divisible by 2, 210 = 2 * 105, so 210^2 = 4 * 105^2. Therefore, n = 4 * 105^2 / 4 = 105^2. Therefore, b = 210 * n = 210 * 105^2. Let me check gcd(a, b). a = 210 * 4, b = 210 * 105^2. The gcd is 210 * gcd(4, 105^2). Since 105^2 is 3^2 * 5^2 * 7^2, which is coprime with 4. Therefore, gcd(4, 105^2) = 1. Therefore, gcd(a, b) = 210 * 1 = 210. Similarly, lcm(a, b) = 210 * lcm(4, 105^2) = 210 * 4 * 105^2 = 210 * 4 * 105^2. But 210 = 2 * 105, so 210 * 4 * 105^2 = 2 * 105 * 4 * 105^2 = 8 * 105^3. However, 210^3 = (2 * 3 * 5 * 7)^3 = 2^3 * 3^3 * 5^3 * 7^3. On the other hand, 8 * 105^3 = 8 * (3 * 5 * 7)^3 = 2^3 * 3^3 * 5^3 * 7^3, which is equal to 210^3. Therefore, that works.

Similarly, for other values of a, the same logic applies. Therefore, all 8 values of a satisfy the given conditions. Therefore, the answer is 8.

Therefore, after all these checks and reasoning, I believe the number of possible values for a is 8.

**Final Answer**
\boxed{8}

To solve the problem, we are given:

- $ \gcd(a, b) = 210 $
- $ \text{lcm}(a, b) = 210^3 $
- $ a < b $
- $ a $ and $ b $ are positive integers

We are to find how many **possible values** of $ a $ satisfy these conditions.

---

### Step 1: Use the GCD-LCM Identity

We use the identity:

$$
\gcd(a, b) \cdot \text{lcm}(a, b) = a \cdot b
$$

Substituting the given values:

$$
210 \cdot 210^3 = a \cdot b \Rightarrow a \cdot b = 210^4
$$

---

### Step 2: Express $ a $ and $ b $ in Terms of Their GCD

Let $ d = \gcd(a, b) = 210 $. Then we can write:

$$
a = d \cdot m = 210m, \quad b = d \cdot n = 210n
$$

Since $ \gcd(a, b) = d $, it follows that $ \gcd(m, n) = 1 $, and:

$$
a \cdot b = (210m)(210n) = 210^2 \cdot m \cdot n = 210^4 \Rightarrow m \cdot n = 210^2
$$

So now the problem reduces to finding the number of **coprime pairs** $ (m, n) $ such that:

- $ m \cdot n = 210^2 $
- $ \gcd(m, n) = 1 $
- $ m < n $

Each such pair corresponds to a unique value of $ a = 210m $, and we are to count how many such $ a $ exist.

---

### Step 3: Prime Factorization of $ 210 $

We factorize $ 210 $:

$$
210 = 2 \cdot 3 \cdot 5 \cdot 7
\Rightarrow 210^2 = (2 \cdot 3 \cdot 5 \cdot 7)^2 = 2^2 \cdot 3^2 \cdot 5^2 \cdot 7^2
$$

So $ m \cdot n = 2^2 \cdot 3^2 \cdot 5^2 \cdot 7^2 $, and since $ \gcd(m, n) = 1 $, each prime factor must be assigned entirely to either $ m $ or $ n $, but not both.

Thus, for each of the 4 distinct primes (2, 3, 5, 7), we have 2 choices: assign it to $ m $ or to $ n $. This gives:

$$
2^4 = 16 \text{ total assignments}
$$

Each assignment corresponds to a unique pair $ (m, n) $, and since $ m \cdot n = 210^2 $, each such pair is unique.

However, since $ m < n $, we only want the **ordered pairs** where $ m < n $. Since the total number of unordered pairs is $ 16 / 2 = 8 $, and each unordered pair corresponds to exactly one ordered pair with $ m < n $, we conclude:

$$
\text{Number of valid } (m, n) \text{ pairs with } m < n = 8
$$

---

### Step 4: Count the Number of Valid $ a $

Each such pair $ (m, n) $ with $ m < n $ gives a unique value of $ a = 210m $. Since there are 8 such pairs, there are **8 possible values** of $ a $.

---

### Final Answer

$$
\boxed{8}
$$

Solution:
Note that the prime factorization of $210$ is $2\cdot 3\cdot 5\cdot 7$, and so the prime factorization of $210^3$ is $2^3\cdot 3^3\cdot 5^3\cdot 7^3$.

Given that $\gcd(a,b)=210$ and $\mathop{\text{lcm}}[a,b]=210^3$, we must have $a=2^k\cdot 3^\ell\cdot 5^m\cdot 7^n$ and $b=2^p\cdot 3^q\cdot 5^r\cdot 7^s$ where each of the ordered pairs $(k,p),(\ell,q),(m,r),(n,s)$ is either $(1,3)$ or $(3,1)$. Therefore, if we ignore the condition $a<b$, there are independently two choices for each of $k$, $\ell$, $m$, and $n$, and these choices determine both of the numbers $a$ and $b$. We have $2\cdot 2\cdot 2\cdot 2=16$ ways to make all four choices.

However, these $16$ sets of choices will generate each possible pair of values for $a$ and $b$ in both possible orders. Half of these choices will satisfy $a<b$ and half will satisfy $a>b$. So, imposing the condition $a<b$, we see that there are $\frac{16}{2}=\boxed{8}$ possible choices for $a$.