Explanation:
Okay, so I need to find all real numbers x such that -4 < x⁴ + 4x² < 21. Hmm, let me think about how to approach this.

First, the inequality is a compound inequality, meaning I have to solve both parts separately and then find the intersection of the solutions. Let me write that down:

-4 < x⁴ + 4x² and x⁴ + 4x² < 21.

Starting with the first inequality: -4 < x⁴ + 4x².

Wait, x⁴ and x² are both even powers, so they are always non-negative, right? That means x⁴ + 4x² is always greater than or equal to zero. But the left side of the inequality is -4, which is negative. So if the left side is always non-negative, then -4 < something non-negative is always true. Therefore, the first inequality is always satisfied for all real numbers x.

So, the first part doesn't restrict the values of x. Therefore, I only need to focus on the second inequality: x⁴ + 4x² < 21.

Alright, so solving x⁴ + 4x² < 21. Let me rewrite that as x⁴ + 4x² - 21 < 0. Maybe I can factor this or find its roots?

Let me consider substituting y = x². Then the inequality becomes y² + 4y - 21 < 0. That seems easier to handle. Let me solve the quadratic inequality in terms of y.

First, find the roots of y² + 4y - 21 = 0. Using the quadratic formula:

y = [-4 ± √(16 + 84)] / 2 = [-4 ± √100]/2 = [-4 ± 10]/2.

Calculating the two roots:

First root: (-4 + 10)/2 = 6/2 = 3.

Second root: (-4 - 10)/2 = -14/2 = -7.

So the quadratic y² + 4y - 21 factors as (y - 3)(y + 7) = 0? Wait, let me check that. If the roots are 3 and -7, then the quadratic is (y - 3)(y + 7) = y² + 7y - 3y -21 = y² + 4y -21. Yes, that's correct.

So the inequality y² + 4y -21 < 0 becomes (y - 3)(y + 7) < 0.

To solve this quadratic inequality, I need to determine where the product is negative. The roots are at y = -7 and y = 3. Since the coefficient of y² is positive, the parabola opens upwards. Therefore, the quadratic is negative between the roots. So the solution is -7 < y < 3.

But remember that y = x², and x² is always greater than or equal to 0. Therefore, y cannot be negative. So the condition -7 < y < 3 becomes 0 ≤ y < 3, because y can't be less than 0.

Therefore, substituting back, we have 0 ≤ x² < 3.

Solving for x, we take square roots. Since x² < 3, then |x| < √3. Therefore, x is between -√3 and √3.

But also, since x² ≥ 0, which is always true, so the entire solution for the second inequality is -√3 < x < √3.

But wait, let me check if I did that correctly. Let me recap:

Original inequality: x⁴ + 4x² < 21.

Substitute y = x²: y² + 4y -21 < 0.

Solutions for y: between -7 and 3. But since y = x² ≥ 0, then y must be between 0 and 3. Therefore, x² < 3 => |x| < √3. So x ∈ (-√3, √3).

Therefore, combining with the first inequality which is always true, the solution set is x ∈ (-√3, √3).

But wait, let me check if there are any mistakes here. Let me test some values. For example, take x = 0. Then x⁴ + 4x² = 0 + 0 = 0, which is between -4 and 21. Good. Take x = √3. Then x² = 3, so x⁴ + 4x² = 9 + 12 = 21. But the inequality is strict: x⁴ + 4x² < 21. Therefore, x = √3 is not included. Similarly for x = -√3. So the interval is open.

What about x = 2? Let's check. x = 2, x² = 4, x⁴ = 16. So 16 + 16 = 32. Which is greater than 21, so it's not in the solution set. Which is consistent with our interval (-√3, √3) since √3 ≈ 1.732, so 2 is outside.

What about x = 1? x² = 1, x⁴ = 1. So 1 + 4 = 5 < 21. Correct.

What about x approaching √3? Let's say x = √3 - ε, where ε is small. Then x² ≈ 3 - 2√3 ε + ε², so x⁴ + 4x² ≈ (3 - 2√3 ε)^2 + 4*(3 - 2√3 ε) ≈ 9 - 12√3 ε + ... + 12 - 8√3 ε ≈ 21 - 20√3 ε, which is less than 21. So approaching √3 from the left, the expression approaches 21 from below. So the interval is correct.

But let me check if there are any other constraints. The original problem was -4 < x⁴ + 4x² < 21. Since x⁴ + 4x² is always ≥ 0, the left inequality is always true. Therefore, the entire solution is just the solution to the right inequality, which is (-√3, √3).

Wait, but let me check if maybe there are some x where x⁴ + 4x² is between -4 and 21. But since x⁴ + 4x² is always ≥ 0, then the left inequality is automatically satisfied. Therefore, the answer is indeed (-√3, √3).

But let me check if maybe there's a mistake in solving the quadratic. Let me re-express the original inequality again.

Original inequality: x⁴ + 4x² < 21.

Let me try to write it as x⁴ + 4x² - 21 < 0.

Alternatively, maybe I can factor the quartic equation. Let me try to factor x⁴ + 4x² - 21.

But since we already did substitution y = x², and found that it factors as (y - 3)(y + 7). Therefore, substituting back, x⁴ + 4x² - 21 = (x² - 3)(x² + 7).

Yes, because (x² - 3)(x² + 7) = x⁴ + 7x² - 3x² -21 = x⁴ + 4x² -21. Correct.

Therefore, the inequality (x² - 3)(x² + 7) < 0.

Now, since x² + 7 is always positive (because x² ≥ 0, so x² + 7 ≥ 7 > 0), the sign of the product depends on (x² - 3). Therefore, the inequality reduces to (x² - 3) < 0, because multiplying by a positive number doesn't change the inequality direction.

Therefore, x² - 3 < 0 => x² < 3 => |x| < √3.

So that's another way to see it. Therefore, the solution is indeed -√3 < x < √3.

Therefore, combining with the first inequality which is always true, the answer is (-√3, √3).

But let me check if maybe there's a mistake here. For example, suppose someone might think that the original inequality could have some other solutions? Let me check with x = 2, which we did before. x⁴ + 4x² = 16 + 16 = 32, which is greater than 21, so not in the solution.

What about x = -√3? Then x⁴ + 4x² = ( (√3)^4 ) + 4*(√3)^2 = 9 + 12 = 21. So the inequality is strict, so x = -√3 and x = √3 are excluded.

Therefore, the interval is open.

But let me check if there's a case where x⁴ + 4x² could be less than -4? Since x⁴ and 4x² are both non-negative, their sum is non-negative. Therefore, x⁴ + 4x² ≥ 0 > -4. Therefore, the left inequality is always true.

Therefore, the entire solution set is indeed (-√3, √3).

But let me check if there's a possibility that maybe when solving the quadratic inequality, I made a mistake.

Original substitution: y = x². Then inequality becomes y² + 4y -21 < 0.

Solutions for equality: y = [-4 ± sqrt(16 + 84)] / 2 = [-4 ± sqrt(100)] / 2 = [-4 ± 10]/2.

So y = (6)/2 = 3 and y = (-14)/2 = -7.

So the quadratic is positive outside the roots and negative between them. Since the coefficient of y² is positive. Therefore, y² + 4y -21 < 0 when y ∈ (-7, 3).

But since y = x² ≥ 0, then the intersection is y ∈ [0, 3). Therefore, x² ∈ [0, 3) => x ∈ (-√3, √3).

Yes, that's correct.

Therefore, the answer is all real numbers between -√3 and √3, not including the endpoints.

So in interval notation, that is (-√3, √3).

But let me check if the problem says "real values of x" so that's fine.

Wait, but maybe I need to write it with decimals? But the question says to express in interval notation, so using √3 is acceptable.

Therefore, the final answer is (-√3, √3).

But let me check once more. Suppose I graph the function f(x) = x⁴ + 4x². It's a quartic function that tends to infinity as x approaches ±∞. It has a minimum at x = 0, where f(0) = 0. Then, as |x| increases, the function increases. Therefore, the equation x⁴ + 4x² = 21 will have two real solutions, symmetric around the y-axis, at x = ±√3. Therefore, between -√3 and √3, the function is below 21, and outside it's above 21. Therefore, the inequality x⁴ + 4x² < 21 holds for x between -√3 and √3.

Therefore, the answer is indeed (-√3, √3).

I think that's solid. I don't see any mistakes in my reasoning. So I think that's the correct answer.

**Final Answer**
The real values of $ x $ that satisfy the inequality are \boxed{(-\sqrt{3}, \sqrt{3})}.

To solve the inequality
$$
-4 < x^4 + 4x^2 < 21,
$$
we analyze it as a compound inequality, breaking it into two parts:

---

### **Part 1: $ -4 < x^4 + 4x^2 $**

Note that $ x^4 $ and $ x^2 $ are both **non-negative** for all real $ x $, so their sum $ x^4 + 4x^2 \geq 0 $. Therefore, the left-hand side of the inequality, $ -4 $, is **strictly less than** any non-negative value. This means the inequality
$$
-4 < x^4 + 4x^2
$$
is **always true** for all real $ x $.

---

### **Part 2: $ x^4 + 4x^2 < 21 $**

We now solve the inequality
$$
x^4 + 4x^2 < 21.
$$
Let’s make a substitution to simplify:
Let $ y = x^2 $, so the inequality becomes
$$
y^2 + 4y < 21.
$$
Rewriting:
$$
y^2 + 4y - 21 < 0.
$$
We solve the corresponding quadratic equation:
$$
y^2 + 4y - 21 = 0.
$$
Using the quadratic formula:
$$
y = \frac{-4 \pm \sqrt{16 + 84}}{2} = \frac{-4 \pm \sqrt{100}}{2} = \frac{-4 \pm 10}{2}.
$$
So the roots are:
$$
y = 3 \quad \text{and} \quad y = -7.
$$
The quadratic $ y^2 + 4y - 21 $ is a parabola opening upwards (since the coefficient of $ y^2 $ is positive), so it is **negative between its roots**. Thus,
$$
-7 < y < 3.
$$
But since $ y = x^2 \geq 0 $, we only consider the intersection of this interval with $ y \geq 0 $, which gives:
$$
0 \leq y < 3.
$$
Substituting back $ y = x^2 $, we get:
$$
0 \leq x^2 < 3.
$$
Taking square roots:
$$
|x| < \sqrt{3} \quad \Rightarrow \quad -\sqrt{3} < x < \sqrt{3}.
$$

---

### **Final Step: Combine Both Inequalities**

From Part 1, we know the inequality is always true.
From Part 2, the inequality is satisfied only when $ x \in (-\sqrt{3}, \sqrt{3}) $.

---

### **Conclusion**

The set of all real values of $ x $ that satisfy the original inequality is:
$$
\boxed{(-\sqrt{3}, \sqrt{3})}
$$

[Assistant]
Explanation:
Okay, so I need to find all real numbers x such that -4 < x⁴ + 4x² < 21. Hmm, let me think about how to approach this.

First, the inequality is a compound inequality, meaning I have to solve both parts separately and then find the intersection of the solutions. Let me write that down:

-4 < x⁴ + 4x² and x⁴ + 4x² < 21.

Starting with the first inequality: -4 < x⁴ + 4x².

Wait, x⁴ and x² are both even powers, so they are always non-negative, right? That means x⁴ + 4x² is always greater than or equal to zero. But the left side of the inequality is -4, which is negative. So if the left side is always non-negative, then -4 < something non-negative is always true. Therefore, the first inequality is always satisfied for all real numbers x.

So, the first part doesn't restrict the values of x. Therefore, I only need to focus on the second inequality: x⁴ + 4x² < 21.

Alright, so solving x⁴ + 4x² < 21. Let me rewrite that as x⁴ + 4x² - 21 < 0. Maybe I can factor this or find its roots?

Let me consider substituting y = x². Then the inequality becomes y² + 4y - 21 < 0. That seems easier to handle. Let me solve the quadratic inequality in terms of y.

First, find the roots of y² + 4y - 21 = 0. Using the quadratic formula:

y = [-4 ± √(16 + 84)] / 2 = [-4 ± √100]/2 = [-4 ± 10]/2.

Calculating the two roots:

First root: (-4 + 10)/2 = 6/2 = 3.

Second root: (-4 - 10)/2 = -14/2 = -7.

So the quadratic y² + 4y - 21 factors as (y - 3)(y + 7) = 0? Wait, let me check that. If the roots are 3 and -7, then the quadratic is (y - 3)(y + 7) = y² + 7y - 3y -21 = y² + 4y -21. Yes, that's correct.

So the inequality y² + 4y -21 < 0 becomes (y - 3)(y + 7) < 0.

To solve this quadratic inequality, I need to determine where the product is negative. The roots are at y = -7 and y = 3. Since the coefficient of y² is positive, the parabola opens upwards. Therefore, the quadratic is negative between the roots. So the solution is -7 < y < 3.

But remember that y = x², and x² is always greater than or equal to 0. Therefore, y cannot be negative. So the condition -7 < y < 3 becomes 0 ≤ y < 3, because y can't be less than 0.

Therefore, substituting back, we have 0 ≤ x² < 3.

Solving for x, we take square roots. Since x² < 3, then |x| < √3. Therefore, x is between -√3 and √3.

But also, since x² ≥ 0, which is always true, so the entire solution for the second inequality is -√3 < x < √3.

But wait, let me check if I did that correctly. Let me recap:

Original inequality: x⁴ + 4x² < 21.

Substitute y = x²: y² + 4y -21 < 0.

Solutions for y: between -7 and 3. But since y = x² ≥ 0, then y must be between 0 and 3. Therefore, x² < 3 => |x| < √3. So x ∈ (-√3, √3).

Therefore, combining with the first inequality which is always true, the solution set is x ∈ (-√3, √3).

But wait, let me check if there are any mistakes here. Let me test some values. For example, take x = 0. Then x⁴ + 4x² = 0 + 0 = 0, which is between -4 and 21. Good. Take x = √3. Then x² = 3, so x⁴ + 4x² = 9 + 12 = 21. But the inequality is strict: x⁴ + 4x² < 21. Therefore, x = √3 is not included. Similarly for x = -√3. So the interval is open.

What about x = 2? Let's check. x = 2, x² = 4, x⁴ = 16. So 16 + 16 = 32. Which is greater than 21, so it's not in the solution set. Which is consistent with our interval (-√3, √3) since √3 ≈ 1.732, so 2 is outside.

What about x = 1? x² = 1, x⁴ = 1. So 1 + 4 = 5 < 21. Correct.

What about x approaching √3? Let's say x = √3 - ε, where ε is small. Then x² ≈ 3 - 2√3 ε + ε², so x⁴ + 4x² ≈ (3 - 2√3 ε)^2 + 4*(3 - 2√3 ε) ≈ 9 - 12√3 ε + ... + 12 - 8√3 ε ≈ 21 - 20√3 ε, which is less than 21. So approaching √3 from the left, the expression approaches 21 from below. So the interval is correct.

But let me check if there are any other constraints. The original problem was -4 < x⁴ + 4x² < 21. Since x⁴ + 4x² is always ≥ 0, the left inequality is always true. Therefore, the entire solution is just the solution to the right inequality, which is (-√3, √3).

Wait, but let me check if maybe there are some x where x⁴ + 4x² is between -4 and 21. But since x⁴ + 4x² is always ≥ 0, then the left inequality is automatically satisfied. Therefore, the answer is indeed (-√3, √3).

But let me check if maybe there's a mistake in solving the quadratic. Let me re-express the original inequality again.

Original inequality: x⁴ + 4x² < 21.

Let me try to write it as x⁴ + 4x² - 21 < 0.

Alternatively, maybe I can factor the quartic equation. Let me try to factor x⁴ + 4x² - 21.

But since we already did substitution y = x², and found that it factors as (y - 3)(y + 7). Therefore, substituting back, x⁴ + 4x² - 21 = (x² - 3)(x² + 7).

Yes, because (x² - 3)(x² + 7) = x⁴ + 7x² - 3x² -21 = x⁴ + 4x² -21. Correct.

Therefore, the inequality (x² - 3)(x² + 7) < 0.

Now, since x² + 7 is always positive (because x² ≥ 0, so x² + 7 ≥ 7 > 0), the sign of the product depends on (x² - 3). Therefore, the inequality reduces to (x² - 3) < 0, because multiplying by a positive number doesn't change the inequality direction.

Therefore, x² - 3 < 0 => x² < 3 => |x| < √3.

So that's another way to see it. Therefore, the solution is indeed -√3 < x < √3.

Therefore, combining with the first inequality which is always true, the answer is (-√3, √3).

But let me check if maybe there's a mistake here. For example, suppose someone might think that the original inequality could have some other solutions? Let me check with x = 2, which we did before. x⁴ + 4x² = 16 + 16 = 32, which is greater than 21, so not in the solution.

What about x = -√3? Then x⁴ + 4x² = ( (√3)^4 ) + 4*(√3)^2 = 9 + 12 = 21. So the inequality is strict, so x = -√3 and x = √3 are excluded.

Therefore, the interval is open.

But let me check if there's a case where x⁴ + 4x² could be less than -4? Since x⁴ and 4x² are both non-negative, their sum is non-negative. Therefore, x⁴ + 4x² ≥ 0 > -4. Therefore, the left inequality is always true.

Therefore, the entire solution set is indeed (-√3, √3).

But let me check if there's a possibility that maybe when solving the quadratic inequality, I made a mistake.

Original substitution: y = x². Then inequality becomes y² + 4y -21 < 0.

Solutions for equality: y = [-4 ± sqrt(16 + 84)] / 2 = [-4 ± sqrt(100)] / 2 = [-4 ± 10]/2.

So y = (6)/2 = 3 and y = (-14)/2 = -7.

So the quadratic is positive outside the roots and negative between them. Since the coefficient of y² is positive. Therefore, y² + 4y -21 < 0 when y ∈ (-7, 3).

But since y = x² ≥ 0, then the intersection is y ∈ [0, 3). Therefore, x² ∈ [0, 3) => x ∈ (-√3, √3).

Yes, that's correct.

Therefore, the answer is all real numbers between -√3 and √3, not including the endpoints.

So in interval notation, that is (-√3, √3).

But let me check if the problem says "real values of x" so that's fine.

Wait, but maybe I need to write it with decimals? But the question says to express in interval notation, so using √3 is acceptable.

Therefore, the final answer is (-√3, √3).

But let me check once more. Suppose I graph the function f(x) = x⁴ + 4x². It's a quartic function that tends to infinity as x approaches ±∞. It has a minimum at x = 0, where f(0) = 0. Then, as |x| increases, the function increases. Therefore, the equation x⁴ + 4x² = 21 will have two real solutions, symmetric around the y-axis, at x = ±√3. Therefore, between -√3 and √3, the function is below 21, and outside it's above 21. Therefore, the inequality x⁴ + 4x² < 21 holds for x between -√3 and √3.

Therefore, the answer is indeed (-√3, √3).

I think that's solid. I don't see any mistakes in my reasoning. So I think that's the correct answer.

**Final Answer**
The real values of $ x $ that satisfy the inequality are \boxed{(-\sqrt{3}, \sqrt{3})}.

To solve the inequality
$$
-4 < x^4 + 4x^2 < 21,
$$
we analyze it as a compound inequality, breaking it into two parts:

---

### **Part 1: $ -4 < x^4 + 4x^2 $**

Note that $ x^4 $ and $ x^2 $ are both **non-negative** for all real $ x $, so their sum $ x^4 + 4x^2 \geq 0 $. Therefore, the left-hand side of the inequality, $ -4 $, is **strictly less than** any non-negative value. This means the inequality
$$
-4 < x^4 + 4x^2
$$
is **always true** for all real $ x $.

---

### **Part 2: $ x^4 + 4x^2 < 21 $**

We now solve the inequality
$$
x^4 + 4x^2 < 21.
$$
Let’s make a substitution to simplify:
Let $ y = x^2 $, so the inequality becomes
$$
y^2 + 4y < 21.
$$
Rewriting:
$$
y^2 + 4y - 21 < 0.
$$
We solve the corresponding quadratic equation:
$$
y^2 + 4y - 21 = 0.
$$
Using the quadratic formula:
$$
y = \frac{-4 \pm \sqrt{16 + 84}}{2} = \frac{-4 \pm \sqrt{100}}{2} = \frac{-4 \pm 10}{2}.
$$
So the roots are:
$$
y = 3 \quad \text{and} \quad y = -7.
$$
The quadratic $ y^2 + 4y - 21 $ is a parabola opening upwards (since the coefficient of $ y^2 $ is positive), so it is **negative between its roots**. Thus,
$$
-7 < y < 3.
$$
But since $ y = x^2 \geq 0 $, we only consider the intersection of this interval with $ y \geq 0 $, which gives:
$$
0 \leq y < 3.
$$
Substituting back $ y = x^2 $, we get:
$$
0 \leq x^2 < 3.
$$
Taking square roots:
$$
|x| < \sqrt{3} \quad \Rightarrow \quad -\sqrt{3} < x < \sqrt{3}.
$$

---

### **Final Step: Combine Both Inequalities**

From Part 1, we know the inequality is always true.
From Part 2, the inequality is satisfied only when $ x \in (-\sqrt{3}, \sqrt{3}) $.

---

### **Conclusion**

The set of all real values of $ x $ that satisfy the original inequality is:
$$
\boxed{(-\sqrt{3}, \sqrt{3})}
$$

Solution:
Let us first define $y=x^{2}$. We can then plug this value into the inequality and add 4 to $-4$, $x^4+4x^2$, and 21, yielding $$0<y^{2}+4y+4<25.$$We can factor $y^2+4y+4$ to obtain $$0<(y+2)^{2}<25.$$Taking the square root gives us $0<|y+2|<5$, which in turn gives us two intervals for solutions of $y$: $-2<y<3$, or $-7<y<-2$.

However, $y$ must be non-negative since $y=x^{2}$, so we have $0\leq y<3$. This means that $-\sqrt{3}< x<\sqrt{3}$ satisfies the original inequality. In interval notation, this is $\boxed{(-\sqrt{3}, \sqrt{3})}$.


Explanation:
Okay, so I need to figure out how many possible values of x there are when dealing a 54-card deck to x players, with each player getting y cards. The constraints are that x is at least 2 and y is at least 5. Let me start by understanding the problem step by step.

First, the deck has 54 cards. When dealing all the cards to x players, each player gets y cards. So, the total number of cards dealt should be equal to the number of players multiplied by the number of cards each player gets. That gives me the equation:

x * y = 54

So, x and y are positive integers that satisfy this equation. But there are constraints: x ≥ 2 and y ≥ 5. Therefore, I need to find all possible integer values of x (number of players) such that when you divide 54 by x, you get an integer y that is at least 5.

Alternatively, since x and y are positive integers, x must be a divisor of 54. So, first, I should find all the divisors of 54, then check which of those divisors satisfy the conditions that x ≥ 2 and y = 54/x ≥ 5.

Let me start by listing all the divisors of 54. To do that, I can factorize 54 into its prime factors. Let me recall that 54 is 2 * 27, and 27 is 3^3. Therefore, the prime factorization of 54 is 2^1 * 3^3.

The formula for finding the number of divisors is to add 1 to each of the exponents in the prime factorization and then multiply them. So here, exponents are 1 and 3, so (1+1)*(3+1) = 2*4 = 8. Therefore, there are 8 positive divisors of 54. Let me list them:

Starting with 1, then 2, 3, 6, 9, 18, 27, 54. Wait, let me check:

Divisors of 54:

1, 2, 3, 6, 9, 18, 27, 54. Yes, that's 8 divisors. Let me confirm:

- 1 divides 54
- 2 divides 54 (54/2=27)
- 3 divides 54 (54/3=18)
- 6 divides 54 (54/6=9)
- 9 divides 54 (54/9=6)
- 18 divides 54 (54/18=3)
- 27 divides 54 (54/27=2)
- 54 divides 54 (54/54=1)

Yes, that's all. So the divisors are 1, 2, 3, 6, 9, 18, 27, 54.

But the problem says that x is at least 2, so we can eliminate 1. So possible x values are 2, 3, 6, 9, 18, 27, 54. However, we also have the constraint that y must be at least 5. Since y = 54/x, we need 54/x ≥ 5. Let me write that inequality:

54/x ≥ 5

Solving for x:

Multiply both sides by x (assuming x positive, which it is):

54 ≥ 5x

Then divide both sides by 5:

54/5 ≥ x

Which is 10.8 ≥ x. Since x must be an integer, this means x ≤ 10.8. Therefore, x must be less than or equal to 10. Since x is an integer, x ≤ 10.

So combining this with the previous list of possible x values (which are 2, 3, 6, 9, 18, 27, 54), we need to find which of these are less than or equal to 10. Let's check each:

- 2: yes, 2 ≤ 10
- 3: yes
- 6: yes
- 9: yes
- 18: 18 > 10, so no
- 27: no
- 54: no

Therefore, the possible x values are 2, 3, 6, 9. Wait, that's four values. But let me check again. Wait, 54/x must be at least 5. Let me check for each x in the list:

For x=2: y=54/2=27 ≥5: yes

x=3: y=18 ≥5: yes

x=6: y=9 ≥5: yes

x=9: y=6 ≥5: yes

x=18: y=3, which is less than 5: not allowed

x=27: y=2: nope

x=54: y=1: nope

So indeed, only x=2,3,6,9 satisfy both x≥2 and y≥5. Therefore, there are 4 possible values of x. Wait, but let me check if there are other divisors that I might have missed?

Wait, earlier I listed the divisors as 1,2,3,6,9,18,27,54. So that's all. So only those. Therefore, the answer is 4?

But wait, let me think again. Maybe I made a mistake in the inequality?

Original condition: y ≥ 5. Since y = 54/x, so 54/x ≥ 5 => x ≤ 54/5 = 10.8. So x must be ≤10. So x can be up to 10. So divisors of 54 less than or equal to 10 are 2,3,6,9. So that's four values. Therefore, the answer is 4?

But let me check if there are other divisors between 1 and 10. For example, 4? Is 4 a divisor of 54? 54 divided by 4 is 13.5, which is not integer. 5? 54 divided by 5 is 10.8, not integer. 7? 54 divided by 7 is about 7.7, not integer. 8? 54 divided by 8 is 6.75, not integer. So no, there are no other divisors between 1 and 10 except 2,3,6,9. Therefore, the answer is 4?

But wait, let me check again. Wait, the problem says "how many possible values of x are there?" So if x can be 2,3,6,9, that's four values. But let me check if there's a possibility that y can be 5 or more. For example, if x is 2, y is 27, which is okay. If x is 3, y is 18. If x is 6, y is 9. If x is 9, y is 6. All of these are ≥5. If x is 10, then y would be 5.4, which is not integer. So x=10 is not a divisor. So the next possible x is 18, but then y=3 which is too small. Therefore, yes, only 4 values. Therefore, the answer is 4?

Wait, but let me check again. Maybe I need to consider that y must be an integer. Since x must divide 54, so y is automatically integer. Therefore, the only constraints are x divides 54, x ≥2, and y=54/x ≥5. Therefore, the answer is 4. But let me check once more.

Alternatively, maybe I can approach it by considering the possible values of y. Since y must be at least 5, and x must be at least 2. Since x = 54/y, so x must be a positive integer, so y must be a divisor of 54. Wait, no. Wait, if x = 54/y, then y must divide 54? Wait, no. Wait, x and y are integers such that x*y=54. Therefore, both x and y must be divisors of 54. Wait, no. For example, if x is 2, then y is 27, which is a divisor. If x is 3, y is 18, which is a divisor. Similarly, if x is 6, y is 9, which is a divisor. If x is 9, y is 6, which is a divisor. So in this case, since x*y=54, both x and y must be divisors of 54. Therefore, the possible values of y are the divisors of 54 that are at least 5, and then x would be 54 divided by those y's. But since x must be at least 2, then y must be at most 54/2=27. So y is between 5 and 27, inclusive, and divides 54. Let me check.

So possible y's are divisors of 54 that are ≥5 and ≤27. Let me list the divisors of 54 again: 1,2,3,6,9,18,27,54. So divisors ≥5 are 6,9,18,27,54. But y must be ≤27 because x must be ≥2 (since x=54/y, so if y is 54, x=1 which is less than 2). Therefore, y can be 6,9,18,27. Therefore, corresponding x's are 54/6=9, 54/9=6, 54/18=3, 54/27=2. Therefore, x can be 9,6,3,2. Which is the same as before. Therefore, four values. Therefore, the answer is 4. Therefore, the number of possible values of x is 4.

But wait, let me check if there are other divisors between 5 and 27. For example, 5 is not a divisor. 7? No. 8? No. 10? No. 12? 54 divided by 12 is 4.5, not integer. 14? No. 15? No. 16? No. 17? No. 19? No. 20? No. 21? No. 22? No. 23? No. 24? No. 25? No. 26? No. So only 6,9,18,27. Therefore, yes, four values. Therefore, the answer is 4. Therefore, the number of possible x is 4.

But wait, let me check once again. Maybe I made a mistake in considering the constraints. The problem says "each player getting y cards", so y must be at least 5. So y ≥5. So if y is 6, that's okay. If y is 5, then x would be 54/5=10.8, which is not integer. Therefore, y cannot be 5. Therefore, the possible y's are 6,9,18,27, which gives x=9,6,3,2. Therefore, four values. Therefore, the answer is 4. Therefore, the number of possible values of x is 4. Therefore, the answer is \boxed{4}?

Wait, but let me check once more. Let me think if there's another way to approach this problem. For example, considering that x must divide 54, x ≥2, and 54/x ≥5. So x divides 54, x ≥2, and x ≤54/5=10.8. Therefore, x is in the set {2,3,6,9} as divisors of 54 between 2 and 10.8. Which is exactly what I found before. Therefore, 4 values. Therefore, the answer is 4. Therefore, I think that is correct.

But just to be thorough, let me check each possible x:

x=2: y=27 ≥5: valid

x=3: y=18 ≥5: valid

x=4: 54/4=13.5: not integer, invalid

x=5: 54/5=10.8: not integer, invalid

x=6: y=9 ≥5: valid

x=7: 54/7≈7.71: not integer

x=8: 54/8=6.75: not integer

x=9: y=6 ≥5: valid

x=10: 54/10=5.4: not integer

x=11: 54/11≈4.9: less than 5, invalid

So, only x=2,3,6,9 are valid. Therefore, 4 values. Therefore, the answer is 4. Therefore, I think that is correct.

But wait, hold on. Wait, the problem says "how many possible values of x are there?" So if x can be 2,3,6,9, that's four values. Therefore, the answer is 4. Therefore, I think that is correct.

But let me check once again if there's a possibility that I missed a divisor. For example, 54 divided by 1 is 54, but x must be at least 2. Divisors of 54 are 1,2,3,6,9,18,27,54. So between 2 and 10.8, the divisors are 2,3,6,9. So yes, four numbers. Therefore, answer is 4. Therefore, I think that is correct.

But just to make sure, let me think of another approach. Let me think of all possible pairs (x,y) such that x*y=54, x≥2, y≥5. So, for each divisor x of 54 with x≥2, check if y=54/x ≥5. So:

x=2: y=27 ≥5: yes

x=3: y=18 ≥5: yes

x=6: y=9 ≥5: yes

x=9: y=6 ≥5: yes

x=18: y=3 <5: no

x=27: y=2 <5: no

x=54: y=1 <5: no

So only four x's. Therefore, answer is 4. Therefore, I think that is correct.

Therefore, the number of possible values of x is 4. Therefore, the answer is \boxed{4}.

**Final Answer**
\boxed{4}

To determine how many possible values of $ x $ (the number of players) are valid under the given constraints, we begin by analyzing the relationship between the number of players and the number of cards each player receives.

---

### Step 1: Understand the Relationship

We are given a deck of 54 cards, and all cards are dealt to $ x $ players, with each player receiving $ y $ cards. This gives the equation:

$$
x \cdot y = 54
$$

We are told that:

- $ x \geq 2 $
- $ y \geq 5 $

Our goal is to find how many **integer values of $ x $** satisfy these conditions.

---

### Step 2: Find All Divisors of 54

Since $ x \cdot y = 54 $, both $ x $ and $ y $ must be **positive integer divisors** of 54. The divisors of 54 are:

$$
1, 2, 3, 6, 9, 18, 27, 54
$$

We are only interested in values of $ x $ such that:

- $ x \geq 2 $
- $ y = \frac{54}{x} \geq 5 $

---

### Step 3: Apply the Constraints

We now test each divisor of 54 that is at least 2, and check whether the corresponding $ y = \frac{54}{x} $ is at least 5.

| $ x $ | $ y = \frac{54}{x} $ | Valid? |
|--------|------------------------|--------|
| 2      | 27                     | Yes    |
| 3      | 18                     | Yes    |
| 6      | 9                      | Yes    |
| 9      | 6                      | Yes    |
| 18     | 3                      | No     |
| 27     | 2                      | No     |
| 54     | 1                      | No     |

Only the values $ x = 2, 3, 6, 9 $ satisfy both $ x \geq 2 $ and $ y \geq 5 $.

---

### Step 4: Count the Valid Values

From the table, the valid values of $ x $ are:

$$
x = 2, 3, 6, 9
$$

Thus, there are **4** valid values of $ x $.

---

### Final Answer

$$
\boxed{4}
$$

[Assistant]
Explanation:
Okay, so I need to figure out how many possible values of x there are when dealing a 54-card deck to x players, with each player getting y cards. The constraints are that x is at least 2 and y is at least 5. Let me start by understanding the problem step by step.

First, the deck has 54 cards. When dealing all the cards to x players, each player gets y cards. So, the total number of cards dealt should be equal to the number of players multiplied by the number of cards each player gets. That gives me the equation:

x * y = 54

So, x and y are positive integers that satisfy this equation. But there are constraints: x ≥ 2 and y ≥ 5. Therefore, I need to find all possible integer values of x (number of players) such that when you divide 54 by x, you get an integer y that is at least 5.

Alternatively, since x and y are positive integers, x must be a divisor of 54. So, first, I should find all the divisors of 54, then check which of those divisors satisfy the conditions that x ≥ 2 and y = 54/x ≥ 5.

Let me start by listing all the divisors of 54. To do that, I can factorize 54 into its prime factors. Let me recall that 54 is 2 * 27, and 27 is 3^3. Therefore, the prime factorization of 54 is 2^1 * 3^3.

The formula for finding the number of divisors is to add 1 to each of the exponents in the prime factorization and then multiply them. So here, exponents are 1 and 3, so (1+1)*(3+1) = 2*4 = 8. Therefore, there are 8 positive divisors of 54. Let me list them:

Starting with 1, then 2, 3, 6, 9, 18, 27, 54. Wait, let me check:

Divisors of 54:

1, 2, 3, 6, 9, 18, 27, 54. Yes, that's 8 divisors. Let me confirm:

- 1 divides 54
- 2 divides 54 (54/2=27)
- 3 divides 54 (54/3=18)
- 6 divides 54 (54/6=9)
- 9 divides 54 (54/9=6)
- 18 divides 54 (54/18=3)
- 27 divides 54 (54/27=2)
- 54 divides 54 (54/54=1)

Yes, that's all. So the divisors are 1, 2, 3, 6, 9, 18, 27, 54.

But the problem says that x is at least 2, so we can eliminate 1. So possible x values are 2, 3, 6, 9, 18, 27, 54. However, we also have the constraint that y must be at least 5. Since y = 54/x, we need 54/x ≥ 5. Let me write that inequality:

54/x ≥ 5

Solving for x:

Multiply both sides by x (assuming x positive, which it is):

54 ≥ 5x

Then divide both sides by 5:

54/5 ≥ x

Which is 10.8 ≥ x. Since x must be an integer, this means x ≤ 10.8. Therefore, x must be less than or equal to 10. Since x is an integer, x ≤ 10.

So combining this with the previous list of possible x values (which are 2, 3, 6, 9, 18, 27, 54), we need to find which of these are less than or equal to 10. Let's check each:

- 2: yes, 2 ≤ 10
- 3: yes
- 6: yes
- 9: yes
- 18: 18 > 10, so no
- 27: no
- 54: no

Therefore, the possible x values are 2, 3, 6, 9. Wait, that's four values. But let me check again. Wait, 54/x must be at least 5. Let me check for each x in the list:

For x=2: y=54/2=27 ≥5: yes

x=3: y=18 ≥5: yes

x=6: y=9 ≥5: yes

x=9: y=6 ≥5: yes

x=18: y=3, which is less than 5: not allowed

x=27: y=2: nope

x=54: y=1: nope

So indeed, only x=2,3,6,9 satisfy both x≥2 and y≥5. Therefore, there are 4 possible values of x. Wait, but let me check if there are other divisors that I might have missed?

Wait, earlier I listed the divisors as 1,2,3,6,9,18,27,54. So that's all. So only those. Therefore, the answer is 4?

But wait, let me think again. Maybe I made a mistake in the inequality?

Original condition: y ≥ 5. Since y = 54/x, so 54/x ≥ 5 => x ≤ 54/5 = 10.8. So x must be ≤10. So x can be up to 10. So divisors of 54 less than or equal to 10 are 2,3,6,9. So that's four values. Therefore, the answer is 4?

But let me check if there are other divisors between 1 and 10. For example, 4? Is 4 a divisor of 54? 54 divided by 4 is 13.5, which is not integer. 5? 54 divided by 5 is 10.8, not integer. 7? 54 divided by 7 is about 7.7, not integer. 8? 54 divided by 8 is 6.75, not integer. So no, there are no other divisors between 1 and 10 except 2,3,6,9. Therefore, the answer is 4?

But wait, let me check again. Wait, the problem says "how many possible values of x are there?" So if x can be 2,3,6,9, that's four values. But let me check if there's a possibility that y can be 5 or more. For example, if x is 2, y is 27, which is okay. If x is 3, y is 18. If x is 6, y is 9. If x is 9, y is 6. All of these are ≥5. If x is 10, then y would be 5.4, which is not integer. So x=10 is not a divisor. So the next possible x is 18, but then y=3 which is too small. Therefore, yes, only 4 values. Therefore, the answer is 4?

Wait, but let me check again. Maybe I need to consider that y must be an integer. Since x must divide 54, so y is automatically integer. Therefore, the only constraints are x divides 54, x ≥2, and y=54/x ≥5. Therefore, the answer is 4. But let me check once more.

Alternatively, maybe I can approach it by considering the possible values of y. Since y must be at least 5, and x must be at least 2. Since x = 54/y, so x must be a positive integer, so y must be a divisor of 54. Wait, no. Wait, if x = 54/y, then y must divide 54? Wait, no. Wait, x and y are integers such that x*y=54. Therefore, both x and y must be divisors of 54. Wait, no. For example, if x is 2, then y is 27, which is a divisor. If x is 3, y is 18, which is a divisor. Similarly, if x is 6, y is 9, which is a divisor. If x is 9, y is 6, which is a divisor. So in this case, since x*y=54, both x and y must be divisors of 54. Therefore, the possible values of y are the divisors of 54 that are at least 5, and then x would be 54 divided by those y's. But since x must be at least 2, then y must be at most 54/2=27. So y is between 5 and 27, inclusive, and divides 54. Let me check.

So possible y's are divisors of 54 that are ≥5 and ≤27. Let me list the divisors of 54 again: 1,2,3,6,9,18,27,54. So divisors ≥5 are 6,9,18,27,54. But y must be ≤27 because x must be ≥2 (since x=54/y, so if y is 54, x=1 which is less than 2). Therefore, y can be 6,9,18,27. Therefore, corresponding x's are 54/6=9, 54/9=6, 54/18=3, 54/27=2. Therefore, x can be 9,6,3,2. Which is the same as before. Therefore, four values. Therefore, the answer is 4. Therefore, the number of possible values of x is 4.

But wait, let me check if there are other divisors between 5 and 27. For example, 5 is not a divisor. 7? No. 8? No. 10? No. 12? 54 divided by 12 is 4.5, not integer. 14? No. 15? No. 16? No. 17? No. 19? No. 20? No. 21? No. 22? No. 23? No. 24? No. 25? No. 26? No. So only 6,9,18,27. Therefore, yes, four values. Therefore, the answer is 4. Therefore, the number of possible x is 4.

But wait, let me check once again. Maybe I made a mistake in considering the constraints. The problem says "each player getting y cards", so y must be at least 5. So y ≥5. So if y is 6, that's okay. If y is 5, then x would be 54/5=10.8, which is not integer. Therefore, y cannot be 5. Therefore, the possible y's are 6,9,18,27, which gives x=9,6,3,2. Therefore, four values. Therefore, the answer is 4. Therefore, the number of possible values of x is 4. Therefore, the answer is \boxed{4}?

Wait, but let me check once more. Let me think if there's another way to approach this problem. For example, considering that x must divide 54, x ≥2, and 54/x ≥5. So x divides 54, x ≥2, and x ≤54/5=10.8. Therefore, x is in the set {2,3,6,9} as divisors of 54 between 2 and 10.8. Which is exactly what I found before. Therefore, 4 values. Therefore, the answer is 4. Therefore, I think that is correct.

But just to be thorough, let me check each possible x:

x=2: y=27 ≥5: valid

x=3: y=18 ≥5: valid

x=4: 54/4=13.5: not integer, invalid

x=5: 54/5=10.8: not integer, invalid

x=6: y=9 ≥5: valid

x=7: 54/7≈7.71: not integer

x=8: 54/8=6.75: not integer

x=9: y=6 ≥5: valid

x=10: 54/10=5.4: not integer

x=11: 54/11≈4.9: less than 5, invalid

So, only x=2,3,6,9 are valid. Therefore, 4 values. Therefore, the answer is 4. Therefore, I think that is correct.

But wait, hold on. Wait, the problem says "how many possible values of x are there?" So if x can be 2,3,6,9, that's four values. Therefore, the answer is 4. Therefore, I think that is correct.

But let me check once again if there's a possibility that I missed a divisor. For example, 54 divided by 1 is 54, but x must be at least 2. Divisors of 54 are 1,2,3,6,9,18,27,54. So between 2 and 10.8, the divisors are 2,3,6,9. So yes, four numbers. Therefore, answer is 4. Therefore, I think that is correct.

But just to make sure, let me think of another approach. Let me think of all possible pairs (x,y) such that x*y=54, x≥2, y≥5. So, for each divisor x of 54 with x≥2, check if y=54/x ≥5. So:

x=2: y=27 ≥5: yes

x=3: y=18 ≥5: yes

x=6: y=9 ≥5: yes

x=9: y=6 ≥5: yes

x=18: y=3 <5: no

x=27: y=2 <5: no

x=54: y=1 <5: no

So only four x's. Therefore, answer is 4. Therefore, I think that is correct.

Therefore, the number of possible values of x is 4. Therefore, the answer is \boxed{4}.

**Final Answer**
\boxed{4}

To determine how many possible values of $ x $ (the number of players) are valid under the given constraints, we begin by analyzing the relationship between the number of players and the number of cards each player receives.

---

### Step 1: Understand the Relationship

We are given a deck of 54 cards, and all cards are dealt to $ x $ players, with each player receiving $ y $ cards. This gives the equation:

$$
x \cdot y = 54
$$

We are told that:

- $ x \geq 2 $
- $ y \geq 5 $

Our goal is to find how many **integer values of $ x $** satisfy these conditions.

---

### Step 2: Find All Divisors of 54

Since $ x \cdot y = 54 $, both $ x $ and $ y $ must be **positive integer divisors** of 54. The divisors of 54 are:

$$
1, 2, 3, 6, 9, 18, 27, 54
$$

We are only interested in values of $ x $ such that:

- $ x \geq 2 $
- $ y = \frac{54}{x} \geq 5 $

---

### Step 3: Apply the Constraints

We now test each divisor of 54 that is at least 2, and check whether the corresponding $ y = \frac{54}{x} $ is at least 5.

| $ x $ | $ y = \frac{54}{x} $ | Valid? |
|--------|------------------------|--------|
| 2      | 27                     | Yes    |
| 3      | 18                     | Yes    |
| 6      | 9                      | Yes    |
| 9      | 6                      | Yes    |
| 18     | 3                      | No     |
| 27     | 2                      | No     |
| 54     | 1                      | No     |

Only the values $ x = 2, 3, 6, 9 $ satisfy both $ x \geq 2 $ and $ y \geq 5 $.

---

### Step 4: Count the Valid Values

From the table, the valid values of $ x $ are:

$$
x = 2, 3, 6, 9
$$

Thus, there are **4** valid values of $ x $.

---

### Final Answer

$$
\boxed{4}
$$

Solution: We want $xy=54=2 \cdot 3^3$ such that $x$ is at least $2$ and $y$ is at least  $5$. Thus, the possible combinations $(x,y)$ are $(2,27)$, $(3,18)$, $(6,9)$, and $(9,6)$. There are $\boxed{4}$ such combinations.


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 factored as:

1*36, 2*18, 3*12, 4*9, 6*6. But only those 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 two pairs. Therefore, answer is 2. Which matches with 2^2 / 2 = 2. So that works. Therefore, in our original problem, since there are 4 primes, the number of ordered pairs with m < n is 2^4 / 2 = 8. Therefore, the answer is 8? But wait, let me check again.

Wait, in the original problem, m * n = 210^2, which is (2^2 * 3^2 * 5^2 * 7^2). So the primes are 2, 3, 5, 7. Each of these primes can be assigned to m or n. So for each prime, we have two choices. Therefore, total number of assignments is 2^4 = 16. Each assignment gives a unique pair (m, n). However, since m and n are different (as we saw earlier), each pair is counted twice. Therefore, the number of unordered pairs is 8, and the number of ordered pairs with m < n is 8. Therefore, there are 8 possible values for a. Therefore, the answer is 8?

Wait, but let me think again. Let me think of how exactly the assignments work. For each prime, we decide whether to put it in m or in n. For example, if we assign all primes to m, then m = 2^2 * 3^2 * 5^2 * 7^2 = 210^2, and n = 1. Similarly, if we assign all primes to n, then m = 1 and n = 210^2. But since we want m < n, we need to count only those assignments where m < n. How many such assignments are there?

Since for each assignment, either m < n, m > n, or m = n. But m = n is impossible here, as we saw. Therefore, exactly half of the assignments will have m < n, and the other half will have m > n. Therefore, since there are 16 assignments, half of them (8) will have m < n. Therefore, the answer is 8. Therefore, there are 8 possible values for a.

But let me check with another example. Let me take the case where m * n = p^2, where p is prime. Then, the coprime pairs (m, n) would be (1, p^2) and (p^2, 1). So only one unordered pair, which would correspond to 1 ordered pair with m < n. According to the formula, number of primes is 1, so 2^1 / 2 = 1. Correct. So that works.

Another example: m * n = p^2 * q^2, with p and q distinct primes. Then, the number of coprime pairs (m, n) with m < n would be... Let's see. The assignments: for each prime, assign to m or n. So 2^2 = 4 assignments. The unordered pairs are 2. The ordered pairs with m < n are 2. Let me list them:

Assign p and q to m: m = p^2 * q^2, n = 1. So m > n.

Assign p to m, q to n: m = p^2, n = q^2. Since p ≠ q, m and n are different. Which one is bigger? It depends on p and q. Wait, but in our case, since m and n are determined by the assignment, but since we have m * n = p^2 * q^2, and m and n are coprime. So if we assign p to m and q to n, then m = p^2, n = q^2. Similarly, if we assign q to m and p to n, then m = q^2, n = p^2. So depending on which is bigger, p or q. But since p and q are distinct primes, say p < q, then p^2 < q^2. Therefore, the pair (p^2, q^2) would have m < n. Similarly, if we assign both primes to n, then m = 1, n = p^2 * q^2. So m < n. Assign both primes to m, then m = p^2 * q^2, n = 1, so m > n. Therefore, the assignments that result in m < n are:

- Assign both primes to n: m = 1, n = p^2 q^2.

- Assign p to m and q to n: m = p^2, n = q^2.

- Assign q to m and p to n: m = q^2, n = p^2. But since p < q, this would be m > n, so not counted.

Wait, so actually, only two assignments where m < n: assigning both primes to n, and assigning p to m and q to n. Wait, but that's two assignments. But according to the formula, it should be 2^2 / 2 = 2. So that works. Therefore, in this case, there are two ordered pairs with m < n. Therefore, the formula holds. Therefore, in our original problem, with 4 primes, the number of ordered pairs with m < n is 2^4 / 2 = 8. Therefore, the answer is 8.

But let me think again. Wait, in the case where m * n = p^2 * q^2 * r^2 * s^2, with four distinct primes, then the number of coprime pairs (m, n) with m < n is 2^4 / 2 = 8. Therefore, the answer is 8. Therefore, the number of possible values for a is 8.

But let me think if there's another way to approach this. Since m and n are coprime and m * n = 210^2, then m and n are divisors of 210^2, coprime, and their product is 210^2. Therefore, m and n are complementary divisors in the sense that they are coprime and multiply to 210^2. Therefore, the number of such pairs is equal to 2^{k}, where k is the number of distinct prime factors of 210^2. Wait, but 210^2 has the same prime factors as 210, which are 2, 3, 5, 7. So k = 4. Therefore, the number of coprime pairs (m, n) with m * n = 210^2 is 2^4 = 16. But since each pair is counted twice except when m = n, which isn't possible here, so the number of unordered pairs is 8, and ordered pairs with m < n is 8. Therefore, the answer is 8.

But let me check with another approach. Let me think about the exponents. Since m and n are coprime, and m * n = 2^2 * 3^2 * 5^2 * 7^2. For each prime, the entire exponent must go to either m or n. For example, for prime 2, which has exponent 2 in 210^2, we can assign all 2^2 to m or to n. Similarly for 3, 5, 7. Therefore, for each prime, 2 choices. Therefore, total number of ways is 2^4 = 16. Each such way gives a unique pair (m, n). Since m and n are determined by the assignment of primes. Therefore, the number of ordered pairs is 16. But since we want m < n, we need to count how many of these 16 ordered pairs satisfy m < n.

Since for each unordered pair {m, n}, there are two ordered pairs: (m, n) and (n, m). Except when m = n, which isn't possible here. Therefore, the number of ordered pairs with m < n is exactly half of 16, which is 8. Therefore, the answer is 8.

Therefore, the number of possible values for a is 8. Therefore, the answer is \boxed{8}.

Wait, but let me think again. Let me think of the actual values. For example, take the case where m is 1. Then n is 210^2. So a = 210 * 1 = 210. Then, another case where m is 2^2, so m = 4, then n would be (210^2)/4 = (2^2 * 3^2 * 5^2 * 7^2)/2^2 = 3^2 * 5^2 * 7^2. Therefore, n = (3*5*7)^2 = 105^2. Therefore, a = 210 * 4 = 840. Similarly, if m is 3^2, then n is 2^2 * 5^2 * 7^2, so a = 210 * 9 = 1890. Similarly, m could be 5^2, then a = 210 * 25 = 5250. Similarly, m could be 7^2, then a = 210 * 49 = 10290.

Then, there are cases where m is a product of two primes squared. For example, m = 2^2 * 3^2, then n = 5^2 * 7^2. Therefore, a = 210 * (2^2 * 3^2) = 210 * 36 = 7560. Similarly, m could be 2^2 * 5^2, then a = 210 * (4 * 25) = 210 * 100 = 21000. Similarly, m = 2^2 * 7^2, a = 210 * (4 * 49) = 210 * 196 = 41160. Similarly, m = 3^2 * 5^2, a = 210 * (9 * 25) = 210 * 225 = 47250. m = 3^2 * 7^2, a = 210 * (9 * 49) = 210 * 441 = 92610. m = 5^2 * 7^2, a = 210 * (25 * 49) = 210 * 1225 = 257250.

Then, there are cases where m is a product of three primes squared. For example, m = 2^2 * 3^2 * 5^2, then n = 7^2. Therefore, a = 210 * (4 * 9 * 25) = 210 * 900 = 189000. Similarly, m = 2^2 * 3^2 * 7^2, a = 210 * (4 * 9 * 49) = 210 * 1764 = 370440. m = 2^2 * 5^2 * 7^2, a = 210 * (4 * 25 * 49) = 210 * 4900 = 1029000. m = 3^2 * 5^2 * 7^2, a = 210 * (9 * 25 * 49) = 210 * 11025 = 2315250.

Finally, m = 2^2 * 3^2 * 5^2 * 7^2, which is 210^2, so n = 1. Therefore, a = 210 * 210^2 = 210^3. But since a < b, and b = 210 * n = 210 * 1 = 210. But then a = 210^3 and b = 210, which would mean a > b, which contradicts a < b. Therefore, this case is excluded. Similarly, the case where m = 1, n = 210^2, gives a = 210 * 1 = 210, and b = 210 * 210^2 = 210^3, which satisfies a < b.

So, how many such cases are there? Let me count:

- m = 1: 1 case.

- m = p^2 for each prime p: 4 cases (2^2, 3^2, 5^2, 7^2).

- m = p^2 * q^2 for each pair of primes p, q: C(4, 2) = 6 cases.

- m = p^2 * q^2 * r^2 for each triplet of primes p, q, r: C(4, 3) = 4 cases.

- m = p^2 * q^2 * r^2 * s^2: 1 case (all primes), but this gives a > b, so excluded.

Therefore, total number of valid m's (with m < n) is 1 + 4 + 6 + 4 = 15? Wait, but earlier we thought it was 8. Wait, there's a contradiction here. Wait, according to this breakdown, there are 1 + 4 + 6 + 4 = 15? But that can't be. Wait, no, because when we count m's, each of these corresponds to different assignments. Wait, but earlier we thought that there are 8 possible values. So where is the mistake here?

Wait, let me think again. The problem is that when I considered m as products of primes squared, I considered different combinations, but actually, each assignment corresponds to a unique m. However, the number of assignments is 2^4 = 16. But when we exclude the case where m = 210^2 (which gives a > b), we have 15 assignments. But since each unordered pair is counted twice, except when m = n, which isn't possible. Wait, but earlier we said that there are 16 assignments, which correspond to 8 unordered pairs. But according to this breakdown, there are 1 + 4 + 6 + 4 = 15? That can't be. There must be a miscalculation here.

Wait, let me recount. The number of assignments where m is assigned certain primes. Each assignment corresponds to a unique m. For example:

- Assign no primes to m: m = 1, n = 210^2.

- Assign one prime to m: 4 possibilities (each single prime).

- Assign two primes to m: C(4, 2) = 6 possibilities.

- Assign three primes to m: C(4, 3) = 4 possibilities.

- Assign all four primes to m: 1 possibility.

Total: 1 + 4 + 6 + 4 + 1 = 16. So that's correct. Now, when we consider m < n, we need to exclude the case where m = 210^2 (which would be when all primes are assigned to m, leading to n = 1, so m > n). Therefore, the number of assignments where m < n is total assignments (16) minus the case where m > n (which is 1 case: all primes assigned to m) minus the case where m = n (which is 0). Wait, but actually, for each assignment, either m < n or m > n. Since there are 16 assignments, and each unordered pair is counted twice, except when m = n. Therefore, the number of assignments where m < n is equal to the number of assignments where m > n. Therefore, since there are 16 assignments, and none have m = n, then the number of assignments with m < n is 16 / 2 = 8. Therefore, the number of valid m's is 8. However, according to the breakdown above, when we list the assignments:

- Assign no primes to m: m = 1, n = 210^2. This is m < n.

- Assign one prime to m: 4 cases. Each of these would have m = p^2, n = (product of other primes squared). Since m = p^2 and n = (product of other primes squared). Since the product of other primes squared is larger than p^2? Not necessarily. For example, if p is the smallest prime, 2, then n would be 3^2 * 5^2 * 7^2, which is much larger than 2^2. Similarly, if p is 3, then n would be 2^2 * 5^2 * 7^2, which is larger than 3^2. Similarly for 5 and 7. Therefore, all these 4 cases have m < n.

- Assign two primes to m: C(4, 2) = 6 cases. For each of these, m is the product of two primes squared, and n is the product of the other two primes squared. Now, which of these have m < n? It depends on the primes. For example, if m is 2^2 * 3^2, then n is 5^2 * 7^2. Since 2*3 = 6 and 5*7 = 35, so m = 6^2 = 36, n = 35^2 = 1225. So m < n. Similarly, if m is 2^2 * 5^2, then n is 3^2 * 7^2. 2*5 = 10, 3*7 = 21. So m = 10^2 = 100, n = 21^2 = 441. m < n. Similarly, m = 2^2 * 7^2, n = 3^2 * 5^2. 2*7 = 14, 3*5 = 15. So m = 14^2 = 196, n = 15^2 = 225. m < n. Similarly, m = 3^2 * 5^2, n = 2^2 * 7^2. 3*5 = 15, 2*7 = 14. So m = 15^2 = 225, n = 14^2 = 196. So here, m > n. Therefore, this case would not be counted. Similarly, m = 3^2 * 7^2, n = 2^2 * 5^2. 3*7 = 21, 2*5 = 10. m = 21^2 = 441, n = 10^2 = 100. So m > n. Similarly, m = 5^2 * 7^2, n = 2^2 * 3^2. 5*7 = 35, 2*3 = 6. m = 35^2 = 1225, n = 6^2 = 36. So m > n. Therefore, out of the 6 cases where two primes are assigned to m, only the first three cases (where the two primes are the smaller ones) have m < n, and the other three have m > n. Therefore, in this case, only 3 of the 6 assignments result in m < n.

Wait, so this contradicts the previous conclusion that half of the assignments result in m < n. Therefore, there must be an error in my previous reasoning.

Wait, so if we have 6 assignments where two primes are assigned to m, and only 3 of them result in m < n, then the total number of assignments with m < n is:

- 1 (no primes assigned) + 4 (one prime assigned) + 3 (two primes assigned) + ?

Then, what about assignments with three primes assigned to m? Let's check. Assign three primes to m, so m is the product of three primes squared, and n is the product of one prime squared. For example, m = 2^2 * 3^2 * 5^2, n = 7^2. Then m = (2*3*5)^2 = 30^2 = 900, n = 7^2 = 49. So m > n. Similarly, m = 2^2 * 3^2 * 7^2, n = 5^2. m = (2*3*7)^2 = 42^2 = 1764, n = 25. m > n. Similarly, m = 2^2 * 5^2 * 7^2, n = 3^2. m = (2*5*7)^2 = 70^2 = 4900, n = 9. m > n. Similarly, m = 3^2 * 5^2 * 7^2, n = 2^2. m = (3*5*7)^2 = 105^2 = 11025, n = 4. m > n. Therefore, all four assignments where three primes are assigned to m result in m > n. Therefore, they are excluded.

Then, the case where all four primes are assigned to m: m = 210^2, n = 1, which is m > n, excluded.

Therefore, total number of assignments with m < n is:

- 1 (no primes assigned) + 4 (one prime assigned) + 3 (two primes assigned) = 1 + 4 + 3 = 8.

Ah! Therefore, the total is 8. Therefore, the previous reasoning was correct. The mistake was in my initial breakdown where I thought that all assignments with two primes assigned to m would result in m < n, but actually, only half of them do. Therefore, the total number of valid assignments is 8. Therefore, the answer is 8.

Therefore, the number of possible values for a is 8. Therefore, the answer is \boxed{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** there are for $ a $.

---

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

We know 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:

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

with $ \gcd(m, n) = 1 $, since the GCD of $ a $ and $ b $ is $ d $, and we have factored out $ d $.

Then:

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

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 values exist.

---

### Step 3: Prime Factorization of $ 210^2 $

We factor $ 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 $ \gcd(m, n) = 1 $.

This means that **each prime factor must go entirely to either $ m $ or $ n $**, but not both. So for each of the 4 primes (2, 3, 5, 7), we have 2 choices: assign it to $ m $ or to $ n $.

Thus, the total number of such assignments is:

$$
2^4 = 16
$$

Each assignment corresponds to a unique pair $ (m, n) $, and since $ m \cdot n = 210^2 $, and $ \gcd(m, n) = 1 $, each pair is **distinct** and **ordered**.

However, since $ m < n $, we are only interested in the **ordered pairs** where $ m < n $. Since the total number of such pairs is 16, and each unordered pair $ \{m, n\} $ is counted twice (once as $ (m, n) $, once as $ (n, m) $), the number of **unordered** pairs is:

$$
\frac{16}{2} = 8
$$

Therefore, the number of **ordered pairs** $ (m, n) $ with $ m < n $ is also 8.

---

### Step 4: Conclusion

Each such pair $ (m, n) $ with $ m < n $ gives a unique value of $ a = 210m $, and since there are 8 such pairs, there are **8 possible values** for $ 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 factored as:

1*36, 2*18, 3*12, 4*9, 6*6. But only those 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 two pairs. Therefore, answer is 2. Which matches with 2^2 / 2 = 2. So that works. Therefore, in our original problem, since there are 4 primes, the number of ordered pairs with m < n is 2^4 / 2 = 8. Therefore, the answer is 8? But wait, let me check again.

Wait, in the original problem, m * n = 210^2, which is (2^2 * 3^2 * 5^2 * 7^2). So the primes are 2, 3, 5, 7. Each of these primes can be assigned to m or n. So for each prime, we have two choices. Therefore, total number of assignments is 2^4 = 16. Each assignment gives a unique pair (m, n). However, since m and n are different (as we saw earlier), each pair is counted twice. Therefore, the number of unordered pairs is 8, and the number of ordered pairs with m < n is 8. Therefore, there are 8 possible values for a. Therefore, the answer is 8?

Wait, but let me think again. Let me think of how exactly the assignments work. For each prime, we decide whether to put it in m or in n. For example, if we assign all primes to m, then m = 2^2 * 3^2 * 5^2 * 7^2 = 210^2, and n = 1. Similarly, if we assign all primes to n, then m = 1 and n = 210^2. But since we want m < n, we need to count only those assignments where m < n. How many such assignments are there?

Since for each assignment, either m < n, m > n, or m = n. But m = n is impossible here, as we saw. Therefore, exactly half of the assignments will have m < n, and the other half will have m > n. Therefore, since there are 16 assignments, half of them (8) will have m < n. Therefore, the answer is 8. Therefore, there are 8 possible values for a.

But let me check with another example. Let me take the case where m * n = p^2, where p is prime. Then, the coprime pairs (m, n) would be (1, p^2) and (p^2, 1). So only one unordered pair, which would correspond to 1 ordered pair with m < n. According to the formula, number of primes is 1, so 2^1 / 2 = 1. Correct. So that works.

Another example: m * n = p^2 * q^2, with p and q distinct primes. Then, the number of coprime pairs (m, n) with m < n would be... Let's see. The assignments: for each prime, assign to m or n. So 2^2 = 4 assignments. The unordered pairs are 2. The ordered pairs with m < n are 2. Let me list them:

Assign p and q to m: m = p^2 * q^2, n = 1. So m > n.

Assign p to m, q to n: m = p^2, n = q^2. Since p ≠ q, m and n are different. Which one is bigger? It depends on p and q. Wait, but in our case, since m and n are determined by the assignment, but since we have m * n = p^2 * q^2, and m and n are coprime. So if we assign p to m and q to n, then m = p^2, n = q^2. Similarly, if we assign q to m and p to n, then m = q^2, n = p^2. So depending on which is bigger, p or q. But since p and q are distinct primes, say p < q, then p^2 < q^2. Therefore, the pair (p^2, q^2) would have m < n. Similarly, if we assign both primes to n, then m = 1, n = p^2 * q^2. So m < n. Assign both primes to m, then m = p^2 * q^2, n = 1, so m > n. Therefore, the assignments that result in m < n are:

- Assign both primes to n: m = 1, n = p^2 q^2.

- Assign p to m and q to n: m = p^2, n = q^2.

- Assign q to m and p to n: m = q^2, n = p^2. But since p < q, this would be m > n, so not counted.

Wait, so actually, only two assignments where m < n: assigning both primes to n, and assigning p to m and q to n. Wait, but that's two assignments. But according to the formula, it should be 2^2 / 2 = 2. So that works. Therefore, in this case, there are two ordered pairs with m < n. Therefore, the formula holds. Therefore, in our original problem, with 4 primes, the number of ordered pairs with m < n is 2^4 / 2 = 8. Therefore, the answer is 8.

But let me think again. Wait, in the case where m * n = p^2 * q^2 * r^2 * s^2, with four distinct primes, then the number of coprime pairs (m, n) with m < n is 2^4 / 2 = 8. Therefore, the answer is 8. Therefore, the number of possible values for a is 8.

But let me think if there's another way to approach this. Since m and n are coprime and m * n = 210^2, then m and n are divisors of 210^2, coprime, and their product is 210^2. Therefore, m and n are complementary divisors in the sense that they are coprime and multiply to 210^2. Therefore, the number of such pairs is equal to 2^{k}, where k is the number of distinct prime factors of 210^2. Wait, but 210^2 has the same prime factors as 210, which are 2, 3, 5, 7. So k = 4. Therefore, the number of coprime pairs (m, n) with m * n = 210^2 is 2^4 = 16. But since each pair is counted twice except when m = n, which isn't possible here, so the number of unordered pairs is 8, and ordered pairs with m < n is 8. Therefore, the answer is 8.

But let me check with another approach. Let me think about the exponents. Since m and n are coprime, and m * n = 2^2 * 3^2 * 5^2 * 7^2. For each prime, the entire exponent must go to either m or n. For example, for prime 2, which has exponent 2 in 210^2, we can assign all 2^2 to m or to n. Similarly for 3, 5, 7. Therefore, for each prime, 2 choices. Therefore, total number of ways is 2^4 = 16. Each such way gives a unique pair (m, n). Since m and n are determined by the assignment of primes. Therefore, the number of ordered pairs is 16. But since we want m < n, we need to count how many of these 16 ordered pairs satisfy m < n.

Since for each unordered pair {m, n}, there are two ordered pairs: (m, n) and (n, m). Except when m = n, which isn't possible here. Therefore, the number of ordered pairs with m < n is exactly half of 16, which is 8. Therefore, the answer is 8.

Therefore, the number of possible values for a is 8. Therefore, the answer is \boxed{8}.

Wait, but let me think again. Let me think of the actual values. For example, take the case where m is 1. Then n is 210^2. So a = 210 * 1 = 210. Then, another case where m is 2^2, so m = 4, then n would be (210^2)/4 = (2^2 * 3^2 * 5^2 * 7^2)/2^2 = 3^2 * 5^2 * 7^2. Therefore, n = (3*5*7)^2 = 105^2. Therefore, a = 210 * 4 = 840. Similarly, if m is 3^2, then n is 2^2 * 5^2 * 7^2, so a = 210 * 9 = 1890. Similarly, m could be 5^2, then a = 210 * 25 = 5250. Similarly, m could be 7^2, then a = 210 * 49 = 10290.

Then, there are cases where m is a product of two primes squared. For example, m = 2^2 * 3^2, then n = 5^2 * 7^2. Therefore, a = 210 * (2^2 * 3^2) = 210 * 36 = 7560. Similarly, m could be 2^2 * 5^2, then a = 210 * (4 * 25) = 210 * 100 = 21000. Similarly, m = 2^2 * 7^2, a = 210 * (4 * 49) = 210 * 196 = 41160. Similarly, m = 3^2 * 5^2, a = 210 * (9 * 25) = 210 * 225 = 47250. m = 3^2 * 7^2, a = 210 * (9 * 49) = 210 * 441 = 92610. m = 5^2 * 7^2, a = 210 * (25 * 49) = 210 * 1225 = 257250.

Then, there are cases where m is a product of three primes squared. For example, m = 2^2 * 3^2 * 5^2, then n = 7^2. Therefore, a = 210 * (4 * 9 * 25) = 210 * 900 = 189000. Similarly, m = 2^2 * 3^2 * 7^2, a = 210 * (4 * 9 * 49) = 210 * 1764 = 370440. m = 2^2 * 5^2 * 7^2, a = 210 * (4 * 25 * 49) = 210 * 4900 = 1029000. m = 3^2 * 5^2 * 7^2, a = 210 * (9 * 25 * 49) = 210 * 11025 = 2315250.

Finally, m = 2^2 * 3^2 * 5^2 * 7^2, which is 210^2, so n = 1. Therefore, a = 210 * 210^2 = 210^3. But since a < b, and b = 210 * n = 210 * 1 = 210. But then a = 210^3 and b = 210, which would mean a > b, which contradicts a < b. Therefore, this case is excluded. Similarly, the case where m = 1, n = 210^2, gives a = 210 * 1 = 210, and b = 210 * 210^2 = 210^3, which satisfies a < b.

So, how many such cases are there? Let me count:

- m = 1: 1 case.

- m = p^2 for each prime p: 4 cases (2^2, 3^2, 5^2, 7^2).

- m = p^2 * q^2 for each pair of primes p, q: C(4, 2) = 6 cases.

- m = p^2 * q^2 * r^2 for each triplet of primes p, q, r: C(4, 3) = 4 cases.

- m = p^2 * q^2 * r^2 * s^2: 1 case (all primes), but this gives a > b, so excluded.

Therefore, total number of valid m's (with m < n) is 1 + 4 + 6 + 4 = 15? Wait, but earlier we thought it was 8. Wait, there's a contradiction here. Wait, according to this breakdown, there are 1 + 4 + 6 + 4 = 15? But that can't be. Wait, no, because when we count m's, each of these corresponds to different assignments. Wait, but earlier we thought that there are 8 possible values. So where is the mistake here?

Wait, let me think again. The problem is that when I considered m as products of primes squared, I considered different combinations, but actually, each assignment corresponds to a unique m. However, the number of assignments is 2^4 = 16. But when we exclude the case where m = 210^2 (which gives a > b), we have 15 assignments. But since each unordered pair is counted twice, except when m = n, which isn't possible. Wait, but earlier we said that there are 16 assignments, which correspond to 8 unordered pairs. But according to this breakdown, there are 1 + 4 + 6 + 4 = 15? That can't be. There must be a miscalculation here.

Wait, let me recount. The number of assignments where m is assigned certain primes. Each assignment corresponds to a unique m. For example:

- Assign no primes to m: m = 1, n = 210^2.

- Assign one prime to m: 4 possibilities (each single prime).

- Assign two primes to m: C(4, 2) = 6 possibilities.

- Assign three primes to m: C(4, 3) = 4 possibilities.

- Assign all four primes to m: 1 possibility.

Total: 1 + 4 + 6 + 4 + 1 = 16. So that's correct. Now, when we consider m < n, we need to exclude the case where m = 210^2 (which would be when all primes are assigned to m, leading to n = 1, so m > n). Therefore, the number of assignments where m < n is total assignments (16) minus the case where m > n (which is 1 case: all primes assigned to m) minus the case where m = n (which is 0). Wait, but actually, for each assignment, either m < n or m > n. Since there are 16 assignments, and each unordered pair is counted twice, except when m = n. Therefore, the number of assignments where m < n is equal to the number of assignments where m > n. Therefore, since there are 16 assignments, and none have m = n, then the number of assignments with m < n is 16 / 2 = 8. Therefore, the number of valid m's is 8. However, according to the breakdown above, when we list the assignments:

- Assign no primes to m: m = 1, n = 210^2. This is m < n.

- Assign one prime to m: 4 cases. Each of these would have m = p^2, n = (product of other primes squared). Since m = p^2 and n = (product of other primes squared). Since the product of other primes squared is larger than p^2? Not necessarily. For example, if p is the smallest prime, 2, then n would be 3^2 * 5^2 * 7^2, which is much larger than 2^2. Similarly, if p is 3, then n would be 2^2 * 5^2 * 7^2, which is larger than 3^2. Similarly for 5 and 7. Therefore, all these 4 cases have m < n.

- Assign two primes to m: C(4, 2) = 6 cases. For each of these, m is the product of two primes squared, and n is the product of the other two primes squared. Now, which of these have m < n? It depends on the primes. For example, if m is 2^2 * 3^2, then n is 5^2 * 7^2. Since 2*3 = 6 and 5*7 = 35, so m = 6^2 = 36, n = 35^2 = 1225. So m < n. Similarly, if m is 2^2 * 5^2, then n is 3^2 * 7^2. 2*5 = 10, 3*7 = 21. So m = 10^2 = 100, n = 21^2 = 441. m < n. Similarly, m = 2^2 * 7^2, n = 3^2 * 5^2. 2*7 = 14, 3*5 = 15. So m = 14^2 = 196, n = 15^2 = 225. m < n. Similarly, m = 3^2 * 5^2, n = 2^2 * 7^2. 3*5 = 15, 2*7 = 14. So m = 15^2 = 225, n = 14^2 = 196. So here, m > n. Therefore, this case would not be counted. Similarly, m = 3^2 * 7^2, n = 2^2 * 5^2. 3*7 = 21, 2*5 = 10. m = 21^2 = 441, n = 10^2 = 100. So m > n. Similarly, m = 5^2 * 7^2, n = 2^2 * 3^2. 5*7 = 35, 2*3 = 6. m = 35^2 = 1225, n = 6^2 = 36. So m > n. Therefore, out of the 6 cases where two primes are assigned to m, only the first three cases (where the two primes are the smaller ones) have m < n, and the other three have m > n. Therefore, in this case, only 3 of the 6 assignments result in m < n.

Wait, so this contradicts the previous conclusion that half of the assignments result in m < n. Therefore, there must be an error in my previous reasoning.

Wait, so if we have 6 assignments where two primes are assigned to m, and only 3 of them result in m < n, then the total number of assignments with m < n is:

- 1 (no primes assigned) + 4 (one prime assigned) + 3 (two primes assigned) + ?

Then, what about assignments with three primes assigned to m? Let's check. Assign three primes to m, so m is the product of three primes squared, and n is the product of one prime squared. For example, m = 2^2 * 3^2 * 5^2, n = 7^2. Then m = (2*3*5)^2 = 30^2 = 900, n = 7^2 = 49. So m > n. Similarly, m = 2^2 * 3^2 * 7^2, n = 5^2. m = (2*3*7)^2 = 42^2 = 1764, n = 25. m > n. Similarly, m = 2^2 * 5^2 * 7^2, n = 3^2. m = (2*5*7)^2 = 70^2 = 4900, n = 9. m > n. Similarly, m = 3^2 * 5^2 * 7^2, n = 2^2. m = (3*5*7)^2 = 105^2 = 11025, n = 4. m > n. Therefore, all four assignments where three primes are assigned to m result in m > n. Therefore, they are excluded.

Then, the case where all four primes are assigned to m: m = 210^2, n = 1, which is m > n, excluded.

Therefore, total number of assignments with m < n is:

- 1 (no primes assigned) + 4 (one prime assigned) + 3 (two primes assigned) = 1 + 4 + 3 = 8.

Ah! Therefore, the total is 8. Therefore, the previous reasoning was correct. The mistake was in my initial breakdown where I thought that all assignments with two primes assigned to m would result in m < n, but actually, only half of them do. Therefore, the total number of valid assignments is 8. Therefore, the answer is 8.

Therefore, the number of possible values for a is 8. Therefore, the answer is \boxed{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** there are for $ a $.

---

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

We know 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:

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

with $ \gcd(m, n) = 1 $, since the GCD of $ a $ and $ b $ is $ d $, and we have factored out $ d $.

Then:

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

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 values exist.

---

### Step 3: Prime Factorization of $ 210^2 $

We factor $ 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 $ \gcd(m, n) = 1 $.

This means that **each prime factor must go entirely to either $ m $ or $ n $**, but not both. So for each of the 4 primes (2, 3, 5, 7), we have 2 choices: assign it to $ m $ or to $ n $.

Thus, the total number of such assignments is:

$$
2^4 = 16
$$

Each assignment corresponds to a unique pair $ (m, n) $, and since $ m \cdot n = 210^2 $, and $ \gcd(m, n) = 1 $, each pair is **distinct** and **ordered**.

However, since $ m < n $, we are only interested in the **ordered pairs** where $ m < n $. Since the total number of such pairs is 16, and each unordered pair $ \{m, n\} $ is counted twice (once as $ (m, n) $, once as $ (n, m) $), the number of **unordered** pairs is:

$$
\frac{16}{2} = 8
$$

Therefore, the number of **ordered pairs** $ (m, n) $ with $ m < n $ is also 8.

---

### Step 4: Conclusion

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