Explanation:
Okay, let's try to figure out for which real values of x the inequality -4 < x⁴ + 4x² < 21 holds. Hmm, so I need to solve this compound inequality. Let me break it down into two parts: the left part, -4 < x⁴ + 4x², and the right part, x⁴ + 4x² < 21. Then I'll find the intersection of the solutions for both parts.

Starting with the left inequality: -4 < x⁴ + 4x². Let me rearrange this to bring all terms to one side. If I add 4 to both sides, it becomes 0 < x⁴ + 4x² + 4. Wait, that simplifies to 0 < (x² + 2)². Because x⁴ + 4x² + 4 is a perfect square. Let me check: (x² + 2)² = x⁴ + 4x² + 4. Yes, that's correct. So the inequality is 0 < (x² + 2)².

Now, since x² is always non-negative, adding 2 makes x² + 2 always at least 2, right? So x² + 2 is always positive, which means (x² + 2)² is also always positive. Therefore, 0 < (x² + 2)² is always true for all real numbers x. So the left inequality is satisfied for all real numbers. That means the entire solution set for the original compound inequality will depend only on the right inequality: x⁴ + 4x² < 21.

So now I need to solve x⁴ + 4x² < 21. Let me rearrange this inequality to bring all terms to the left side: x⁴ + 4x² - 21 < 0. Now, let me consider this quartic equation. Maybe I can make a substitution to simplify it. Let me set y = x². Then the inequality becomes y² + 4y - 21 < 0.

So solving y² + 4y - 21 < 0. First, let me find the roots of the quadratic equation y² + 4y - 21 = 0. Using the quadratic formula, y = [-4 ± sqrt(16 + 84)] / 2. Let me compute the discriminant: 16 + 84 = 100. So sqrt(100) = 10. Thus, the roots are y = [-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. Expanding (y - 3)(y + 7) = y² + 7y - 3y -21 = y² + 4y -21. Yes, that's correct.

So the inequality is (y - 3)(y + 7) < 0. To solve this quadratic inequality, I need to find 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 two roots. So the solution for y is -7 < y < 3.

But remember that y = x², and x² is always non-negative. Therefore, y cannot be less than 0. So we need to intersect the interval (-7, 3) with y ≥ 0. Therefore, the valid solutions for y are 0 ≤ y < 3.

So translating back to x: y = x² < 3. Therefore, x² < 3. Taking square roots, we get |x| < sqrt(3). Which implies that x is between -sqrt(3) and sqrt(3).

Wait, but hold on. Let me make sure. The original inequality for y was -7 < y < 3. But since y = x², which is always ≥ 0, so the overlap is 0 ≤ y < 3. Therefore, x² < 3. So x is in (-√3, √3).

But let me check if there are any mistakes here. Because when we solved for y, we had -7 < y < 3. But since y must be ≥ 0, the solution is y ∈ [0, 3). Then, yes, x² ∈ [0, 3). Therefore, x ∈ (-√3, √3).

Therefore, combining the results, since the left inequality is always true, the solution to the original compound inequality is x ∈ (-√3, √3).

Wait, but hold on. Let me verify with some test values. Let me check x = 0: x⁴ + 4x² = 0 + 0 = 0. Then 0 is between -4 and 21, so that works.

What about x = sqrt(3)? Then x² = 3, x⁴ = 9. So x⁴ + 4x² = 9 + 12 = 21. But the original inequality is strict: x⁴ + 4x² < 21. So at x = sqrt(3), the value is equal to 21, which is not less than 21. Therefore, sqrt(3) is excluded. Similarly, -sqrt(3) is excluded. So the interval is indeed open at both ends.

But let me check another point outside this interval. Let's take x = 2. Then x² = 4, x⁴ = 16. So 16 + 16 = 32, which is greater than 21, so it's not in the solution set. What about x = 1? x² = 1, x⁴ = 1. 1 + 4 = 5 < 21, so that's good. What about x approaching infinity? As x increases, x⁴ dominates, so the expression x⁴ + 4x² will go to infinity, which is greater than 21. So indeed, the values of x beyond sqrt(3) will not satisfy the inequality.

But just to make sure, let me also check if there are any other points where maybe x⁴ + 4x² equals -4? Wait, the original left inequality is -4 < x⁴ + 4x². But since x⁴ + 4x² is always non-negative (as x⁴ and 4x² are both non-negative), the left inequality is automatically true for all real numbers. Because the smallest value of x⁴ + 4x² is when x = 0, which is 0. So 0 is greater than -4. Therefore, the entire inequality reduces to x⁴ + 4x² < 21, which we solved as x ∈ (-√3, √3).

So the final answer should be the interval from -√3 to √3, not including the endpoints. Therefore, in interval notation, that's (-√3, √3).

But let me check once more. Let me solve x⁴ + 4x² < 21. So x⁴ + 4x² - 21 < 0. Let me consider the quartic function f(x) = x⁴ + 4x² - 21. We can analyze its graph. Since it's a quartic with positive leading coefficient, it tends to infinity as x approaches both positive and negative infinity. The function has a minimum at some point. Let me find its critical points. Taking derivative: f’(x) = 4x³ + 8x. Setting this equal to zero: 4x³ + 8x = 0 ⇒ 4x(x² + 2) = 0. Solutions are x = 0 or x² + 2 = 0 ⇒ x = ±i√2. Since we are dealing with real numbers, the only critical point is at x = 0.

So the function f(x) has a critical point at x = 0. Let me compute f(0): 0 + 0 -21 = -21. So the minimum value of the function is -21 at x = 0. Wait, but how does this relate to solving f(x) < 0? Well, since the function tends to infinity as x approaches ±infty, and there is a minimum at x=0 of -21. Therefore, the equation f(x) = 0 will have two real roots, symmetric around the y-axis. Which is consistent with our previous result where the solutions for y were between -7 and 3, and since y = x², the solutions for x are between -sqrt(3) and sqrt(3).

Therefore, the inequality f(x) < 0 is satisfied between the two roots, which are at x = -sqrt(3) and x = sqrt(3). So that confirms the interval.

Therefore, combining all of this, the original compound inequality is satisfied for all real numbers between -√3 and √3.

Hence, the answer is (-√3, √3) in interval notation.

**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'll break it into two parts and analyze each separately.

---

### Step 1: Analyze the Left Inequality
$$
-4 < x^4 + 4x^2
$$

Let's move all terms to one side:
$$
x^4 + 4x^2 + 4 > 0
$$

Recognize this as a perfect square:
$$
(x^2 + 2)^2 > 0
$$

The expression $(x^2 + 2)^2$ is a square of a real number, which is always **non-negative**, and equals 0 **only when** $ x = 0 $ because $ x^2 + 2 \geq 2 > 0 $ for all real $ x $.

Therefore,
$$
(x^2 + 2)^2 > 0 \quad \text{is true for all real } x \text{ except } x = 0.
$$

So the **left inequality** is **always true** for all real $ x $, except possibly at $ x = 0 $. However, the original left part is strictly greater than $-4$, and since $ x^4 + 4x^2 \geq 0 $, the inequality is satisfied.

---

### Step 2: Analyze the Right Inequality
$$
x^4 + 4x^2 < 21
$$

Bring all terms to one side:
$$
x^4 + 4x^2 - 21 < 0
$$

Let’s apply a substitution to make it easier: Let $ y = x^2 $, so that:
$$
y^2 + 4y - 21 < 0
$$

Solve the quadratic inequality:
$$
y^2 + 4y - 21 = 0
$$

Use the quadratic formula to find the roots:
$$
y = \frac{-4 \pm \sqrt{16 + 84}}{2} = \frac{-4 \pm \sqrt{100}}{2} = \frac{-4 \pm 10}{2}
$$

So,
$$
y = \frac{6}{2} = 3 \quad \text{or} \quad y = \frac{-14}{2} = -7
$$

Now the inequality becomes:
$$
y^2 + 4y - 21 < 0 \quad \Rightarrow \quad (y - 3)(y + 7) < 0
$$

This is true when $ y \in (-7, 3) $

But remember that $ y = x^2 $, and $ x^2 \geq 0 $, so:
$$
y \in [0, 3)
$$

Thus:
$$
x^2 \in [0, 3) \quad \Rightarrow \quad |x| < \sqrt{3}
$$

So the solution set for this part is:
$$
x \in (-\sqrt{3}, \sqrt{3})
$$

---

### Step 3: Combine Both Inequalities

- The **left inequality** is always true for all real $ x $, since $ x^4 + 4x^2 \geq 0 > -4 $ for all real $ x $.
- The **right inequality** is only true for $ x \in (-\sqrt{3}, \sqrt{3}) $.

Therefore, the **final solution set** to the original compound inequality is:
$$
x \in (-\sqrt{3}, \sqrt{3})
$$

---

### Final Answer

$$
\boxed{(-\sqrt{3}, \sqrt{3})}
$$

[Assistant]
Explanation:
Okay, let's try to figure out for which real values of x the inequality -4 < x⁴ + 4x² < 21 holds. Hmm, so I need to solve this compound inequality. Let me break it down into two parts: the left part, -4 < x⁴ + 4x², and the right part, x⁴ + 4x² < 21. Then I'll find the intersection of the solutions for both parts.

Starting with the left inequality: -4 < x⁴ + 4x². Let me rearrange this to bring all terms to one side. If I add 4 to both sides, it becomes 0 < x⁴ + 4x² + 4. Wait, that simplifies to 0 < (x² + 2)². Because x⁴ + 4x² + 4 is a perfect square. Let me check: (x² + 2)² = x⁴ + 4x² + 4. Yes, that's correct. So the inequality is 0 < (x² + 2)².

Now, since x² is always non-negative, adding 2 makes x² + 2 always at least 2, right? So x² + 2 is always positive, which means (x² + 2)² is also always positive. Therefore, 0 < (x² + 2)² is always true for all real numbers x. So the left inequality is satisfied for all real numbers. That means the entire solution set for the original compound inequality will depend only on the right inequality: x⁴ + 4x² < 21.

So now I need to solve x⁴ + 4x² < 21. Let me rearrange this inequality to bring all terms to the left side: x⁴ + 4x² - 21 < 0. Now, let me consider this quartic equation. Maybe I can make a substitution to simplify it. Let me set y = x². Then the inequality becomes y² + 4y - 21 < 0.

So solving y² + 4y - 21 < 0. First, let me find the roots of the quadratic equation y² + 4y - 21 = 0. Using the quadratic formula, y = [-4 ± sqrt(16 + 84)] / 2. Let me compute the discriminant: 16 + 84 = 100. So sqrt(100) = 10. Thus, the roots are y = [-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. Expanding (y - 3)(y + 7) = y² + 7y - 3y -21 = y² + 4y -21. Yes, that's correct.

So the inequality is (y - 3)(y + 7) < 0. To solve this quadratic inequality, I need to find 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 two roots. So the solution for y is -7 < y < 3.

But remember that y = x², and x² is always non-negative. Therefore, y cannot be less than 0. So we need to intersect the interval (-7, 3) with y ≥ 0. Therefore, the valid solutions for y are 0 ≤ y < 3.

So translating back to x: y = x² < 3. Therefore, x² < 3. Taking square roots, we get |x| < sqrt(3). Which implies that x is between -sqrt(3) and sqrt(3).

Wait, but hold on. Let me make sure. The original inequality for y was -7 < y < 3. But since y = x², which is always ≥ 0, so the overlap is 0 ≤ y < 3. Therefore, x² < 3. So x is in (-√3, √3).

But let me check if there are any mistakes here. Because when we solved for y, we had -7 < y < 3. But since y must be ≥ 0, the solution is y ∈ [0, 3). Then, yes, x² ∈ [0, 3). Therefore, x ∈ (-√3, √3).

Therefore, combining the results, since the left inequality is always true, the solution to the original compound inequality is x ∈ (-√3, √3).

Wait, but hold on. Let me verify with some test values. Let me check x = 0: x⁴ + 4x² = 0 + 0 = 0. Then 0 is between -4 and 21, so that works.

What about x = sqrt(3)? Then x² = 3, x⁴ = 9. So x⁴ + 4x² = 9 + 12 = 21. But the original inequality is strict: x⁴ + 4x² < 21. So at x = sqrt(3), the value is equal to 21, which is not less than 21. Therefore, sqrt(3) is excluded. Similarly, -sqrt(3) is excluded. So the interval is indeed open at both ends.

But let me check another point outside this interval. Let's take x = 2. Then x² = 4, x⁴ = 16. So 16 + 16 = 32, which is greater than 21, so it's not in the solution set. What about x = 1? x² = 1, x⁴ = 1. 1 + 4 = 5 < 21, so that's good. What about x approaching infinity? As x increases, x⁴ dominates, so the expression x⁴ + 4x² will go to infinity, which is greater than 21. So indeed, the values of x beyond sqrt(3) will not satisfy the inequality.

But just to make sure, let me also check if there are any other points where maybe x⁴ + 4x² equals -4? Wait, the original left inequality is -4 < x⁴ + 4x². But since x⁴ + 4x² is always non-negative (as x⁴ and 4x² are both non-negative), the left inequality is automatically true for all real numbers. Because the smallest value of x⁴ + 4x² is when x = 0, which is 0. So 0 is greater than -4. Therefore, the entire inequality reduces to x⁴ + 4x² < 21, which we solved as x ∈ (-√3, √3).

So the final answer should be the interval from -√3 to √3, not including the endpoints. Therefore, in interval notation, that's (-√3, √3).

But let me check once more. Let me solve x⁴ + 4x² < 21. So x⁴ + 4x² - 21 < 0. Let me consider the quartic function f(x) = x⁴ + 4x² - 21. We can analyze its graph. Since it's a quartic with positive leading coefficient, it tends to infinity as x approaches both positive and negative infinity. The function has a minimum at some point. Let me find its critical points. Taking derivative: f’(x) = 4x³ + 8x. Setting this equal to zero: 4x³ + 8x = 0 ⇒ 4x(x² + 2) = 0. Solutions are x = 0 or x² + 2 = 0 ⇒ x = ±i√2. Since we are dealing with real numbers, the only critical point is at x = 0.

So the function f(x) has a critical point at x = 0. Let me compute f(0): 0 + 0 -21 = -21. So the minimum value of the function is -21 at x = 0. Wait, but how does this relate to solving f(x) < 0? Well, since the function tends to infinity as x approaches ±infty, and there is a minimum at x=0 of -21. Therefore, the equation f(x) = 0 will have two real roots, symmetric around the y-axis. Which is consistent with our previous result where the solutions for y were between -7 and 3, and since y = x², the solutions for x are between -sqrt(3) and sqrt(3).

Therefore, the inequality f(x) < 0 is satisfied between the two roots, which are at x = -sqrt(3) and x = sqrt(3). So that confirms the interval.

Therefore, combining all of this, the original compound inequality is satisfied for all real numbers between -√3 and √3.

Hence, the answer is (-√3, √3) in interval notation.

**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'll break it into two parts and analyze each separately.

---

### Step 1: Analyze the Left Inequality
$$
-4 < x^4 + 4x^2
$$

Let's move all terms to one side:
$$
x^4 + 4x^2 + 4 > 0
$$

Recognize this as a perfect square:
$$
(x^2 + 2)^2 > 0
$$

The expression $(x^2 + 2)^2$ is a square of a real number, which is always **non-negative**, and equals 0 **only when** $ x = 0 $ because $ x^2 + 2 \geq 2 > 0 $ for all real $ x $.

Therefore,
$$
(x^2 + 2)^2 > 0 \quad \text{is true for all real } x \text{ except } x = 0.
$$

So the **left inequality** is **always true** for all real $ x $, except possibly at $ x = 0 $. However, the original left part is strictly greater than $-4$, and since $ x^4 + 4x^2 \geq 0 $, the inequality is satisfied.

---

### Step 2: Analyze the Right Inequality
$$
x^4 + 4x^2 < 21
$$

Bring all terms to one side:
$$
x^4 + 4x^2 - 21 < 0
$$

Let’s apply a substitution to make it easier: Let $ y = x^2 $, so that:
$$
y^2 + 4y - 21 < 0
$$

Solve the quadratic inequality:
$$
y^2 + 4y - 21 = 0
$$

Use the quadratic formula to find the roots:
$$
y = \frac{-4 \pm \sqrt{16 + 84}}{2} = \frac{-4 \pm \sqrt{100}}{2} = \frac{-4 \pm 10}{2}
$$

So,
$$
y = \frac{6}{2} = 3 \quad \text{or} \quad y = \frac{-14}{2} = -7
$$

Now the inequality becomes:
$$
y^2 + 4y - 21 < 0 \quad \Rightarrow \quad (y - 3)(y + 7) < 0
$$

This is true when $ y \in (-7, 3) $

But remember that $ y = x^2 $, and $ x^2 \geq 0 $, so:
$$
y \in [0, 3)
$$

Thus:
$$
x^2 \in [0, 3) \quad \Rightarrow \quad |x| < \sqrt{3}
$$

So the solution set for this part is:
$$
x \in (-\sqrt{3}, \sqrt{3})
$$

---

### Step 3: Combine Both Inequalities

- The **left inequality** is always true for all real $ x $, since $ x^4 + 4x^2 \geq 0 > -4 $ for all real $ x $.
- The **right inequality** is only true for $ x \in (-\sqrt{3}, \sqrt{3}) $.

Therefore, the **final solution set** to the original compound inequality is:
$$
x \in (-\sqrt{3}, \sqrt{3})
$$

---

### Final Answer

$$
\boxed{(-\sqrt{3}, \sqrt{3})}
$$

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, each 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 the cards to x players, each player gets y cards. So, the total number of cards dealt should be equal to the total number of cards in the deck. That means:

x * y = 54

So, x and y are positive integers such that their product is 54. Also, the problem states that x is at least 2 and y is at least 5. Therefore, we need to find all pairs (x, y) where x ≥ 2, y ≥ 5, and x * y = 54. Then, the question asks for the number of possible values of x.

So essentially, I need to find all divisors x of 54 such that x ≥ 2 and when you divide 54 by x, the result y is at least 5. Therefore, for each divisor x of 54 (with x ≥ 2), check if 54/x ≥ 5. Then count how many such x's there are.

Alternatively, since x * y = 54, and y ≥ 5, that implies that x ≤ 54 / 5 = 10.8. Since x must be an integer, x ≤ 10. Also, since x ≥ 2, the possible values of x are integers from 2 up to 10, but only those that divide 54. Wait, but maybe I need to check that.

Wait, but maybe I need to think in terms of factors of 54. Let me recall the factors of 54.

First, prime factorization of 54: 54 = 2 * 3^3.

Therefore, the divisors of 54 can be found by taking exponents of primes:

For 2: exponent can be 0 or 1.

For 3: exponent can be 0, 1, 2, or 3.

Therefore, total number of divisors is (1+1)*(3+1)= 2*4=8. Let me list them:

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

So these are the divisors of 54. Now, the problem says that x must be at least 2, so we can exclude 1. So possible x values from the divisors are 2, 3, 6, 9, 18, 27, 54. But then we also have the constraint that y must be at least 5. Since y = 54 / x, so 54 / x ≥ 5 => x ≤ 54 / 5 = 10.8. Since x must be an integer, x ≤ 10. Therefore, from the list of divisors above, which are 2, 3, 6, 9, 18, 27, 54, we need to take only those x's that are less than or equal to 10. So let's check:

- 2: yes, 2 ≤ 10
- 3: yes
- 6: yes
- 9: yes
- 18: 18 > 10, exclude
- 27: same, exclude
- 54: exclude

So the possible x values are 2, 3, 6, 9. That's 4 values? Wait, but hold on, let me make sure.

Wait, maybe there are other divisors? Wait, the divisors of 54 are only those 8 numbers. So only 2, 3, 6, 9, 18, 27, 54. So from those, the ones less than or equal to 10 are 2, 3, 6, 9. So that is 4 values. But wait, let me check for each x:

For x=2: y=54/2=27 ≥5, okay.

x=3: y=18 ≥5, okay.

x=6: y=9 ≥5, okay.

x=9: y=6 ≥5, okay.

x=18: y=3, which is less than 5, so invalid.

Similarly x=27: y=2, invalid.

x=54: y=1, invalid.

So only 4 possible values of x. But hold on, is that all?

Wait, but maybe I need to think again. Because maybe x doesn't have to be a divisor of 54? Wait, no. Wait, the problem says you deal all of the cards to x players, with each player getting y cards. So the total number of cards is x*y, which must equal 54. Therefore, x must divide 54. Therefore, x must be a divisor of 54. Therefore, the only possible x's are divisors of 54. Therefore, only those 8 divisors. So we already checked them. So only 4 values of x (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 4 values.

But wait, let me think again. Wait, perhaps there are other possibilities where x is not a divisor of 54, but y is still integer? No, because x*y=54, so if x is not a divisor of 54, then y would not be integer. For example, suppose x=4. Then y=54/4=13.5, which is not an integer. Therefore, since y must be an integer (number of cards per person), x must divide 54. So x must be a divisor. Therefore, our previous list is correct.

Wait, but hold on. Let me check if there are more divisors between 2 and 10. Wait, the divisors of 54 are 1,2,3,6,9,18,27,54. So between 2 and 10, we have 2,3,6,9. So four divisors. Therefore the answer is 4?

Wait, but the problem says "how many possible values of x are there?" So is 4 the answer? But let me check again.

Alternatively, perhaps I need to consider that y must be at least 5, so for each x, check if y=54/x is integer and y≥5. So maybe I can approach this by first finding all pairs (x,y) such that x*y=54, x≥2, y≥5.

So let me list all possible pairs:

Start with x=2: y=27. Okay, since y≥5.

x=3: y=18. Okay.

x=6: y=9. Okay.

x=9: y=6. Okay.

x=18: y=3. Not okay.

x=27: y=2. Not okay.

x=54: y=1. Not okay.

So only four possible x's. So answer is 4? But let me check once more. Maybe there are other divisors? Let me check 54 divided by numbers between 2 and 10. For example, is 54 divisible by 4? 54 divided by 4 is 13.5, no. 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 only 2,3,6,9. Therefore, 4 values. So answer is 4?

But wait, wait. Let me check if the question counts different x's even if they result in same y? No, the question is about how many possible values of x, so each x is unique. So answer is 4?

But I feel like maybe I'm missing something. Let me think again. Perhaps there are more divisors?

Wait, 54: divisors are 1, 2, 3, 6, 9, 18, 27, 54. Yes, that's correct. Because:

Starting with 1, 2, 3, then 6 (2*3), 9 (3^2), 18 (2*9), 27 (3^3), 54 (2*27). So that's all. So no others.

Alternatively, maybe the problem allows for different distributions? For example, maybe some players get more cards than others? But the problem says "each player getting y cards". So each player gets exactly y cards. Therefore, the total number of cards is x*y, so x must divide 54. So my previous reasoning holds. Therefore, only 4 possible values of x.

Wait, but maybe I need to check the problem again to make sure.

Original problem: "I have a deck of 54 cards, and I deal all of the cards to x players, with each player getting y cards. If x is at least 2 and y is at least 5, then how many possible values of x are there?"

So deal all cards to x players, each getting y cards. So total cards is x*y=54. So yeah, x must divide 54, and y=54/x, which must be ≥5. So x ≤ 54/5=10.8, so x ≤10. So possible x's are divisors of 54 between 2 and 10 inclusive. Which are 2,3,6,9. So 4 values. Therefore answer is 4?

But maybe I need to check once again. Let me think of x=4. If x=4, y=13.5. Not allowed. x=5, y=10.8. Not allowed. x=7, y≈7.7. Not allowed. x=8, y=6.75. Not allowed. x=10, y=5.4. Not allowed. So only when x is 2,3,6,9, y is integer and ≥5. Therefore, answer is 4. So \boxed{4}?

But wait, let me check if x=1 is allowed? No, because x is at least 2. So no. Therefore, answer is 4. Wait, but I remember sometimes problems consider that when you deal cards, you can't have fractional cards, so y has to be integer. Which is already considered here.

Alternatively, maybe there's a mistake in considering that y has to be integer. But since the problem says "each player getting y cards", so y must be integer. Therefore, x must divide 54.

Therefore, I think my answer is 4. Wait, but maybe the answer is different? Let me check once more.

Wait, maybe I need to list all the divisors again. Divisors of 54:

Start with 1, 2, 3, 6, 9, 18, 27, 54. So that's it. So x ≥2, so starting from 2. Now, for each of these, check y ≥5.

x=2: y=27 ≥5: valid.

x=3: y=18 ≥5: valid.

x=6: y=9 ≥5: valid.

x=9: y=6 ≥5: valid.

x=18: y=3 <5: invalid.

x=27: y=2 <5: invalid.

x=54: y=1 <5: invalid.

Therefore, only four values. Therefore, answer is 4. So \boxed{4}?

Wait, but I'm confused because sometimes, maybe there are other factors if you consider different groupings? Wait, but no, because the problem says dealing all cards to x players, each getting y cards. Therefore, x*y=54. So unless there's a way to have different y per player, but the problem says "each player getting y cards", so same number for each. Therefore, only divisors.

But let me check for example, if x=4, even though 4 doesn't divide 54, but maybe you can deal 54 cards to 4 players with each getting 13 or 14 cards? But the problem says "each player getting y cards", so y must be same for all. Therefore, y must be integer. Therefore, x must divide 54. Hence, no.

Therefore, answer is 4. But let me check if the original question is maybe about something else. Wait, the original question says "how many possible values of x are there?" So if there are four possible values, then answer is 4. Wait, but maybe I made a mistake in considering that x must be less than or equal to 10. Let me check that again.

Since y must be at least 5, so y ≥5. Therefore, x=54/y ≤ 54/5=10.8. Therefore, since x must be integer, x ≤10. So x is integer between 2 and 10 inclusive, and divides 54. The divisors of 54 in that range are 2,3,6,9. So that is four. Therefore, answer is 4. So I think this is correct.

But maybe the answer is different. Let me check with an example.

Suppose x=2: 2 players, each gets 27 cards. Okay.

x=3: 3 players, 18 each.

x=6: 6 players, 9 each.

x=9: 9 players, 6 each.

x=18: 18 players, 3 each. But y=3 <5, invalid.

Similarly, other higher x's are invalid. So yes, 4 values.

Alternatively, maybe if y is allowed to be more than 5? But no, the problem says y is at least 5, so y can be 5 or more. So if y is 5, then x=54/5=10.8, which is not integer. Therefore, x=10.8 is not allowed. Therefore, there is no x which gives y=5. So the next possible y is 6, which gives x=9. So that's already included. Therefore, all valid y's are 6,9,18,27. So corresponding x's are 9,6,3,2. Therefore, 4 values. Therefore, answer is 4.

Therefore, I think the correct answer is 4. But let me check once again if there's any other possible divisors between 2 and 10. Let me list numbers from 2 to 10 and check if they divide 54.

2 divides 54: yes.

3 divides 54: yes.

4: 54 divided by 4 is 13.5: no.

5: 54/5=10.8: no.

6: yes.

7: 54/7≈7.71: no.

8: 54/8=6.75: no.

9: yes.

10: 54/10=5.4: no.

Therefore, only 2,3,6,9. So four numbers. Therefore, answer is 4. Therefore, I think the answer is 4.

But just to be thorough, maybe the problem is considering that x and y can be any integers such that x*y=54, without requiring that x divides 54? But that would be impossible because if x*y=54, then x must divide 54. Because 54 divided by x is y, which must be integer. Therefore, x must be a divisor.

Hence, I think my answer is correct. Therefore, the number of possible values of x is 4.

**Final Answer**
\boxed{4}

To solve the problem, we're given a deck of **54 cards** being dealt to **$ x $** players, with each player receiving exactly **$ y $** cards. The key constraints are:

- $ x \geq 2 $
- $ y \geq 5 $
- All cards are dealt, so $ x \cdot y = 54 $

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

---

### Step 1: Set Up the Equation

From the condition that all cards are dealt, we have the equation:

$$
x \cdot y = 54
$$

Solving for $ y $, we get:

$$
y = \frac{54}{x}
$$

Since $ y $ must be an **integer**, $ x $ must be a **divisor of 54**. This ensures $ y $ is an integer (as the number of cards per player cannot be a fraction).

---

### Step 2: Find the Divisors of 54

The prime factorization of 54 is:

$$
54 = 2 \cdot 3^3
$$

From this, we can compute the **divisors of 54**:

$$
\text{Divisors of 54} = \{1, 2, 3, 6, 9, 18, 27, 54\}
$$

We are only concerned with values of $ x $ such that:

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

These conditions filter valid $ x $ values.

---

### Step 3: Evaluate Each Divisor

We now evaluate each divisor $ x $ to check if $ y = 54/x \geq 5 $:

| $ x $ | $ y = 54/x $ | Valid? |
|--------|----------------|--------|
| 1      | —              | No     |
| 2      | 27             | Yes    |
| 3      | 18             | Yes    |
| 6      | 9              | Yes    |
| 9      | 6              | Yes    |
| 18     | 3              | No     |
| 27     | 2              | No     |
| 54     | 1              | No     |

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

- $ x = 2 $ → $ y = 27 $
- $ x = 3 $ → $ y = 18 $
- $ x = 6 $ → $ y = 9 $
- $ x = 9 $ → $ y = 6 $

These are the only $ x $ values that satisfy both:

- $ x \geq 2 $
- $ y \geq 5 $
- $ x \cdot y = 54 $

---

### Final Count

There are **4 valid values** of $ x $ that satisfy the conditions.

---

### 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, each 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 the cards to x players, each player gets y cards. So, the total number of cards dealt should be equal to the total number of cards in the deck. That means:

x * y = 54

So, x and y are positive integers such that their product is 54. Also, the problem states that x is at least 2 and y is at least 5. Therefore, we need to find all pairs (x, y) where x ≥ 2, y ≥ 5, and x * y = 54. Then, the question asks for the number of possible values of x.

So essentially, I need to find all divisors x of 54 such that x ≥ 2 and when you divide 54 by x, the result y is at least 5. Therefore, for each divisor x of 54 (with x ≥ 2), check if 54/x ≥ 5. Then count how many such x's there are.

Alternatively, since x * y = 54, and y ≥ 5, that implies that x ≤ 54 / 5 = 10.8. Since x must be an integer, x ≤ 10. Also, since x ≥ 2, the possible values of x are integers from 2 up to 10, but only those that divide 54. Wait, but maybe I need to check that.

Wait, but maybe I need to think in terms of factors of 54. Let me recall the factors of 54.

First, prime factorization of 54: 54 = 2 * 3^3.

Therefore, the divisors of 54 can be found by taking exponents of primes:

For 2: exponent can be 0 or 1.

For 3: exponent can be 0, 1, 2, or 3.

Therefore, total number of divisors is (1+1)*(3+1)= 2*4=8. Let me list them:

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

So these are the divisors of 54. Now, the problem says that x must be at least 2, so we can exclude 1. So possible x values from the divisors are 2, 3, 6, 9, 18, 27, 54. But then we also have the constraint that y must be at least 5. Since y = 54 / x, so 54 / x ≥ 5 => x ≤ 54 / 5 = 10.8. Since x must be an integer, x ≤ 10. Therefore, from the list of divisors above, which are 2, 3, 6, 9, 18, 27, 54, we need to take only those x's that are less than or equal to 10. So let's check:

- 2: yes, 2 ≤ 10
- 3: yes
- 6: yes
- 9: yes
- 18: 18 > 10, exclude
- 27: same, exclude
- 54: exclude

So the possible x values are 2, 3, 6, 9. That's 4 values? Wait, but hold on, let me make sure.

Wait, maybe there are other divisors? Wait, the divisors of 54 are only those 8 numbers. So only 2, 3, 6, 9, 18, 27, 54. So from those, the ones less than or equal to 10 are 2, 3, 6, 9. So that is 4 values. But wait, let me check for each x:

For x=2: y=54/2=27 ≥5, okay.

x=3: y=18 ≥5, okay.

x=6: y=9 ≥5, okay.

x=9: y=6 ≥5, okay.

x=18: y=3, which is less than 5, so invalid.

Similarly x=27: y=2, invalid.

x=54: y=1, invalid.

So only 4 possible values of x. But hold on, is that all?

Wait, but maybe I need to think again. Because maybe x doesn't have to be a divisor of 54? Wait, no. Wait, the problem says you deal all of the cards to x players, with each player getting y cards. So the total number of cards is x*y, which must equal 54. Therefore, x must divide 54. Therefore, x must be a divisor of 54. Therefore, the only possible x's are divisors of 54. Therefore, only those 8 divisors. So we already checked them. So only 4 values of x (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 4 values.

But wait, let me think again. Wait, perhaps there are other possibilities where x is not a divisor of 54, but y is still integer? No, because x*y=54, so if x is not a divisor of 54, then y would not be integer. For example, suppose x=4. Then y=54/4=13.5, which is not an integer. Therefore, since y must be an integer (number of cards per person), x must divide 54. So x must be a divisor. Therefore, our previous list is correct.

Wait, but hold on. Let me check if there are more divisors between 2 and 10. Wait, the divisors of 54 are 1,2,3,6,9,18,27,54. So between 2 and 10, we have 2,3,6,9. So four divisors. Therefore the answer is 4?

Wait, but the problem says "how many possible values of x are there?" So is 4 the answer? But let me check again.

Alternatively, perhaps I need to consider that y must be at least 5, so for each x, check if y=54/x is integer and y≥5. So maybe I can approach this by first finding all pairs (x,y) such that x*y=54, x≥2, y≥5.

So let me list all possible pairs:

Start with x=2: y=27. Okay, since y≥5.

x=3: y=18. Okay.

x=6: y=9. Okay.

x=9: y=6. Okay.

x=18: y=3. Not okay.

x=27: y=2. Not okay.

x=54: y=1. Not okay.

So only four possible x's. So answer is 4? But let me check once more. Maybe there are other divisors? Let me check 54 divided by numbers between 2 and 10. For example, is 54 divisible by 4? 54 divided by 4 is 13.5, no. 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 only 2,3,6,9. Therefore, 4 values. So answer is 4?

But wait, wait. Let me check if the question counts different x's even if they result in same y? No, the question is about how many possible values of x, so each x is unique. So answer is 4?

But I feel like maybe I'm missing something. Let me think again. Perhaps there are more divisors?

Wait, 54: divisors are 1, 2, 3, 6, 9, 18, 27, 54. Yes, that's correct. Because:

Starting with 1, 2, 3, then 6 (2*3), 9 (3^2), 18 (2*9), 27 (3^3), 54 (2*27). So that's all. So no others.

Alternatively, maybe the problem allows for different distributions? For example, maybe some players get more cards than others? But the problem says "each player getting y cards". So each player gets exactly y cards. Therefore, the total number of cards is x*y, so x must divide 54. So my previous reasoning holds. Therefore, only 4 possible values of x.

Wait, but maybe I need to check the problem again to make sure.

Original problem: "I have a deck of 54 cards, and I deal all of the cards to x players, with each player getting y cards. If x is at least 2 and y is at least 5, then how many possible values of x are there?"

So deal all cards to x players, each getting y cards. So total cards is x*y=54. So yeah, x must divide 54, and y=54/x, which must be ≥5. So x ≤ 54/5=10.8, so x ≤10. So possible x's are divisors of 54 between 2 and 10 inclusive. Which are 2,3,6,9. So 4 values. Therefore answer is 4?

But maybe I need to check once again. Let me think of x=4. If x=4, y=13.5. Not allowed. x=5, y=10.8. Not allowed. x=7, y≈7.7. Not allowed. x=8, y=6.75. Not allowed. x=10, y=5.4. Not allowed. So only when x is 2,3,6,9, y is integer and ≥5. Therefore, answer is 4. So \boxed{4}?

But wait, let me check if x=1 is allowed? No, because x is at least 2. So no. Therefore, answer is 4. Wait, but I remember sometimes problems consider that when you deal cards, you can't have fractional cards, so y has to be integer. Which is already considered here.

Alternatively, maybe there's a mistake in considering that y has to be integer. But since the problem says "each player getting y cards", so y must be integer. Therefore, x must divide 54.

Therefore, I think my answer is 4. Wait, but maybe the answer is different? Let me check once more.

Wait, maybe I need to list all the divisors again. Divisors of 54:

Start with 1, 2, 3, 6, 9, 18, 27, 54. So that's it. So x ≥2, so starting from 2. Now, for each of these, check y ≥5.

x=2: y=27 ≥5: valid.

x=3: y=18 ≥5: valid.

x=6: y=9 ≥5: valid.

x=9: y=6 ≥5: valid.

x=18: y=3 <5: invalid.

x=27: y=2 <5: invalid.

x=54: y=1 <5: invalid.

Therefore, only four values. Therefore, answer is 4. So \boxed{4}?

Wait, but I'm confused because sometimes, maybe there are other factors if you consider different groupings? Wait, but no, because the problem says dealing all cards to x players, each getting y cards. Therefore, x*y=54. So unless there's a way to have different y per player, but the problem says "each player getting y cards", so same number for each. Therefore, only divisors.

But let me check for example, if x=4, even though 4 doesn't divide 54, but maybe you can deal 54 cards to 4 players with each getting 13 or 14 cards? But the problem says "each player getting y cards", so y must be same for all. Therefore, y must be integer. Therefore, x must divide 54. Hence, no.

Therefore, answer is 4. But let me check if the original question is maybe about something else. Wait, the original question says "how many possible values of x are there?" So if there are four possible values, then answer is 4. Wait, but maybe I made a mistake in considering that x must be less than or equal to 10. Let me check that again.

Since y must be at least 5, so y ≥5. Therefore, x=54/y ≤ 54/5=10.8. Therefore, since x must be integer, x ≤10. So x is integer between 2 and 10 inclusive, and divides 54. The divisors of 54 in that range are 2,3,6,9. So that is four. Therefore, answer is 4. So I think this is correct.

But maybe the answer is different. Let me check with an example.

Suppose x=2: 2 players, each gets 27 cards. Okay.

x=3: 3 players, 18 each.

x=6: 6 players, 9 each.

x=9: 9 players, 6 each.

x=18: 18 players, 3 each. But y=3 <5, invalid.

Similarly, other higher x's are invalid. So yes, 4 values.

Alternatively, maybe if y is allowed to be more than 5? But no, the problem says y is at least 5, so y can be 5 or more. So if y is 5, then x=54/5=10.8, which is not integer. Therefore, x=10.8 is not allowed. Therefore, there is no x which gives y=5. So the next possible y is 6, which gives x=9. So that's already included. Therefore, all valid y's are 6,9,18,27. So corresponding x's are 9,6,3,2. Therefore, 4 values. Therefore, answer is 4.

Therefore, I think the correct answer is 4. But let me check once again if there's any other possible divisors between 2 and 10. Let me list numbers from 2 to 10 and check if they divide 54.

2 divides 54: yes.

3 divides 54: yes.

4: 54 divided by 4 is 13.5: no.

5: 54/5=10.8: no.

6: yes.

7: 54/7≈7.71: no.

8: 54/8=6.75: no.

9: yes.

10: 54/10=5.4: no.

Therefore, only 2,3,6,9. So four numbers. Therefore, answer is 4. Therefore, I think the answer is 4.

But just to be thorough, maybe the problem is considering that x and y can be any integers such that x*y=54, without requiring that x divides 54? But that would be impossible because if x*y=54, then x must divide 54. Because 54 divided by x is y, which must be integer. Therefore, x must be a divisor.

Hence, I think my answer is correct. Therefore, the number of possible values of x is 4.

**Final Answer**
\boxed{4}

To solve the problem, we're given a deck of **54 cards** being dealt to **$ x $** players, with each player receiving exactly **$ y $** cards. The key constraints are:

- $ x \geq 2 $
- $ y \geq 5 $
- All cards are dealt, so $ x \cdot y = 54 $

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

---

### Step 1: Set Up the Equation

From the condition that all cards are dealt, we have the equation:

$$
x \cdot y = 54
$$

Solving for $ y $, we get:

$$
y = \frac{54}{x}
$$

Since $ y $ must be an **integer**, $ x $ must be a **divisor of 54**. This ensures $ y $ is an integer (as the number of cards per player cannot be a fraction).

---

### Step 2: Find the Divisors of 54

The prime factorization of 54 is:

$$
54 = 2 \cdot 3^3
$$

From this, we can compute the **divisors of 54**:

$$
\text{Divisors of 54} = \{1, 2, 3, 6, 9, 18, 27, 54\}
$$

We are only concerned with values of $ x $ such that:

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

These conditions filter valid $ x $ values.

---

### Step 3: Evaluate Each Divisor

We now evaluate each divisor $ x $ to check if $ y = 54/x \geq 5 $:

| $ x $ | $ y = 54/x $ | Valid? |
|--------|----------------|--------|
| 1      | —              | No     |
| 2      | 27             | Yes    |
| 3      | 18             | Yes    |
| 6      | 9              | Yes    |
| 9      | 6              | Yes    |
| 18     | 3              | No     |
| 27     | 2              | No     |
| 54     | 1              | No     |

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

- $ x = 2 $ → $ y = 27 $
- $ x = 3 $ → $ y = 18 $
- $ x = 6 $ → $ y = 9 $
- $ x = 9 $ → $ y = 6 $

These are the only $ x $ values that satisfy both:

- $ x \geq 2 $
- $ y \geq 5 $
- $ x \cdot y = 54 $

---

### Final Count

There are **4 valid values** of $ x $ that satisfy the conditions.

---

### 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, let's try to tackle this problem. So, we have two positive integers a and b. Their greatest common divisor (gcd) is 210, and their least common multiple (lcm) is 210 cubed, which is 210³. Also, it's given that a is less than b. We need to find how many possible values there are for a.

First, I remember that for any two positive integers a and b, the product of their gcd and lcm is equal to the product of the numbers themselves. So, mathematically, that would be:

gcd(a, b) × lcm[a, b] = a × b

Given that gcd(a, b) is 210 and lcm[a, b] is 210³, substituting these into the equation gives:

210 × 210³ = a × b

Simplifying the left side, 210 × 210³ is 210⁴. Therefore:

a × b = 210⁴

So, now we know that the product of a and b is 210⁴. But also, their gcd is 210. Let me recall that when dealing with gcd and lcm, it's helpful to express the numbers in terms of their prime factors.

First, let me factorize 210. 210 is 2 × 3 × 5 × 7. So, 210 is the product of the first four prime numbers. Therefore, 210³ would be (2 × 3 × 5 × 7)³, and 210⁴ is (2 × 3 × 5 × 7)⁴.

Now, since the gcd of a and b is 210, that tells me that both a and b must be multiples of 210. Moreover, the lcm of a and b is 210³, which is a multiple of 210 as well.

Let me think about how to represent a and b in terms of their prime factors. Let me denote:

Let’s suppose that a = 210 × m and b = 210 × n, where m and n are positive integers. Since the gcd(a, b) is 210, this implies that gcd(m, n) must be 1. Because if there was a common divisor greater than 1 between m and n, then that would multiply with 210 to give a larger gcd.

Similarly, the lcm[a, b] can be expressed as 210 × lcm(m, n). Since lcm[a, b] is given as 210³, then:

210 × lcm(m, n) = 210³ ⇒ lcm(m, n) = 210²

Therefore, now the problem reduces to finding the number of pairs (m, n) where m and n are positive integers with gcd(m, n) = 1 and lcm(m, n) = 210². Then, since a = 210 × m and b = 210 × n, and a < b, we need to count the number of such m's where m < n.

But first, let's focus on the properties of m and n. Since gcd(m, n) = 1 and lcm(m, n) = 210², which is (2 × 3 × 5 × 7)².

Now, since m and n are coprime, their lcm is just the product of m and n. Wait, is that right? Wait, no. Wait, the lcm of two numbers is equal to the product of the numbers divided by their gcd. So, since gcd(m, n) = 1, then lcm(m, n) = m × n / 1 = m × n. Therefore, m × n = lcm(m, n) = 210².

Therefore, m and n are coprime positive integers such that m × n = 210².

Therefore, we need to find the number of coprime pairs (m, n) with m × n = 210², and then since a < b, each such pair where m ≠ n will correspond to one value of a (since a = 210 × m, and b = 210 × n, and since m < n implies a < b). However, if m = n, then a = b, but the problem states that a < b, so we need to exclude the case where m = n.

But first, let's check if m and n can be equal. If m = n, then since m × n = 210², we have m² = 210² ⇒ m = 210. But since m and n are coprime, if m = n = 210, then gcd(m, n) = 210 ≠ 1. Which contradicts the requirement that gcd(m, n) = 1. Therefore, m and n cannot be equal. Therefore, all pairs (m, n) must have m ≠ n, so each pair corresponds to exactly one a where a < b.

Therefore, the problem reduces to finding the number of coprime pairs (m, n) with m × n = 210², and then since each such pair corresponds to a unique a (with a = 210m, and since m < n, a < b).

But how many such coprime pairs (m, n) are there?

Given that m × n = 210² and gcd(m, n) = 1, then m and n must be coprime factors of 210².

But since m and n are coprime and multiply to 210², each prime factor of 210² must be entirely in m or entirely in n.

Since 210 is 2 × 3 × 5 × 7, so 210² is (2 × 3 × 5 × 7)² = 2² × 3² × 5² × 7².

So, the prime factors of 210² are 2², 3², 5², 7².

So, since m and n are coprime, each prime factor must go entirely to m or entirely to n. So, for each prime factor, we have two choices: assign it to m or to n.

However, since m and n are positive integers, and since the product is fixed, each prime's exponents are fixed.

But since m and n are coprime, for each prime in the factorization of 210², we can assign the entire exponent of that prime to either m or n.

Therefore, for each of the four primes (2, 3, 5, 7), we have two choices: assign the prime's exponents to m or to n.

Therefore, the total number of coprime pairs (m, n) is 2⁴ = 16. But since order matters here (i.e., (m, n) and (n, m) are different ordered pairs), but since we are looking for unordered pairs? Wait, no. Wait, actually, since we need pairs where m × n = 210² and gcd(m, n) = 1, each pair is determined uniquely by the assignment of primes to m and n.

But since m and n are positive integers, and since the product is fixed, each assignment corresponds to exactly one pair (m, n). However, since (m, n) and (n, m) are distinct unless m = n, which we already saw can't happen here. Therefore, the total number of ordered pairs is 16. But since we need unordered pairs, but in our problem, since a < b, each unordered pair {m, n} corresponds to exactly one ordered pair (m, n) where m < n.

Therefore, the number of unordered pairs would be half of the number of ordered pairs. But since the total number of ordered pairs is 16, the number of unordered pairs is 8. However, since in our problem, we are considering ordered pairs where m < n, so the number of such pairs is 8.

But wait, let me check again.

Each coprime pair (m, n) with m × n = 210² corresponds to assigning each prime factor to m or n. Since for each prime, we have 2 choices, total 2⁴ = 16 ordered pairs. However, since for each unordered pair {m, n}, there are two ordered pairs (m, n) and (n, m), except when m = n. But as we saw earlier, m = n is impossible here. Therefore, the number of unordered pairs is 16 / 2 = 8.

Therefore, there are 8 unordered pairs {m, n} with m × n = 210² and gcd(m, n) = 1, which correspond to 8 ordered pairs (m, n) with m < n. Therefore, since each such ordered pair gives a unique a = 210 × m, the number of possible a's is 8? Wait, but hold on. Wait, but let me think again.

Wait, actually, if we have 16 ordered pairs (m, n), each with m × n = 210² and gcd(m, n) = 1. Since m and n can be in any order, but since a < b, which is equivalent to m < n (since a = 210m and b = 210n, and 210 is positive). Therefore, the number of possible a's is equal to the number of m's such that m < n where m × n = 210² and gcd(m, n) = 1.

But since each such pair is uniquely determined by m, and each m corresponds to exactly one n, then the number of such m's is equal to half of the total number of ordered pairs. Which would be 8. Therefore, the answer would be 8? But wait, is this correct? Let me check with an example.

Alternatively, perhaps there is a different approach. Let me think again.

Since m × n = 210² and gcd(m, n) = 1, then m and n are coprime divisors of 210². The number of such coprime pairs (m, n) is equal to the number of ways to partition the set of prime factors of 210² between m and n. Since each prime factor has to go entirely to m or entirely to n, as they are coprime.

The prime factors are 2, 3, 5, 7, each with exponent 2 in 210². But since m and n are coprime, each prime can be in m or in n. Therefore, for each of the 4 primes, 2 choices, so 2^4 = 16 ordered pairs. So that's correct.

Now, since we need m < n, how many such ordered pairs are there? Since for each unordered pair {m, n}, there are two ordered pairs, except when m = n, which isn't possible here. Therefore, the number of ordered pairs with m < n is 16 / 2 = 8. Hence, there are 8 possible values of m, each giving a unique a. Therefore, the answer should be 8?

But wait, let me check with an example. Let me take a smaller case. Suppose instead of 210², let's take something simpler. Let's say that gcd(a, b) = 2, lcm(a, b) = 8. Then, similar logic: a*b = gcd(a,b)*lcm(a,b) = 2*8=16. Then, since gcd(a,b)=2, we can write a=2m, b=2n, gcd(m,n)=1, and lcm(a,b)=2*lcm(m,n)=8 ⇒ lcm(m,n)=4. Then m*n = lcm(m,n)*gcd(m,n) = 4*1=4. So m and n are coprime positive integers with m*n=4. The coprime pairs (m,n) are (1,4) and (4,1). So there are 2 ordered pairs, which correspond to 1 unordered pair. Therefore, the number of possible a's where a < b would be 1. Let me check:

Possible a's: a=2*1=2, and b=2*4=8. So only one possible a, which is 2. But let's see if there are others. Suppose m=2 and n=2? Then gcd(m,n)=2≠1. So invalid. So only the pairs (1,4) and (4,1). Therefore, the answer would be 1. Which is 2/2=1. So that works. Therefore, in that case, the formula holds. Therefore, for our problem, 16 ordered pairs, 8 unordered pairs, so answer is 8?

But wait, hold on. Let me try with another example. Let me take gcd(a,b)=6 and lcm(a,b)=6^2=36. Then a*b=6*36=216. So, writing a=6m, b=6n, gcd(m,n)=1, lcm(a,b)=6*lcm(m,n)=36 ⇒ lcm(m,n)=6. Then m*n=lcm(m,n)*gcd(m,n)=6*1=6. So m*n=6, gcd(m,n)=1. The coprime pairs (m,n) with m*n=6: (1,6), (2,3), (3,2), (6,1). So ordered pairs: 4. Unordered pairs: 2. So number of a's where a < b is 2. Let's check:

a=6*1=6, b=6*6=36.

a=6*2=12, b=6*3=18.

Are there any other pairs? For example, if m=1 and n=6, or m=2 and n=3. So two possible a's: 6 and 12. Hence, answer is 2. Which is 4 ordered pairs divided by 2, gives 2. So formula holds.

Therefore, returning to our original problem. The number of ordered pairs is 16, so number of unordered pairs is 8, hence the number of possible a's is 8.

Wait, but let me think again. Wait, but the problem says "how many possible values are there for a". Each a corresponds to a unique m. Since m can be any of the divisors of 210² such that m and n = 210²/m are coprime, and m < n. So the number of such m's is equal to the number of such pairs.

Alternatively, since each m is a divisor of 210², and for each m, if we can ensure that m is coprime with 210²/m, then m is a valid m.

But since m and n are coprime, then m must be a divisor of 210², and for each prime in the factorization of 210², the prime is entirely in m or entirely in n. Therefore, the number of such m's is equal to 2^4 = 16, but since m and n are distinct, we divide by 2, getting 8. Hence, the answer is 8.

But hold on, let me think if there's another way to compute this. Since m must be a divisor of 210², and coprime with n = 210²/m. So m must be a divisor of 210² such that m is squareful? Wait, no. Wait, m can be any divisor of 210², but with the condition that m and n are coprime. Which implies that m is a product of some subset of the prime factors of 210², each raised to their respective exponents. Wait, but 210² is (2×3×5×7)², so exponents of primes are 2 for each. But since m and n must be coprime, m can have any combination of the primes, but each prime can only be in m or in n, not both. So for each prime, you choose whether it is in m or in n. Therefore, the number of m's is equal to 2^4=16. However, each m corresponds to a unique n. However, since m can be in any order, so when we count the number of m's where m < n, it's half of 16, which is 8.

Therefore, the answer should be 8. But wait, let me check with the earlier example where 210 was replaced with 2, and the answer was 1.

Wait, but maybe there's something wrong here. Let me think again.

Suppose that m is a divisor of 210² with gcd(m, 210²/m)=1. Then m must be squareful? Wait, no, m can be any divisor. Wait, for example, let's take m=1. Then n=210²/1=210², and gcd(1, 210²)=1. So that works. Similarly, m=2², which is 4. Then n=210² / 4. Since 210² is (2×3×5×7)², dividing by 4 gives (3×5×7)², so n=(3×5×7)². Now, gcd(m, n) = gcd(4, (3×5×7)²) = 1, because 4 is 2² and the other number has no factor of 2. Similarly, for m=3², n= (2×5×7)², gcd(m,n)=1. Similarly for m=5², m=7², and any combinations.

Therefore, the number of m's is equal to the number of subsets of the set {2, 3, 5, 7} where each prime is included or not, with the exponents being the full exponents in 210². Wait, but actually, since m can have any combination of the primes, each prime can be present or not. So, the number of m's is 2^4=16. But since m and n are different, then for each m ≠ n, they form a pair, so the number of pairs where m < n is 8. Therefore, the answer is 8.

But wait, let me check with actual numbers. Let's think of the possible m's.

Each m corresponds to a subset of the primes {2, 3, 5, 7}. For example:

- The empty set: m=1 (since none of the primes are included), then n=210².

- The subset {2}: m=2²=4, n=(3×5×7)²= (105)².

- The subset {3}: m=3²=9, n=(2×5×7)²=70².

- The subset {5}: m=5²=25, n=(2×3×7)²=42².

- The subset {7}: m=7²=49, n=(2×3×5)²=30².

- The subset {2,3}: m=2²×3²= (2×3)²=36, n=(5×7)²=35².

- The subset {2,5}: m=2²×5²=100, n=(3×7)²=21².

- The subset {2,7}: m=2²×7²=196, n=(3×5)²=15².

- The subset {3,5}: m=3²×5²=225, n=(2×7)²=14².

- The subset {3,7}: m=3²×7²=441, n=(2×5)²=10².

- The subset {5,7}: m=5²×7²=1225, n=(2×3)²=6².

- The subset {2,3,5}: m=2²×3²×5²= (2×3×5)²=30², n=7²=49.

- The subset {2,3,7}: m=2²×3²×7²= (2×3×7)²=42², n=5²=25.

- The subset {2,5,7}: m=2²×5²×7²= (2×5×7)²=70², n=3²=9.

- The subset {3,5,7}: m=3²×5²×7²= (3×5×7)²=105², n=2²=4.

- The subset {2,3,5,7}: m=210², n=1.

So, total of 16 m's. Now, since we need m < n, we can check which of these m's are less than their corresponding n.

Let me list them:

1. m=1, n=210²: 1 < 210² ✔

2. m=4, n=105². Since 105² is 11025, and 4 < 11025 ✔

3. m=9, n=70²=4900. 9 < 4900 ✔

4. m=25, n=42²=1764. 25 < 1764 ✔

5. m=49, n=30²=900. 49 < 900 ✔

6. m=36, n=35²=1225. 36 < 1225 ✔

7. m=100, n=21²=441. 100 < 441 ✔

8. m=196, n=15²=225. 196 < 225 ✔

9. m=225, n=14²=196. 225 > 196 ❌

10. m=441, n=10²=100. 441 > 100 ❌

11. m=1225, n=6²=36. 1225 > 36 ❌

12. m=30²=900, n=49. 900 > 49 ❌

13. m=42²=1764, n=25. 1764 > 25 ❌

14. m=70²=4900, n=9. 4900 > 9 ❌

15. m=105²=11025, n=4. 11025 > 4 ❌

16. m=210², n=1. 210² > 1 ❌

So, out of these 16 m's, only the first 8 (from 1 to 8) satisfy m < n. The rest have m > n. Therefore, there are exactly 8 values of m where m < n, hence 8 possible values of a. Therefore, the answer is 8.

But wait, let me check the 8th one. The 8th m is 196, n=225. 196 is 14², and 225 is 15². 14²=196, 15²=225. Yes, 196 < 225. So that's valid.

Then, the remaining ones (from 9 to 16) have m > n. Therefore, the total number of valid a's is 8.

Therefore, the answer is 8.

Wait, but let me think again if there's another way to compute this. Since the number of such m's is equal to the number of subsets of the set {2,3,5,7}, which is 2^4=16. Each subset corresponds to a unique m. However, since for each subset, its complement corresponds to another m, and when you take the subset and its complement, one will be less than the other. Except when the subset is equal to its complement, but since there are 4 primes, which is even, there is no subset equal to its complement. Because for a subset to be equal to its complement, the subset would have to contain exactly half the elements. But since 4 is even, there are subsets of size 2. However, for each subset, the complement is a different subset. But in our case, since we have 4 primes, the number of subsets is 16, so the number of unordered pairs is 8. Therefore, the answer is 8.

Therefore, I think the answer is 8. However, let me check once more.

Alternatively, since each m must be a divisor of 210², and coprime with n = 210²/m. Which implies that m is a squareful? Not necessarily, but in our case, m is a product of primes each raised to their maximum exponent in 210². Wait, no. Wait, m can be any combination of primes, each with exponent 0 or 2. Wait, no. Wait, let me clarify.

Wait, 210² is (2×3×5×7)² = 2²×3²×5²×7². So when we take m as a divisor of 210², and we want m and n = 210²/m to be coprime, then m must have exponents only for some primes, and n will have exponents for the remaining primes, with each prime having exponent 0 or 2 in m.

But since m and n are coprime, no prime can be in both m and n. Therefore, for each prime in 210², we decide whether to include it in m (with exponent 2) or in n (with exponent 2). Therefore, m can be thought of as the product over some subset of {2, 3, 5, 7} each squared. So for example, if the subset is {2,3}, then m = 2²×3², and n = 5²×7².

Therefore, the number of such m's is the number of subsets of {2, 3, 5, 7}, which is 2^4 = 16. Each subset corresponds to a unique m. However, when considering m < n, we need to count the subsets where the corresponding m is less than n.

But since for each subset S, the complement subset S' corresponds to n, and the product m is the product of primes in S squared, and n is the product of primes in S' squared. Then, comparing m and n would depend on the sizes of S and S', or the actual products.

But in our earlier enumeration, we saw that for subsets of size 0 (empty set) or 1 or 2, etc., some of them result in m < n and others m > n.

However, the key point is that exactly half of the subsets will have m < n and the other half will have m > n. Since there are 16 subsets, and they come in pairs where one is S and the other is S', and for each pair, exactly one of m < n or m > n holds. Therefore, the number of subsets where m < n is 16 / 2 = 8, which is the answer.

Therefore, the answer is 8.

But just to be thorough, let me count the number of subsets S where the product of primes in S squared is less than the product of primes in S' squared.

Note that the product of primes in S squared is (product of primes in S)^2, same with S'. Therefore, comparing m and n is equivalent to comparing (product of primes in S)^2 and (product of primes in S')^2, which is the same as comparing product of primes in S and product of primes in S'.

Therefore, the problem reduces to counting how many subsets S of {2, 3, 5, 7} have product less than the product of the complement subset.

Note that the complement subset S' has product equal to (product of all primes) / (product of primes in S).

Let me denote P = 2×3×5×7 = 210.

So the product of primes in S is, let's say, x, and the product of primes in S' is P / x. Then, we want x < P / x ⇒ x² < P ⇒ x < sqrt(P).

Therefore, the number of subsets S where product of primes in S is less than sqrt(P) is equal to the number of subsets S where m < n.

So, let's compute sqrt(P) where P = 210. sqrt(210) ≈ 14.49. Therefore, we need the subsets S whose product of primes is less than 14.49.

The primes are 2, 3, 5, 7.

So, let's compute the product of primes for each subset:

- Empty set: product = 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

Now, we need to find the subsets where the product is less than sqrt(210) ≈ 14.49. Let's list those:

- Empty set: 1 < 14.49 ✔

- {2}: 2 < 14.49 ✔

- {3}: 3 < 14.49 ✔

- {5}: 5 < 14.49 ✔

- {7}: 7 < 14.49 ✔

- {2, 3}: 6 < 14.49 ✔

- {2, 5}: 10 < 14.49 ✔

- {2, 7}: 14 < 14.49 ✔ (since 14 is less than 14.49)

- {3, 5}: 15 > 14.49 ❌

- {3, 7}: 21 > 14.49 ❌

- {5, 7}: 35 > 14.49 ❌

- {2, 3, 5}: 30 > 14.49 ❌

And so on. Therefore, only the first 8 subsets (from the empty set up to {2,7}) have product less than sqrt(210). Therefore, there are 8 subsets where x < P/x, which corresponds to m < n. Therefore, 8 values of a.

Therefore, this confirms the previous result.

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

**Final Answer**
\boxed{8}

To determine how many possible values there are for $ a $ given that:

- $ \gcd(a, b) = 210 $
- $ \text{lcm}(a, b) = 210^3 $
- $ a < b $, and $ a, b \in \mathbb{Z}^+ $

we begin with the identity:

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

Substituting the known values:

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

Let’s factorize $ 210 = 2 \cdot 3 \cdot 5 \cdot 7 $, so:

$$
210^4 = (2 \cdot 3 \cdot 5 \cdot 7)^4 = 2^4 \cdot 3^4 \cdot 5^4 \cdot 7^4
$$

Let’s now define:

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

Then:

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

Also, since $ \gcd(a, b) = 210 $, this implies $ \gcd(m, n) = 1 $.

We now have the problem: **find the number of positive integers $ m $ such that** $ \gcd(m, n) = 1 $ and $ m \cdot n = 210^2 $, **with $ m < n $**.

---

### Step 1: Factorization and Coprime Constraints

We know that the prime factorization of $ 210^2 $ is:

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

For $ m \cdot n = 210^2 $ and $ \gcd(m, n) = 1 $, we must assign each of the prime factors to **either $ m $ or $ n $** entirely — not both — because if a prime is in both, the gcd would be at least that prime. So for each of the 4 primes (2, 3, 5, 7), we can assign them to either $ m $ or $ n $.

Thus, the number of such ordered pairs $ (m, n) $ is:

$$
2^4 = 16
$$

Each such assignment corresponds to a unique ordered pair $ (m, n) $, where $ m $ and $ n $ are coprime, and their product is $ 210^2 $.

---

### Step 2: Counting Valid $ a $ with $ a < b $

Since $ a = 210 \cdot m $ and $ b = 210 \cdot n $, and we require $ a < b $, this corresponds to $ m < n $. Each ordered pair $ (m, n) $ with $ m \ne n $ gives a unique pair $ (a, b) $ with $ a < b $.

From the 16 ordered pairs $ (m, n) $, exactly half will satisfy $ m < n $, because for every ordered pair $ (m, n) $, the reverse $ (n, m) $ is also valid and disjoint. So:

$$
\text{Number of valid } (a, b) \text{ with } a < b = \frac{16}{2} = 8
$$

---

### Final Answer

Thus, there are exactly $ \boxed{8} $ possible values for $ a $ that satisfy all the given conditions.

$$
\boxed{8}
$$

[Assistant]
Explanation:
Okay, let's try to tackle this problem. So, we have two positive integers a and b. Their greatest common divisor (gcd) is 210, and their least common multiple (lcm) is 210 cubed, which is 210³. Also, it's given that a is less than b. We need to find how many possible values there are for a.

First, I remember that for any two positive integers a and b, the product of their gcd and lcm is equal to the product of the numbers themselves. So, mathematically, that would be:

gcd(a, b) × lcm[a, b] = a × b

Given that gcd(a, b) is 210 and lcm[a, b] is 210³, substituting these into the equation gives:

210 × 210³ = a × b

Simplifying the left side, 210 × 210³ is 210⁴. Therefore:

a × b = 210⁴

So, now we know that the product of a and b is 210⁴. But also, their gcd is 210. Let me recall that when dealing with gcd and lcm, it's helpful to express the numbers in terms of their prime factors.

First, let me factorize 210. 210 is 2 × 3 × 5 × 7. So, 210 is the product of the first four prime numbers. Therefore, 210³ would be (2 × 3 × 5 × 7)³, and 210⁴ is (2 × 3 × 5 × 7)⁴.

Now, since the gcd of a and b is 210, that tells me that both a and b must be multiples of 210. Moreover, the lcm of a and b is 210³, which is a multiple of 210 as well.

Let me think about how to represent a and b in terms of their prime factors. Let me denote:

Let’s suppose that a = 210 × m and b = 210 × n, where m and n are positive integers. Since the gcd(a, b) is 210, this implies that gcd(m, n) must be 1. Because if there was a common divisor greater than 1 between m and n, then that would multiply with 210 to give a larger gcd.

Similarly, the lcm[a, b] can be expressed as 210 × lcm(m, n). Since lcm[a, b] is given as 210³, then:

210 × lcm(m, n) = 210³ ⇒ lcm(m, n) = 210²

Therefore, now the problem reduces to finding the number of pairs (m, n) where m and n are positive integers with gcd(m, n) = 1 and lcm(m, n) = 210². Then, since a = 210 × m and b = 210 × n, and a < b, we need to count the number of such m's where m < n.

But first, let's focus on the properties of m and n. Since gcd(m, n) = 1 and lcm(m, n) = 210², which is (2 × 3 × 5 × 7)².

Now, since m and n are coprime, their lcm is just the product of m and n. Wait, is that right? Wait, no. Wait, the lcm of two numbers is equal to the product of the numbers divided by their gcd. So, since gcd(m, n) = 1, then lcm(m, n) = m × n / 1 = m × n. Therefore, m × n = lcm(m, n) = 210².

Therefore, m and n are coprime positive integers such that m × n = 210².

Therefore, we need to find the number of coprime pairs (m, n) with m × n = 210², and then since a < b, each such pair where m ≠ n will correspond to one value of a (since a = 210 × m, and b = 210 × n, and since m < n implies a < b). However, if m = n, then a = b, but the problem states that a < b, so we need to exclude the case where m = n.

But first, let's check if m and n can be equal. If m = n, then since m × n = 210², we have m² = 210² ⇒ m = 210. But since m and n are coprime, if m = n = 210, then gcd(m, n) = 210 ≠ 1. Which contradicts the requirement that gcd(m, n) = 1. Therefore, m and n cannot be equal. Therefore, all pairs (m, n) must have m ≠ n, so each pair corresponds to exactly one a where a < b.

Therefore, the problem reduces to finding the number of coprime pairs (m, n) with m × n = 210², and then since each such pair corresponds to a unique a (with a = 210m, and since m < n, a < b).

But how many such coprime pairs (m, n) are there?

Given that m × n = 210² and gcd(m, n) = 1, then m and n must be coprime factors of 210².

But since m and n are coprime and multiply to 210², each prime factor of 210² must be entirely in m or entirely in n.

Since 210 is 2 × 3 × 5 × 7, so 210² is (2 × 3 × 5 × 7)² = 2² × 3² × 5² × 7².

So, the prime factors of 210² are 2², 3², 5², 7².

So, since m and n are coprime, each prime factor must go entirely to m or entirely to n. So, for each prime factor, we have two choices: assign it to m or to n.

However, since m and n are positive integers, and since the product is fixed, each prime's exponents are fixed.

But since m and n are coprime, for each prime in the factorization of 210², we can assign the entire exponent of that prime to either m or n.

Therefore, for each of the four primes (2, 3, 5, 7), we have two choices: assign the prime's exponents to m or to n.

Therefore, the total number of coprime pairs (m, n) is 2⁴ = 16. But since order matters here (i.e., (m, n) and (n, m) are different ordered pairs), but since we are looking for unordered pairs? Wait, no. Wait, actually, since we need pairs where m × n = 210² and gcd(m, n) = 1, each pair is determined uniquely by the assignment of primes to m and n.

But since m and n are positive integers, and since the product is fixed, each assignment corresponds to exactly one pair (m, n). However, since (m, n) and (n, m) are distinct unless m = n, which we already saw can't happen here. Therefore, the total number of ordered pairs is 16. But since we need unordered pairs, but in our problem, since a < b, each unordered pair {m, n} corresponds to exactly one ordered pair (m, n) where m < n.

Therefore, the number of unordered pairs would be half of the number of ordered pairs. But since the total number of ordered pairs is 16, the number of unordered pairs is 8. However, since in our problem, we are considering ordered pairs where m < n, so the number of such pairs is 8.

But wait, let me check again.

Each coprime pair (m, n) with m × n = 210² corresponds to assigning each prime factor to m or n. Since for each prime, we have 2 choices, total 2⁴ = 16 ordered pairs. However, since for each unordered pair {m, n}, there are two ordered pairs (m, n) and (n, m), except when m = n. But as we saw earlier, m = n is impossible here. Therefore, the number of unordered pairs is 16 / 2 = 8.

Therefore, there are 8 unordered pairs {m, n} with m × n = 210² and gcd(m, n) = 1, which correspond to 8 ordered pairs (m, n) with m < n. Therefore, since each such ordered pair gives a unique a = 210 × m, the number of possible a's is 8? Wait, but hold on. Wait, but let me think again.

Wait, actually, if we have 16 ordered pairs (m, n), each with m × n = 210² and gcd(m, n) = 1. Since m and n can be in any order, but since a < b, which is equivalent to m < n (since a = 210m and b = 210n, and 210 is positive). Therefore, the number of possible a's is equal to the number of m's such that m < n where m × n = 210² and gcd(m, n) = 1.

But since each such pair is uniquely determined by m, and each m corresponds to exactly one n, then the number of such m's is equal to half of the total number of ordered pairs. Which would be 8. Therefore, the answer would be 8? But wait, is this correct? Let me check with an example.

Alternatively, perhaps there is a different approach. Let me think again.

Since m × n = 210² and gcd(m, n) = 1, then m and n are coprime divisors of 210². The number of such coprime pairs (m, n) is equal to the number of ways to partition the set of prime factors of 210² between m and n. Since each prime factor has to go entirely to m or entirely to n, as they are coprime.

The prime factors are 2, 3, 5, 7, each with exponent 2 in 210². But since m and n are coprime, each prime can be in m or in n. Therefore, for each of the 4 primes, 2 choices, so 2^4 = 16 ordered pairs. So that's correct.

Now, since we need m < n, how many such ordered pairs are there? Since for each unordered pair {m, n}, there are two ordered pairs, except when m = n, which isn't possible here. Therefore, the number of ordered pairs with m < n is 16 / 2 = 8. Hence, there are 8 possible values of m, each giving a unique a. Therefore, the answer should be 8?

But wait, let me check with an example. Let me take a smaller case. Suppose instead of 210², let's take something simpler. Let's say that gcd(a, b) = 2, lcm(a, b) = 8. Then, similar logic: a*b = gcd(a,b)*lcm(a,b) = 2*8=16. Then, since gcd(a,b)=2, we can write a=2m, b=2n, gcd(m,n)=1, and lcm(a,b)=2*lcm(m,n)=8 ⇒ lcm(m,n)=4. Then m*n = lcm(m,n)*gcd(m,n) = 4*1=4. So m and n are coprime positive integers with m*n=4. The coprime pairs (m,n) are (1,4) and (4,1). So there are 2 ordered pairs, which correspond to 1 unordered pair. Therefore, the number of possible a's where a < b would be 1. Let me check:

Possible a's: a=2*1=2, and b=2*4=8. So only one possible a, which is 2. But let's see if there are others. Suppose m=2 and n=2? Then gcd(m,n)=2≠1. So invalid. So only the pairs (1,4) and (4,1). Therefore, the answer would be 1. Which is 2/2=1. So that works. Therefore, in that case, the formula holds. Therefore, for our problem, 16 ordered pairs, 8 unordered pairs, so answer is 8?

But wait, hold on. Let me try with another example. Let me take gcd(a,b)=6 and lcm(a,b)=6^2=36. Then a*b=6*36=216. So, writing a=6m, b=6n, gcd(m,n)=1, lcm(a,b)=6*lcm(m,n)=36 ⇒ lcm(m,n)=6. Then m*n=lcm(m,n)*gcd(m,n)=6*1=6. So m*n=6, gcd(m,n)=1. The coprime pairs (m,n) with m*n=6: (1,6), (2,3), (3,2), (6,1). So ordered pairs: 4. Unordered pairs: 2. So number of a's where a < b is 2. Let's check:

a=6*1=6, b=6*6=36.

a=6*2=12, b=6*3=18.

Are there any other pairs? For example, if m=1 and n=6, or m=2 and n=3. So two possible a's: 6 and 12. Hence, answer is 2. Which is 4 ordered pairs divided by 2, gives 2. So formula holds.

Therefore, returning to our original problem. The number of ordered pairs is 16, so number of unordered pairs is 8, hence the number of possible a's is 8.

Wait, but let me think again. Wait, but the problem says "how many possible values are there for a". Each a corresponds to a unique m. Since m can be any of the divisors of 210² such that m and n = 210²/m are coprime, and m < n. So the number of such m's is equal to the number of such pairs.

Alternatively, since each m is a divisor of 210², and for each m, if we can ensure that m is coprime with 210²/m, then m is a valid m.

But since m and n are coprime, then m must be a divisor of 210², and for each prime in the factorization of 210², the prime is entirely in m or entirely in n. Therefore, the number of such m's is equal to 2^4 = 16, but since m and n are distinct, we divide by 2, getting 8. Hence, the answer is 8.

But hold on, let me think if there's another way to compute this. Since m must be a divisor of 210², and coprime with n = 210²/m. So m must be a divisor of 210² such that m is squareful? Wait, no. Wait, m can be any divisor of 210², but with the condition that m and n are coprime. Which implies that m is a product of some subset of the prime factors of 210², each raised to their respective exponents. Wait, but 210² is (2×3×5×7)², so exponents of primes are 2 for each. But since m and n must be coprime, m can have any combination of the primes, but each prime can only be in m or in n, not both. So for each prime, you choose whether it is in m or in n. Therefore, the number of m's is equal to 2^4=16. However, each m corresponds to a unique n. However, since m can be in any order, so when we count the number of m's where m < n, it's half of 16, which is 8.

Therefore, the answer should be 8. But wait, let me check with the earlier example where 210 was replaced with 2, and the answer was 1.

Wait, but maybe there's something wrong here. Let me think again.

Suppose that m is a divisor of 210² with gcd(m, 210²/m)=1. Then m must be squareful? Wait, no, m can be any divisor. Wait, for example, let's take m=1. Then n=210²/1=210², and gcd(1, 210²)=1. So that works. Similarly, m=2², which is 4. Then n=210² / 4. Since 210² is (2×3×5×7)², dividing by 4 gives (3×5×7)², so n=(3×5×7)². Now, gcd(m, n) = gcd(4, (3×5×7)²) = 1, because 4 is 2² and the other number has no factor of 2. Similarly, for m=3², n= (2×5×7)², gcd(m,n)=1. Similarly for m=5², m=7², and any combinations.

Therefore, the number of m's is equal to the number of subsets of the set {2, 3, 5, 7} where each prime is included or not, with the exponents being the full exponents in 210². Wait, but actually, since m can have any combination of the primes, each prime can be present or not. So, the number of m's is 2^4=16. But since m and n are different, then for each m ≠ n, they form a pair, so the number of pairs where m < n is 8. Therefore, the answer is 8.

But wait, let me check with actual numbers. Let's think of the possible m's.

Each m corresponds to a subset of the primes {2, 3, 5, 7}. For example:

- The empty set: m=1 (since none of the primes are included), then n=210².

- The subset {2}: m=2²=4, n=(3×5×7)²= (105)².

- The subset {3}: m=3²=9, n=(2×5×7)²=70².

- The subset {5}: m=5²=25, n=(2×3×7)²=42².

- The subset {7}: m=7²=49, n=(2×3×5)²=30².

- The subset {2,3}: m=2²×3²= (2×3)²=36, n=(5×7)²=35².

- The subset {2,5}: m=2²×5²=100, n=(3×7)²=21².

- The subset {2,7}: m=2²×7²=196, n=(3×5)²=15².

- The subset {3,5}: m=3²×5²=225, n=(2×7)²=14².

- The subset {3,7}: m=3²×7²=441, n=(2×5)²=10².

- The subset {5,7}: m=5²×7²=1225, n=(2×3)²=6².

- The subset {2,3,5}: m=2²×3²×5²= (2×3×5)²=30², n=7²=49.

- The subset {2,3,7}: m=2²×3²×7²= (2×3×7)²=42², n=5²=25.

- The subset {2,5,7}: m=2²×5²×7²= (2×5×7)²=70², n=3²=9.

- The subset {3,5,7}: m=3²×5²×7²= (3×5×7)²=105², n=2²=4.

- The subset {2,3,5,7}: m=210², n=1.

So, total of 16 m's. Now, since we need m < n, we can check which of these m's are less than their corresponding n.

Let me list them:

1. m=1, n=210²: 1 < 210² ✔

2. m=4, n=105². Since 105² is 11025, and 4 < 11025 ✔

3. m=9, n=70²=4900. 9 < 4900 ✔

4. m=25, n=42²=1764. 25 < 1764 ✔

5. m=49, n=30²=900. 49 < 900 ✔

6. m=36, n=35²=1225. 36 < 1225 ✔

7. m=100, n=21²=441. 100 < 441 ✔

8. m=196, n=15²=225. 196 < 225 ✔

9. m=225, n=14²=196. 225 > 196 ❌

10. m=441, n=10²=100. 441 > 100 ❌

11. m=1225, n=6²=36. 1225 > 36 ❌

12. m=30²=900, n=49. 900 > 49 ❌

13. m=42²=1764, n=25. 1764 > 25 ❌

14. m=70²=4900, n=9. 4900 > 9 ❌

15. m=105²=11025, n=4. 11025 > 4 ❌

16. m=210², n=1. 210² > 1 ❌

So, out of these 16 m's, only the first 8 (from 1 to 8) satisfy m < n. The rest have m > n. Therefore, there are exactly 8 values of m where m < n, hence 8 possible values of a. Therefore, the answer is 8.

But wait, let me check the 8th one. The 8th m is 196, n=225. 196 is 14², and 225 is 15². 14²=196, 15²=225. Yes, 196 < 225. So that's valid.

Then, the remaining ones (from 9 to 16) have m > n. Therefore, the total number of valid a's is 8.

Therefore, the answer is 8.

Wait, but let me think again if there's another way to compute this. Since the number of such m's is equal to the number of subsets of the set {2,3,5,7}, which is 2^4=16. Each subset corresponds to a unique m. However, since for each subset, its complement corresponds to another m, and when you take the subset and its complement, one will be less than the other. Except when the subset is equal to its complement, but since there are 4 primes, which is even, there is no subset equal to its complement. Because for a subset to be equal to its complement, the subset would have to contain exactly half the elements. But since 4 is even, there are subsets of size 2. However, for each subset, the complement is a different subset. But in our case, since we have 4 primes, the number of subsets is 16, so the number of unordered pairs is 8. Therefore, the answer is 8.

Therefore, I think the answer is 8. However, let me check once more.

Alternatively, since each m must be a divisor of 210², and coprime with n = 210²/m. Which implies that m is a squareful? Not necessarily, but in our case, m is a product of primes each raised to their maximum exponent in 210². Wait, no. Wait, m can be any combination of primes, each with exponent 0 or 2. Wait, no. Wait, let me clarify.

Wait, 210² is (2×3×5×7)² = 2²×3²×5²×7². So when we take m as a divisor of 210², and we want m and n = 210²/m to be coprime, then m must have exponents only for some primes, and n will have exponents for the remaining primes, with each prime having exponent 0 or 2 in m.

But since m and n are coprime, no prime can be in both m and n. Therefore, for each prime in 210², we decide whether to include it in m (with exponent 2) or in n (with exponent 2). Therefore, m can be thought of as the product over some subset of {2, 3, 5, 7} each squared. So for example, if the subset is {2,3}, then m = 2²×3², and n = 5²×7².

Therefore, the number of such m's is the number of subsets of {2, 3, 5, 7}, which is 2^4 = 16. Each subset corresponds to a unique m. However, when considering m < n, we need to count the subsets where the corresponding m is less than n.

But since for each subset S, the complement subset S' corresponds to n, and the product m is the product of primes in S squared, and n is the product of primes in S' squared. Then, comparing m and n would depend on the sizes of S and S', or the actual products.

But in our earlier enumeration, we saw that for subsets of size 0 (empty set) or 1 or 2, etc., some of them result in m < n and others m > n.

However, the key point is that exactly half of the subsets will have m < n and the other half will have m > n. Since there are 16 subsets, and they come in pairs where one is S and the other is S', and for each pair, exactly one of m < n or m > n holds. Therefore, the number of subsets where m < n is 16 / 2 = 8, which is the answer.

Therefore, the answer is 8.

But just to be thorough, let me count the number of subsets S where the product of primes in S squared is less than the product of primes in S' squared.

Note that the product of primes in S squared is (product of primes in S)^2, same with S'. Therefore, comparing m and n is equivalent to comparing (product of primes in S)^2 and (product of primes in S')^2, which is the same as comparing product of primes in S and product of primes in S'.

Therefore, the problem reduces to counting how many subsets S of {2, 3, 5, 7} have product less than the product of the complement subset.

Note that the complement subset S' has product equal to (product of all primes) / (product of primes in S).

Let me denote P = 2×3×5×7 = 210.

So the product of primes in S is, let's say, x, and the product of primes in S' is P / x. Then, we want x < P / x ⇒ x² < P ⇒ x < sqrt(P).

Therefore, the number of subsets S where product of primes in S is less than sqrt(P) is equal to the number of subsets S where m < n.

So, let's compute sqrt(P) where P = 210. sqrt(210) ≈ 14.49. Therefore, we need the subsets S whose product of primes is less than 14.49.

The primes are 2, 3, 5, 7.

So, let's compute the product of primes for each subset:

- Empty set: product = 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

Now, we need to find the subsets where the product is less than sqrt(210) ≈ 14.49. Let's list those:

- Empty set: 1 < 14.49 ✔

- {2}: 2 < 14.49 ✔

- {3}: 3 < 14.49 ✔

- {5}: 5 < 14.49 ✔

- {7}: 7 < 14.49 ✔

- {2, 3}: 6 < 14.49 ✔

- {2, 5}: 10 < 14.49 ✔

- {2, 7}: 14 < 14.49 ✔ (since 14 is less than 14.49)

- {3, 5}: 15 > 14.49 ❌

- {3, 7}: 21 > 14.49 ❌

- {5, 7}: 35 > 14.49 ❌

- {2, 3, 5}: 30 > 14.49 ❌

And so on. Therefore, only the first 8 subsets (from the empty set up to {2,7}) have product less than sqrt(210). Therefore, there are 8 subsets where x < P/x, which corresponds to m < n. Therefore, 8 values of a.

Therefore, this confirms the previous result.

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

**Final Answer**
\boxed{8}

To determine how many possible values there are for $ a $ given that:

- $ \gcd(a, b) = 210 $
- $ \text{lcm}(a, b) = 210^3 $
- $ a < b $, and $ a, b \in \mathbb{Z}^+ $

we begin with the identity:

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

Substituting the known values:

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

Let’s factorize $ 210 = 2 \cdot 3 \cdot 5 \cdot 7 $, so:

$$
210^4 = (2 \cdot 3 \cdot 5 \cdot 7)^4 = 2^4 \cdot 3^4 \cdot 5^4 \cdot 7^4
$$

Let’s now define:

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

Then:

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

Also, since $ \gcd(a, b) = 210 $, this implies $ \gcd(m, n) = 1 $.

We now have the problem: **find the number of positive integers $ m $ such that** $ \gcd(m, n) = 1 $ and $ m \cdot n = 210^2 $, **with $ m < n $**.

---

### Step 1: Factorization and Coprime Constraints

We know that the prime factorization of $ 210^2 $ is:

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

For $ m \cdot n = 210^2 $ and $ \gcd(m, n) = 1 $, we must assign each of the prime factors to **either $ m $ or $ n $** entirely — not both — because if a prime is in both, the gcd would be at least that prime. So for each of the 4 primes (2, 3, 5, 7), we can assign them to either $ m $ or $ n $.

Thus, the number of such ordered pairs $ (m, n) $ is:

$$
2^4 = 16
$$

Each such assignment corresponds to a unique ordered pair $ (m, n) $, where $ m $ and $ n $ are coprime, and their product is $ 210^2 $.

---

### Step 2: Counting Valid $ a $ with $ a < b $

Since $ a = 210 \cdot m $ and $ b = 210 \cdot n $, and we require $ a < b $, this corresponds to $ m < n $. Each ordered pair $ (m, n) $ with $ m \ne n $ gives a unique pair $ (a, b) $ with $ a < b $.

From the 16 ordered pairs $ (m, n) $, exactly half will satisfy $ m < n $, because for every ordered pair $ (m, n) $, the reverse $ (n, m) $ is also valid and disjoint. So:

$$
\text{Number of valid } (a, b) \text{ with } a < b = \frac{16}{2} = 8
$$

---

### Final Answer

Thus, there are exactly $ \boxed{8} $ possible values for $ a $ that satisfy all the given conditions.

$$
\boxed{8}
$$

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 $a$.