Okay, let's tackle this problem step by step. The question is: If \( x^3 \) is a positive factor of \( 10! \), how many possible integer values of \( x \) are there? 

First, I need to understand what the problem is asking. It says that \( x^3 \) must be a factor of \( 10! \). So, \( x^3 \) divides \( 10! \) without leaving a remainder. My goal is to find all positive integers \( x \) such that this condition holds. Then, count how many such integers there are.

To approach this, I recall that if a number divides another number, then the prime factorization of the divisor must have exponents that are less than or equal to those in the dividend. So, maybe I should start by finding the prime factorization of \( 10! \), then see how \( x^3 \) can be a factor of that.

Let me start by finding the prime factors of \( 10! \). The prime numbers less than or equal to 10 are 2, 3, 5, 7. 

Calculating the exponents for each prime in \( 10! \):

For prime 2:
\( \lfloor \frac{10}{2} \rfloor + \lfloor \frac{10}{4} \rfloor + \lfloor \frac{10}{8} \rfloor = 5 + 2 + 1 = 8 \)

For prime 3:
\( \lfloor \frac{10}{3} \rfloor + \lfloor \frac{10}{9} \rfloor = 3 + 1 = 4 \)

For prime 5:
\( \lfloor \frac{10}{5} \rfloor = 2 \)

For prime 7:
\( \lfloor \frac{10}{7} \rfloor = 1 \)

So, the prime factorization of \( 10! \) is \( 2^8 \times 3^4 \times 5^2 \times 7^1 \).

Now, \( x^3 \) must divide \( 10! \). Let's denote the prime factorization of \( x \) as \( 2^a \times 3^b \times 5^c \times 7^d \). Then, \( x^3 \) would be \( 2^{3a} \times 3^{3b} \times 5^{3c} \times 7^{3d} \).

For \( x^3 \) to divide \( 10! \), the exponents in \( x^3 \) must be less than or equal to those in \( 10! \). Therefore, we have the following inequalities:

1. \( 3a \leq 8 \) (for prime 2)
2. \( 3b \leq 4 \) (for prime 3)
3. \( 3c \leq 2 \) (for prime 5)
4. \( 3d \leq 1 \) (for prime 7)

We need to find non-negative integers \( a, b, c, d \) satisfying these inequalities. Then, the number of possible \( x \) is the product of the number of possible values for each exponent \( a, b, c, d \).

Let's solve each inequality one by one.

Starting with the first inequality: \( 3a \leq 8 \). Dividing both sides by 3 gives \( a \leq \frac{8}{3} \approx 2.666 \). Since \( a \) must be an integer, the maximum value \( a \) can take is 2. So, possible values for \( a \) are 0, 1, 2. That's 3 possibilities.

Next, the second inequality: \( 3b \leq 4 \). Dividing by 3 gives \( b \leq \frac{4}{3} \approx 1.333 \). So, \( b \) can be 0, 1. That's 2 possibilities.

Third inequality: \( 3c \leq 2 \). Dividing by 3 gives \( c \leq \frac{2}{3} \approx 0.666 \). Since \( c \) is an integer, the only possible value is 0. So, only 1 possibility here.

Fourth inequality: \( 3d \leq 1 \). Dividing by 3 gives \( d \leq \frac{1}{3} \approx 0.333 \). Again, since \( d \) must be an integer, the only possible value is 0. So, 1 possibility here.

Therefore, multiplying the number of possibilities for each exponent:

Total \( x \) values = 3 (for a) * 2 (for b) * 1 (for c) * 1 (for d) = 3*2*1*1 = 6.

Wait, but hold on. Let me verify this. If each exponent is independent, then yes, multiplying the possibilities. But let me check with an example. For instance, take \( a=2 \), \( b=1 \), \( c=0 \), \( d=0 \). Then \( x = 2^2 * 3^1 = 4*3 = 12 \). Then \( x^3 = 12^3 = 1728 \). Let's check if 1728 divides 10! which is 3628800. 3628800 ÷ 1728 = 2100. So yes, it divides.

Another example: \( a=0 \), \( b=0 \), \( c=0 \), \( d=0 \). Then \( x=1 \), and \( x^3=1 \), which divides any number, including 10!.

Wait, but according to the calculation above, the total number of possible x is 6. Let me list them out to make sure.

Each combination of exponents:

For a: 0,1,2

For b: 0,1

For c: 0

For d: 0

So, possible combinations:

1. a=0, b=0: x=1

2. a=0, b=1: x=3

3. a=1, b=0: x=2

4. a=1, b=1: x=2*3=6

5. a=2, b=0: x=4

6. a=2, b=1: x=4*3=12

So, x can be 1, 3, 2, 6, 4, 12. That's 6 values. Wait, but hold on. Is that all? Let me check if there are more possibilities. For example, can c or d be non-zero? For c, since 3c ≤2, c can only be 0. Similarly, d can only be 0. So, c and d are fixed at 0. So, the possible x's are as listed above. So, 6 values. Therefore, the answer should be 6.

But wait, let me cross-validate this with another approach. Maybe by considering that x must be a factor of the cube root of 10!.

But the cube root of 10! is not an integer, so that might complicate things. Alternatively, perhaps there's a way to compute the number of x's by considering exponents as I did before.

Alternatively, since x^3 divides 10!, then x must divide the radical of 10! raised to the power of 3. Wait, maybe that's not helpful. Alternatively, the exponents in x must be such that when multiplied by 3, they don't exceed the exponents in 10!.

But that's exactly what I did. So, the approach is correct.

Therefore, the number of possible x is 6.

But let me check once again. Maybe I made a mistake in the inequalities.

First inequality: 3a ≤8 → a ≤8/3 ≈2.666 → a=0,1,2. Correct.

Second inequality: 3b ≤4 → b ≤4/3≈1.333 → b=0,1. Correct.

Third inequality: 3c ≤2 → c ≤2/3≈0.666 → c=0. Correct.

Fourth inequality: 3d ≤1 → d ≤1/3≈0.333 → d=0. Correct.

So, exponents a: 3 options, b: 2 options, c:1, d:1. 3*2*1*1=6. So, yes, 6 possible x's.

But let me verify with an example. Let's take x=12. Then x^3=1728. 10! is 3628800. 3628800 divided by 1728 is 2100, which is an integer. So, 1728 divides 3628800.

Another example: x=6. x^3=216. 3628800 ÷216=16800. Integer, so yes.

x=4: x^3=64. 3628800 ÷64=56700. Integer.

x=3: 27. 3628800 ÷27=134400.

x=2: 8. 3628800 ÷8=453600.

x=1: 1. 3628800 ÷1=3628800.

All of these are integers, so x=1,2,3,4,6,12 are valid. Therefore, 6 values. So, the answer is 6.

But wait, the question says "positive integer values of x". So, x must be positive. So, all the x's we found are positive integers. Therefore, the count is 6.

Therefore, the final answer is 6.
Therefore, the number of possible integer values of $x$ is $\boxed{6}$.
</think>