Okay, let's see. The problem is asking for how many real values of x make the expression sqrt(120 - sqrt(x)) an integer. Hmm. Alright, let's break this down step by step.

First, since the square root is involved, we need to make sure that the expression inside the square root is non-negative. So, 120 - sqrt(x) must be greater than or equal to zero. That gives us the inequality 120 - sqrt(x) ≥ 0. Let me write that down:

120 - sqrt(x) ≥ 0  
=> sqrt(x) ≤ 120  
=> x ≤ (120)^2  
=> x ≤ 14400.

But since sqrt(x) is defined only for x ≥ 0, the domain of x is [0, 14400]. So x has to be between 0 and 14400, inclusive.

Now, the problem states that sqrt(120 - sqrt(x)) is an integer. Let's denote this integer as n. So we can write:

sqrt(120 - sqrt(x)) = n  
=> 120 - sqrt(x) = n²  
=> sqrt(x) = 120 - n²  
=> x = (120 - n²)².

But we need to ensure that sqrt(x) is real and non-negative, which it will be as long as 120 - n² ≥ 0. So 120 - n² ≥ 0  
=> n² ≤ 120  
=> n ≤ sqrt(120)  
Since n is an integer, the maximum possible value of n is the floor of sqrt(120). Let's compute sqrt(120):

sqrt(121) is 11, so sqrt(120) is slightly less than 11. Therefore, the maximum integer n can be is 10. Since n is a non-negative integer (because square roots are non-negative), n can take values from 0 up to 10 inclusive. Let's check that.

Wait, but n is sqrt(120 - sqrt(x)), which is a square root. So n must also be a non-negative integer. Therefore, n can be 0,1,2,...,10. Let's verify for each n from 0 to 10 whether x is a real number in the domain [0,14400].

But wait, when n=0:  
sqrt(120 - sqrt(x)) = 0  
=> 120 - sqrt(x) = 0  
=> sqrt(x) = 120  
=> x = 120² = 14400. Which is within the domain. So that's valid.

For n=1:  
sqrt(120 - sqrt(x)) =1  
=> 120 - sqrt(x) =1  
=> sqrt(x)=119  
=> x=119²=14161. Which is also within the domain.

Similarly, for n=2:  
sqrt(120 - sqrt(x))=2  
=> 120 - sqrt(x)=4  
=> sqrt(x)=116  
=> x=116²=13456. Valid.

Continuing this way, each n from 0 to 10 will give a corresponding x. But wait, let's check the last value, n=10:

sqrt(120 - sqrt(x))=10  
=> 120 - sqrt(x)=100  
=> sqrt(x)=20  
=> x=400. Which is within [0,14400]. So that's valid.

So n can be 0,1,2,...,10. That's 11 values. Therefore, there are 11 real values of x that satisfy the equation. So the answer should be 11. But wait, let me double-check.

Wait, but n is an integer, but sqrt(x) must also be real. So for each n, sqrt(x)=120 -n² must be non-negative. So 120 -n² ≥0. So n² ≤120. As we saw, n can be 0 to 10. So 11 integers. Each integer n gives exactly one x, which is (120 -n²)². Therefore, 11 real values of x. Therefore, the answer is 11. Hmm.

Wait, but let me check if there are any constraints I missed. For example, when n=0, x=14400. That's valid. For n=1, x=14161. Also valid. All the way up to n=10, x=400. So each x is unique because each n gives a different x. Since (120 -n²) is different for each n in 0 to 10, their squares will also be different. So each x is distinct. Therefore, the number of real x is 11.

But let me verify with another approach. Let's suppose that sqrt(120 - sqrt(x)) = n, where n is an integer. Then 120 - sqrt(x) must be a perfect square. Let's denote m = sqrt(x). Then 120 - m = n², so m = 120 -n². Since m must be non-negative, 120 -n² ≥0 => n² ≤120. So n can be 0,1,2,...,10 as before. Then x = m² = (120 -n²)². Each n gives a different x, so 11 solutions. Therefore, the answer is 11.

But wait, let me check if there are any other possible values of n. For example, negative integers. But n is sqrt(120 - sqrt(x)), which is a principal (non-negative) square root. Therefore, n must be non-negative. So negative integers are excluded. So n can only be 0 to 10. Therefore, 11 values.

So the answer should be 11. Therefore, the number of real values of x is 11. So the final answer is \boxed{11}.

**Final Answer**
\boxed{11}
Therefore, the number of real values of $x$ is $\boxed{11}$.
</think>