Explanation:
Okay, so I need to figure out for how many real values of x the expression √(120 - √x) is an integer. Hmm, let me start by understanding what the problem is asking.

First, the expression inside the square root must be non-negative because you can't take the square root of a negative number in real numbers. So, the inner square root, √x, must be real, which means x has to be greater than or equal to 0. Also, the entire expression under the outer square root, which is 120 minus √x, must be non-negative. Therefore, 120 - √x ≥ 0. Let me write that down:

120 - √x ≥ 0
Which implies that √x ≤ 120
Then squaring both sides (since both sides are non-negative), x ≤ 120² = 14400.

So x must be between 0 and 14400, inclusive. But x has to be such that √(120 - √x) is an integer. Let me denote that integer as n. So, let me set:

√(120 - √x) = n, where n is a non-negative integer.

Since square roots are non-negative, n must be an integer greater than or equal to 0. Also, since √x must be real, x ≥ 0.

So, squaring both sides of the equation:

120 - √x = n²
Therefore, √x = 120 - n²
Then squaring both sides again:

x = (120 - n²)²

But x has to be real, and since we already have conditions on n for √x to be real, let me think about the constraints on n.

First, √x = 120 - n² must be non-negative, since square roots are non-negative. Therefore:

120 - n² ≥ 0
Which implies n² ≤ 120
Therefore, n can be integers such that n² ≤ 120.

So n is an integer greater than or equal to 0, and n² ≤ 120. Let me find all such integers n.

What's the largest integer n for which n² ≤ 120? Let's compute sqrt(120). Well, sqrt(121) is 11, so sqrt(120) is a bit less than 11. So the largest integer n is 10, since 10² = 100, and 11² = 121, which is too big. Therefore, n can be integers from 0 up to 10 inclusive.

Wait, let me check. Let me compute n² for n=0 up to n=10:

n=0: 0
n=1: 1
n=2: 4
n=3: 9
n=4: 16
n=5: 25
n=6: 36
n=7: 49
n=8: 64
n=9: 81
n=10: 100
n=11: 121 (which is over 120, so n=11 is not allowed)

So n can be integers from 0 to 10 inclusive. Therefore, there are 11 possible integer values for n. However, I need to check if each of these values of n leads to a valid x. Because x is defined as (120 - n²)². But we also have the constraint that √x must be real, which we already considered by requiring that 120 - n² ≥ 0. Since we already have that n is between 0 and 10 inclusive, so 120 - n² is non-negative. Therefore, each n in 0 to 10 inclusive gives a valid x.

But wait, hold on. But x is (120 - n²)². However, since √x is 120 - n², which is non-negative, so x is (non-negative)^2, which is always non-negative. So that's okay. Therefore, each integer n from 0 to 10 inclusive gives a unique x. Therefore, there are 11 such values?

But wait, let me check if different n's can lead to same x. For example, suppose that for different n's, (120 - n²)² is same. Is that possible? Let me think. Suppose that for two different integers n and m, we have 120 - n² = 120 - m², or 120 - n² = -(120 - m²). But since 120 - n² and 120 - m² are both non-negative (since n and m are between 0 and 10), so 120 - n² = 120 - m² implies n² = m². Since n and m are non-negative integers, that would imply n = m. Therefore, each n gives unique x. Therefore, there are 11 different values of x.

But wait, but hold on. Wait, let me think again. Let me check for n and -n? But n is non-negative integer, so n can't be negative. Therefore, all n are non-negative, so each n corresponds to unique x. So, 11 different values?

But wait, hold on. Wait, but maybe some of these x's are the same? For example, if for some n and m, (120 - n²)² = (120 - m²)². Which would mean that 120 - n² = 120 - m² or 120 - n² = -(120 - m²). But again, since 120 - n² and 120 - m² are non-negative, so 120 - n² = -(120 - m²) would imply 120 - n² is negative, which is not allowed. Therefore, only 120 - n² = 120 - m², which gives n² = m², which as before gives n = m. Therefore, all x's are unique for each n. Therefore, number of real x is 11?

Wait, but let me check with an example. Let's take n=0. Then x=(120 - 0)^2=120²=14400. For n=1, x=(120 - 1)^2=119²=14161. For n=2, x=(120 - 4)=116²=13456. Continuing, each n gives a different x. Since 120 - n² is different for each n from 0 to 10, so when squared, they are different. Because if 120 - n² = 120 - m², then n = m. So, yes, unique x. So there are 11 values. Therefore, the answer is 11?

Wait, but hold on. Wait, the problem says "for how many real values of x". However, maybe there is something wrong here. Let me check again.

Wait, suppose n is an integer such that √(120 - √x) = n. So, solving for x gives x = (120 - n²)^2, but also √x must be real, which requires that √x = 120 - n² ≥ 0. Therefore, as before, n can be 0,1,...,10. Each gives a unique x, so 11 values. Therefore, answer is 11?

But wait, but the problem says "real values of x". However, maybe there are multiple x's for the same n? Wait, no, because x is uniquely determined by n. Because once n is fixed, x is fixed as (120 - n²)^2. So each n gives exactly one x. Therefore, with n from 0 to 10 inclusive, that's 11 values.

Wait, but let me think again. Let me check for n=10. Then √x = 120 - 10² = 120 - 100 = 20. Then x = 20² = 400. For n=9, √x = 120 - 81 = 39, so x = 39² = 1521. For n=8, √x = 120 - 64 = 56, x = 56² = 3136. So each n gives different x. Therefore, yes, 11 values.

But hold on. Let me check if there's another way to interpret the question. The problem says "for how many real values of x is √(120 - √x) an integer". So, maybe there are other x's where √(120 - √x) is integer? Wait, but we derived that if √(120 - √x) is integer, then x must be equal to (120 - n²)^2 for some integer n, with n between 0 and 10. Therefore, there are exactly 11 such x's. So answer is 11? But let me check for n=0. Let me substitute x=14400 into the original expression.

√(120 - √14400) = √(120 - 120) = √0 = 0, which is integer. For n=1, x=119²=14161, then √x=119, so 120 - 119 =1, √1=1, which is integer. Similarly for others, seems okay. So all these x's work.

But maybe there's a mistake here. Wait, but let me check if n can be negative? Wait, n is defined as √(120 - √x). Since square roots are non-negative, n has to be non-negative integer, so n ≥0. Therefore, n can't be negative. Therefore, n ranges from 0 to 10 inclusive, giving 11 values. Therefore, the answer should be 11?

Wait, but maybe there's something about the square roots? For example, √x is defined as the principal square root, which is non-negative, so √x is non-negative. Therefore, 120 - √x must be non-negative. Therefore, √x ≤ 120. Therefore, x ≤ 120²=14400. And x has to be non-negative. So the domain of x is [0, 14400].

But for each x in that domain, √(120 - √x) is real. But we want it to be integer. So, the problem reduces to finding integers n such that 120 - √x = n², so √x = 120 - n². Therefore, since √x is non-negative, 120 - n² ≥ 0, so n² ≤ 120. Therefore, n can be integers from 0 to 10.

Therefore, for each such n, we have a unique x, which is (120 - n²)^2. So, total number of x's is equal to the number of integers n with n² ≤ 120. Which is 11 integers (n=0,...,10). Therefore, the answer is 11.

Wait, but the question says "real values of x". But x is determined uniquely by n. So, each n gives exactly one real x. Therefore, 11 real numbers. So answer is 11?

Wait, but let me check if maybe different n's can lead to same x? Suppose that for two different n1 and n2, (120 - n1²)^2 = (120 - n2²)^2. Then, either 120 - n1² = 120 - n2², which implies n1² = n2², so n1 = n2 (since they are non-negative). Or 120 - n1² = - (120 - n2²). Which would imply 120 - n1² = -120 + n2². Then, 240 = n1² + n2². So, if there exists n1, n2 in 0,...,10 such that n1² + n2² = 240. Let me check if that's possible.

Maximum value of n² is 10²=100. So, two numbers each up to 100. So 100 + 100=200 < 240. Therefore, impossible. Therefore, there's no solution. Therefore, no two different n's can give same x. Therefore, there are 11 different x's.

Therefore, answer is 11.

Wait, but I'm confused because sometimes problems like this can have more solutions, so maybe I made a mistake here. Let me check with n=0 and n= something else?

Wait, for example, if n=0, x=(120 - 0)^2=14400. If there is another n such that (120 - n²)^2=14400, then 120 -n²=±120. But 120 -n²=120 implies n²=0, so n=0. Or 120 -n²=-120, which would mean n²=240, but n²=240 would require n≈15.49, which is outside our range (since n can be maximum 10). Therefore, only n=0. Therefore, unique. Similarly, for other x's. Therefore, no overlap. Therefore, 11 values.

So maybe the answer is 11? But let me check again.

Alternatively, maybe there's another interpretation. The problem says "for how many real values of x". Maybe there is something wrong with my reasoning? For example, maybe x must satisfy that √x is real, which is already considered. So x must be between 0 and 14400. But for each n from 0 to 10 inclusive, x=(120 -n²)^2, and these x's are all in [0,14400]. For example, when n=10, x=(120 -100)^2=20²=400, which is within 0 to 14400. When n=0, x=120²=14400. So yes.

But let me check if for example, x=0. Is that allowed? If x=0, then √x=0, so 120 - √x=120, √120 is not integer. Wait, but according to our previous reasoning, x=(120 -n²)^2. If x=0, then (120 -n²)^2=0, so 120 -n²=0, which implies n²=120, but n must be integer. However, 120 is not a perfect square. Therefore, x=0 is not among our solutions. Therefore, our solutions are only those x's where 120 -n² is integer, and n is integer. Therefore, x is (integer)^2. But in our case, 120 -n² is integer, so x is square of integer.

But in our case, the values of x are (120 -n²)^2 where n is integer between 0 and 10. Therefore, each x is unique. Therefore, 11 values.

Wait, but let me check n= sqrt(120 - √x). But n must be integer. Therefore, the problem is equivalent to finding x such that 120 - √x is a perfect square, because √(120 - √x) is integer implies that 120 - √x is a perfect square. Let me denote that 120 - √x = k², where k is non-negative integer. Therefore, √x = 120 - k², so x=(120 -k²)^2. Therefore, same as before. Therefore, k is non-negative integer such that 120 -k² ≥0, so k² ≤120. So k can be 0,...,10. Hence 11 values. Hence answer is 11.

But let me check for example, if k=10, then x=(120 - 100)^2=20²=400. Then √x=20, so 120 - 20=100, √100=10, which is integer. Correct. Similarly, for k=11, which we said is not allowed, 120 - 121= -1, which would give √x=-1, which is invalid. Therefore, k can't be 11.

So, seems correct. Therefore, answer is 11.

Wait, but I feel like sometimes problems like these have more solutions because of different roots or something. Wait, but in this case, we have only one equation. Let me think. Let me consider that maybe there are multiple n's leading to same x? But as we saw, no. Because if (120 -n²)^2 = (120 -m²)^2, then either 120 -n² = 120 -m² or 120 -n² = m² -120. But since 120 -n² is non-negative, the second case would mean that m² -120 is non-negative, so m² ≥120. But m is an integer between 0 and 10, so m² ≤100. Hence, m² -120 ≤ -20 <0. Hence, impossible. Therefore, only possible if 120 -n² = 120 -m², so n=m. Therefore, unique. So 11 values.

Therefore, answer should be 11. Hmm. Wait, but let me check with actual numbers.

Let me list all possible n from 0 to 10 and compute x:

n=0: x=(120 - 0)^2 = 14400
n=1: (120 - 1)^2 = 119² = 14161
n=2: 118² = 13924
n=3: 117² = 13689
n=4: 116² = 13456
n=5: 115² = 13225
n=6: 114² = 12996
n=7: 113² = 12769
n=8: 112² = 12544
n=9: 111² = 12321
n=10: 110² = 12100

So that's 11 different x's. Each of these x's is unique. Therefore, 11 values. Therefore, the answer is 11.

But wait, the problem says "real values of x". Are there any other real numbers x that could satisfy the condition? For example, suppose x is such that √(120 - √x) is integer. But according to our reasoning, that would require that √x = 120 - n², so x=(120 - n²)^2. So, unless there is some other way...

Wait, suppose someone says that √(120 - √x) is integer, but maybe √x is not equal to 120 - n²? Wait, no. Let me think again.

Let me denote y = √x. Then, since x ≥0, y ≥0. The expression becomes √(120 - y) is integer. Let me denote that √(120 - y) = k, where k is integer ≥0. Then, 120 - y = k², so y = 120 - k². Since y ≥0, 120 -k² ≥0, so k² ≤120. Therefore, k can be 0,...,10 as before. Then, since y = √x = 120 -k², then x = y² = (120 -k²)^2. Therefore, same result. Therefore, there are exactly 11 values of x. Therefore, answer is 11.

Therefore, I think my answer is correct. But wait, but the original problem says "for how many real values of x". So, maybe the answer is 11? But maybe the problem is from an Olympiad or something where sometimes the answer is different. Wait, but let me check if for example, when n=0, x=14400. Then √x=120, so √(120 - √x)=√(120 -120)=√0=0, which is integer. Correct. Similarly, for n=10, x=110²=12100, √x=110, so √(120 - 110)=√10. Wait, wait! Wait, hold on. Wait, this is a problem. Wait, earlier, I thought that when n=10, √x=120 -n²=120 -100=20. Therefore, x=20²=400. But in the above, if I set k=10, then y=√x=120 -k²=120 -100=20, so x=20²=400. But in my previous list, for n=10, I had x=110²=12100. Wait, this is a contradiction! Wait, where did I make the mistake here?

Wait, let me go back. Wait, earlier, when I thought of k= n, I said for n=10, x=(120 -n²)^2. But when n=10, that's (120 -10²)= 120 -100=20. Then x=20²=400. But in my list above, I mistakenly wrote n=10: 110²=12100. Wait, what's going on here?

Wait, this seems to be an error in my previous list. Let me check again.

Earlier, I thought that for n=0 to 10, x=(120 - n²)^2. But when I listed them, I incorrectly wrote for n=10, (120 - 10²)²? Wait, no, in my previous list I said:

n=10: 110²=12100. Wait, but 120 -n² for n=10 is 120 -100=20. Therefore, x=20²=400. Therefore, my previous list was wrong. So where did I get 110² from? That was a mistake.

Wait, so my previous list was wrong. Let me correct it.

So let me redo the list. Let me compute x=(120 - n²)^2 for n=0,...,10.

For n=0: 120 - 0 = 120. So x=120²=14400
For n=1: 120 -1=119, x=119²=14161
For n=2: 120 -4=116, x=116²=13456
For n=3: 120 -9=111, x=111²=12321
Wait, hold on. Wait, 120 -3²=120-9=111? Wait, 120-9 is 111? No, 120-9=111? Wait, 120-9 is 111? Wait, 120 - 9 is 111? Wait, 120-10 is 110, so 120-9=111. Yes. So x=111²=12321.
For n=4: 120 - 16 = 104. So x=104²=10816
For n=5: 120 -25=95, x=95²=9025
n=6: 120 - 36=84, x=84²=7056
n=7: 120 -49=71, x=71²=5041
n=8: 120 -64=56, x=56²=3136
n=9: 120 -81=39, x=39²=1521
n=10: 120 -100=20, x=20²=400

So my previous list had errors in the numbers. I had mistakenly written for n=10, 110², which is wrong. Actually, for n=10, it is 20²=400. Similarly, for n=3, earlier I had written 117², but it's actually 111².

So the correct list is:

n=0: 14400
n=1: 14161
n=2: 13456
n=3: 12321
n=4: 10816
n=5: 9025
n=6: 7056
n=7: 5041
n=8: 3136
n=9: 1521
n=10: 400

Therefore, these are 11 different x's. Each is unique. For example, let me check if any of these numbers repeat. For example, 400 is unique, 1521 is unique, etc. Since each of them is square of different numbers (120 -n²), which are different numbers, so their squares are different. Hence, 11 different x's.

Therefore, my initial conclusion was correct, but I had a mistake in the intermediate step where I incorrectly calculated the values for n=10 and n=3. But that doesn't affect the conclusion that there are 11 values.

Therefore, the answer is 11.

But let me check once more with n=10. If x=400, then √x=20. Then 120 - √x=100. Then √100=10, which is integer. Correct.

Similarly, for n=3, x=111²=12321. √x=111. Then 120 - 111=9. √9=3, which is integer. Correct.

Therefore, my mistake was in the list, but the conclusion that there are 11 x's is correct.

Therefore, the answer is 11.

Wait, but let me check another example. For n=7: x=71²=5041. Then √x=71. 120 -71=49. √49=7, which is integer. Correct.

Therefore, all the values work. So, there are 11 real values. Therefore, the answer is 11.

But why did I think earlier that there might be something wrong? Maybe because of miscalculating the x's. But no, after checking, everything is correct.

Therefore, I think the correct answer is 11.

**Final Answer**
\boxed{11}

To determine how many **real values of $ x $** satisfy the condition that
$$
\sqrt{120 - \sqrt{x}} \text{ is an integer},
$$
we proceed step by step.

---

### Step 1: Let $ \sqrt{120 - \sqrt{x}} = n $, where $ n $ is a non-negative integer.

Since square roots are always non-negative, $ n \geq 0 $. Squaring both sides, we get:

$$
120 - \sqrt{x} = n^2 \quad \Rightarrow \quad \sqrt{x} = 120 - n^2.
$$

Since the square root of $ x $ must be real, it must also be non-negative, so:

$$
120 - n^2 \geq 0 \quad \Rightarrow \quad n^2 \leq 120.
$$

Thus, $ n $ can be any **integer** such that $ n^2 \leq 120 $.

---

### Step 2: Find all valid non-negative integers $ n $ such that $ n^2 \leq 120 $

We know:

$$
\sqrt{120} \approx 10.95, \quad \text{so the largest integer } n \text{ such that } n^2 \leq 120 \text{ is } n = 10.
$$

Hence, $ n $ ranges from $ 0 $ to $ 10 $, inclusive. That gives us **11** possible values for $ n $.

---

### Step 3: For each such $ n $, compute $ x $

From the equation:

$$
x = (\sqrt{120 - \sqrt{x}})^2 = (120 - n^2)^2.
$$

So for each $ n $ in $ \{0, 1, 2, \ldots, 10\} $, we compute:

$$
x = (120 - n^2)^2.
$$

Now, each of these values gives a unique real $ x $, and all such $ x $ are in the range $ [0, 120^2 = 14400] $, satisfying all required conditions.

---

### Step 4: Ensure no duplicates among calculated $ x $

Note that for **each** $ n \in \{0, 1, 2, \ldots, 10\} $, we compute a unique value of $ x $, because:

- The square of a number is **unique for distinct inputs**
- $ 120 - n^2 $ is **unique** for each $ n $
- Therefore, $ x = (120 - n^2)^2 $ is **unique** for each $ n $

Hence, the mapping from $ n $ to $ x $ is injective (one-to-one), and thus the list of values for $ x $ contains exactly 11 **distinct real** values.

---

### Final Conclusion

There are **11** real values of $ x $ such that $ \sqrt{120 - \sqrt{x}} $ is an integer.

$$
\boxed{11}
$$

[Assistant]
Explanation:
Okay, so I need to figure out for how many real values of x the expression √(120 - √x) is an integer. Hmm, let me start by understanding what the problem is asking.

First, the expression inside the square root must be non-negative because you can't take the square root of a negative number in real numbers. So, the inner square root, √x, must be real, which means x has to be greater than or equal to 0. Also, the entire expression under the outer square root, which is 120 minus √x, must be non-negative. Therefore, 120 - √x ≥ 0. Let me write that down:

120 - √x ≥ 0
Which implies that √x ≤ 120
Then squaring both sides (since both sides are non-negative), x ≤ 120² = 14400.

So x must be between 0 and 14400, inclusive. But x has to be such that √(120 - √x) is an integer. Let me denote that integer as n. So, let me set:

√(120 - √x) = n, where n is a non-negative integer.

Since square roots are non-negative, n must be an integer greater than or equal to 0. Also, since √x must be real, x ≥ 0.

So, squaring both sides of the equation:

120 - √x = n²
Therefore, √x = 120 - n²
Then squaring both sides again:

x = (120 - n²)²

But x has to be real, and since we already have conditions on n for √x to be real, let me think about the constraints on n.

First, √x = 120 - n² must be non-negative, since square roots are non-negative. Therefore:

120 - n² ≥ 0
Which implies n² ≤ 120
Therefore, n can be integers such that n² ≤ 120.

So n is an integer greater than or equal to 0, and n² ≤ 120. Let me find all such integers n.

What's the largest integer n for which n² ≤ 120? Let's compute sqrt(120). Well, sqrt(121) is 11, so sqrt(120) is a bit less than 11. So the largest integer n is 10, since 10² = 100, and 11² = 121, which is too big. Therefore, n can be integers from 0 up to 10 inclusive.

Wait, let me check. Let me compute n² for n=0 up to n=10:

n=0: 0
n=1: 1
n=2: 4
n=3: 9
n=4: 16
n=5: 25
n=6: 36
n=7: 49
n=8: 64
n=9: 81
n=10: 100
n=11: 121 (which is over 120, so n=11 is not allowed)

So n can be integers from 0 to 10 inclusive. Therefore, there are 11 possible integer values for n. However, I need to check if each of these values of n leads to a valid x. Because x is defined as (120 - n²)². But we also have the constraint that √x must be real, which we already considered by requiring that 120 - n² ≥ 0. Since we already have that n is between 0 and 10 inclusive, so 120 - n² is non-negative. Therefore, each n in 0 to 10 inclusive gives a valid x.

But wait, hold on. But x is (120 - n²)². However, since √x is 120 - n², which is non-negative, so x is (non-negative)^2, which is always non-negative. So that's okay. Therefore, each integer n from 0 to 10 inclusive gives a unique x. Therefore, there are 11 such values?

But wait, let me check if different n's can lead to same x. For example, suppose that for different n's, (120 - n²)² is same. Is that possible? Let me think. Suppose that for two different integers n and m, we have 120 - n² = 120 - m², or 120 - n² = -(120 - m²). But since 120 - n² and 120 - m² are both non-negative (since n and m are between 0 and 10), so 120 - n² = 120 - m² implies n² = m². Since n and m are non-negative integers, that would imply n = m. Therefore, each n gives unique x. Therefore, there are 11 different values of x.

But wait, but hold on. Wait, let me think again. Let me check for n and -n? But n is non-negative integer, so n can't be negative. Therefore, all n are non-negative, so each n corresponds to unique x. So, 11 different values?

But wait, hold on. Wait, but maybe some of these x's are the same? For example, if for some n and m, (120 - n²)² = (120 - m²)². Which would mean that 120 - n² = 120 - m² or 120 - n² = -(120 - m²). But again, since 120 - n² and 120 - m² are non-negative, so 120 - n² = -(120 - m²) would imply 120 - n² is negative, which is not allowed. Therefore, only 120 - n² = 120 - m², which gives n² = m², which as before gives n = m. Therefore, all x's are unique for each n. Therefore, number of real x is 11?

Wait, but let me check with an example. Let's take n=0. Then x=(120 - 0)^2=120²=14400. For n=1, x=(120 - 1)^2=119²=14161. For n=2, x=(120 - 4)=116²=13456. Continuing, each n gives a different x. Since 120 - n² is different for each n from 0 to 10, so when squared, they are different. Because if 120 - n² = 120 - m², then n = m. So, yes, unique x. So there are 11 values. Therefore, the answer is 11?

Wait, but hold on. Wait, the problem says "for how many real values of x". However, maybe there is something wrong here. Let me check again.

Wait, suppose n is an integer such that √(120 - √x) = n. So, solving for x gives x = (120 - n²)^2, but also √x must be real, which requires that √x = 120 - n² ≥ 0. Therefore, as before, n can be 0,1,...,10. Each gives a unique x, so 11 values. Therefore, answer is 11?

But wait, but the problem says "real values of x". However, maybe there are multiple x's for the same n? Wait, no, because x is uniquely determined by n. Because once n is fixed, x is fixed as (120 - n²)^2. So each n gives exactly one x. Therefore, with n from 0 to 10 inclusive, that's 11 values.

Wait, but let me think again. Let me check for n=10. Then √x = 120 - 10² = 120 - 100 = 20. Then x = 20² = 400. For n=9, √x = 120 - 81 = 39, so x = 39² = 1521. For n=8, √x = 120 - 64 = 56, x = 56² = 3136. So each n gives different x. Therefore, yes, 11 values.

But hold on. Let me check if there's another way to interpret the question. The problem says "for how many real values of x is √(120 - √x) an integer". So, maybe there are other x's where √(120 - √x) is integer? Wait, but we derived that if √(120 - √x) is integer, then x must be equal to (120 - n²)^2 for some integer n, with n between 0 and 10. Therefore, there are exactly 11 such x's. So answer is 11? But let me check for n=0. Let me substitute x=14400 into the original expression.

√(120 - √14400) = √(120 - 120) = √0 = 0, which is integer. For n=1, x=119²=14161, then √x=119, so 120 - 119 =1, √1=1, which is integer. Similarly for others, seems okay. So all these x's work.

But maybe there's a mistake here. Wait, but let me check if n can be negative? Wait, n is defined as √(120 - √x). Since square roots are non-negative, n has to be non-negative integer, so n ≥0. Therefore, n can't be negative. Therefore, n ranges from 0 to 10 inclusive, giving 11 values. Therefore, the answer should be 11?

Wait, but maybe there's something about the square roots? For example, √x is defined as the principal square root, which is non-negative, so √x is non-negative. Therefore, 120 - √x must be non-negative. Therefore, √x ≤ 120. Therefore, x ≤ 120²=14400. And x has to be non-negative. So the domain of x is [0, 14400].

But for each x in that domain, √(120 - √x) is real. But we want it to be integer. So, the problem reduces to finding integers n such that 120 - √x = n², so √x = 120 - n². Therefore, since √x is non-negative, 120 - n² ≥ 0, so n² ≤ 120. Therefore, n can be integers from 0 to 10.

Therefore, for each such n, we have a unique x, which is (120 - n²)^2. So, total number of x's is equal to the number of integers n with n² ≤ 120. Which is 11 integers (n=0,...,10). Therefore, the answer is 11.

Wait, but the question says "real values of x". But x is determined uniquely by n. So, each n gives exactly one real x. Therefore, 11 real numbers. So answer is 11?

Wait, but let me check if maybe different n's can lead to same x? Suppose that for two different n1 and n2, (120 - n1²)^2 = (120 - n2²)^2. Then, either 120 - n1² = 120 - n2², which implies n1² = n2², so n1 = n2 (since they are non-negative). Or 120 - n1² = - (120 - n2²). Which would imply 120 - n1² = -120 + n2². Then, 240 = n1² + n2². So, if there exists n1, n2 in 0,...,10 such that n1² + n2² = 240. Let me check if that's possible.

Maximum value of n² is 10²=100. So, two numbers each up to 100. So 100 + 100=200 < 240. Therefore, impossible. Therefore, there's no solution. Therefore, no two different n's can give same x. Therefore, there are 11 different x's.

Therefore, answer is 11.

Wait, but I'm confused because sometimes problems like this can have more solutions, so maybe I made a mistake here. Let me check with n=0 and n= something else?

Wait, for example, if n=0, x=(120 - 0)^2=14400. If there is another n such that (120 - n²)^2=14400, then 120 -n²=±120. But 120 -n²=120 implies n²=0, so n=0. Or 120 -n²=-120, which would mean n²=240, but n²=240 would require n≈15.49, which is outside our range (since n can be maximum 10). Therefore, only n=0. Therefore, unique. Similarly, for other x's. Therefore, no overlap. Therefore, 11 values.

So maybe the answer is 11? But let me check again.

Alternatively, maybe there's another interpretation. The problem says "for how many real values of x". Maybe there is something wrong with my reasoning? For example, maybe x must satisfy that √x is real, which is already considered. So x must be between 0 and 14400. But for each n from 0 to 10 inclusive, x=(120 -n²)^2, and these x's are all in [0,14400]. For example, when n=10, x=(120 -100)^2=20²=400, which is within 0 to 14400. When n=0, x=120²=14400. So yes.

But let me check if for example, x=0. Is that allowed? If x=0, then √x=0, so 120 - √x=120, √120 is not integer. Wait, but according to our previous reasoning, x=(120 -n²)^2. If x=0, then (120 -n²)^2=0, so 120 -n²=0, which implies n²=120, but n must be integer. However, 120 is not a perfect square. Therefore, x=0 is not among our solutions. Therefore, our solutions are only those x's where 120 -n² is integer, and n is integer. Therefore, x is (integer)^2. But in our case, 120 -n² is integer, so x is square of integer.

But in our case, the values of x are (120 -n²)^2 where n is integer between 0 and 10. Therefore, each x is unique. Therefore, 11 values.

Wait, but let me check n= sqrt(120 - √x). But n must be integer. Therefore, the problem is equivalent to finding x such that 120 - √x is a perfect square, because √(120 - √x) is integer implies that 120 - √x is a perfect square. Let me denote that 120 - √x = k², where k is non-negative integer. Therefore, √x = 120 - k², so x=(120 -k²)^2. Therefore, same as before. Therefore, k is non-negative integer such that 120 -k² ≥0, so k² ≤120. So k can be 0,...,10. Hence 11 values. Hence answer is 11.

But let me check for example, if k=10, then x=(120 - 100)^2=20²=400. Then √x=20, so 120 - 20=100, √100=10, which is integer. Correct. Similarly, for k=11, which we said is not allowed, 120 - 121= -1, which would give √x=-1, which is invalid. Therefore, k can't be 11.

So, seems correct. Therefore, answer is 11.

Wait, but I feel like sometimes problems like these have more solutions because of different roots or something. Wait, but in this case, we have only one equation. Let me think. Let me consider that maybe there are multiple n's leading to same x? But as we saw, no. Because if (120 -n²)^2 = (120 -m²)^2, then either 120 -n² = 120 -m² or 120 -n² = m² -120. But since 120 -n² is non-negative, the second case would mean that m² -120 is non-negative, so m² ≥120. But m is an integer between 0 and 10, so m² ≤100. Hence, m² -120 ≤ -20 <0. Hence, impossible. Therefore, only possible if 120 -n² = 120 -m², so n=m. Therefore, unique. So 11 values.

Therefore, answer should be 11. Hmm. Wait, but let me check with actual numbers.

Let me list all possible n from 0 to 10 and compute x:

n=0: x=(120 - 0)^2 = 14400
n=1: (120 - 1)^2 = 119² = 14161
n=2: 118² = 13924
n=3: 117² = 13689
n=4: 116² = 13456
n=5: 115² = 13225
n=6: 114² = 12996
n=7: 113² = 12769
n=8: 112² = 12544
n=9: 111² = 12321
n=10: 110² = 12100

So that's 11 different x's. Each of these x's is unique. Therefore, 11 values. Therefore, the answer is 11.

But wait, the problem says "real values of x". Are there any other real numbers x that could satisfy the condition? For example, suppose x is such that √(120 - √x) is integer. But according to our reasoning, that would require that √x = 120 - n², so x=(120 - n²)^2. So, unless there is some other way...

Wait, suppose someone says that √(120 - √x) is integer, but maybe √x is not equal to 120 - n²? Wait, no. Let me think again.

Let me denote y = √x. Then, since x ≥0, y ≥0. The expression becomes √(120 - y) is integer. Let me denote that √(120 - y) = k, where k is integer ≥0. Then, 120 - y = k², so y = 120 - k². Since y ≥0, 120 -k² ≥0, so k² ≤120. Therefore, k can be 0,...,10 as before. Then, since y = √x = 120 -k², then x = y² = (120 -k²)^2. Therefore, same result. Therefore, there are exactly 11 values of x. Therefore, answer is 11.

Therefore, I think my answer is correct. But wait, but the original problem says "for how many real values of x". So, maybe the answer is 11? But maybe the problem is from an Olympiad or something where sometimes the answer is different. Wait, but let me check if for example, when n=0, x=14400. Then √x=120, so √(120 - √x)=√(120 -120)=√0=0, which is integer. Correct. Similarly, for n=10, x=110²=12100, √x=110, so √(120 - 110)=√10. Wait, wait! Wait, hold on. Wait, this is a problem. Wait, earlier, I thought that when n=10, √x=120 -n²=120 -100=20. Therefore, x=20²=400. But in the above, if I set k=10, then y=√x=120 -k²=120 -100=20, so x=20²=400. But in my previous list, for n=10, I had x=110²=12100. Wait, this is a contradiction! Wait, where did I make the mistake here?

Wait, let me go back. Wait, earlier, when I thought of k= n, I said for n=10, x=(120 -n²)^2. But when n=10, that's (120 -10²)= 120 -100=20. Then x=20²=400. But in my list above, I mistakenly wrote n=10: 110²=12100. Wait, what's going on here?

Wait, this seems to be an error in my previous list. Let me check again.

Earlier, I thought that for n=0 to 10, x=(120 - n²)^2. But when I listed them, I incorrectly wrote for n=10, (120 - 10²)²? Wait, no, in my previous list I said:

n=10: 110²=12100. Wait, but 120 -n² for n=10 is 120 -100=20. Therefore, x=20²=400. Therefore, my previous list was wrong. So where did I get 110² from? That was a mistake.

Wait, so my previous list was wrong. Let me correct it.

So let me redo the list. Let me compute x=(120 - n²)^2 for n=0,...,10.

For n=0: 120 - 0 = 120. So x=120²=14400
For n=1: 120 -1=119, x=119²=14161
For n=2: 120 -4=116, x=116²=13456
For n=3: 120 -9=111, x=111²=12321
Wait, hold on. Wait, 120 -3²=120-9=111? Wait, 120-9 is 111? No, 120-9=111? Wait, 120-9 is 111? Wait, 120 - 9 is 111? Wait, 120-10 is 110, so 120-9=111. Yes. So x=111²=12321.
For n=4: 120 - 16 = 104. So x=104²=10816
For n=5: 120 -25=95, x=95²=9025
n=6: 120 - 36=84, x=84²=7056
n=7: 120 -49=71, x=71²=5041
n=8: 120 -64=56, x=56²=3136
n=9: 120 -81=39, x=39²=1521
n=10: 120 -100=20, x=20²=400

So my previous list had errors in the numbers. I had mistakenly written for n=10, 110², which is wrong. Actually, for n=10, it is 20²=400. Similarly, for n=3, earlier I had written 117², but it's actually 111².

So the correct list is:

n=0: 14400
n=1: 14161
n=2: 13456
n=3: 12321
n=4: 10816
n=5: 9025
n=6: 7056
n=7: 5041
n=8: 3136
n=9: 1521
n=10: 400

Therefore, these are 11 different x's. Each is unique. For example, let me check if any of these numbers repeat. For example, 400 is unique, 1521 is unique, etc. Since each of them is square of different numbers (120 -n²), which are different numbers, so their squares are different. Hence, 11 different x's.

Therefore, my initial conclusion was correct, but I had a mistake in the intermediate step where I incorrectly calculated the values for n=10 and n=3. But that doesn't affect the conclusion that there are 11 values.

Therefore, the answer is 11.

But let me check once more with n=10. If x=400, then √x=20. Then 120 - √x=100. Then √100=10, which is integer. Correct.

Similarly, for n=3, x=111²=12321. √x=111. Then 120 - 111=9. √9=3, which is integer. Correct.

Therefore, my mistake was in the list, but the conclusion that there are 11 x's is correct.

Therefore, the answer is 11.

Wait, but let me check another example. For n=7: x=71²=5041. Then √x=71. 120 -71=49. √49=7, which is integer. Correct.

Therefore, all the values work. So, there are 11 real values. Therefore, the answer is 11.

But why did I think earlier that there might be something wrong? Maybe because of miscalculating the x's. But no, after checking, everything is correct.

Therefore, I think the correct answer is 11.

**Final Answer**
\boxed{11}

To determine how many **real values of $ x $** satisfy the condition that
$$
\sqrt{120 - \sqrt{x}} \text{ is an integer},
$$
we proceed step by step.

---

### Step 1: Let $ \sqrt{120 - \sqrt{x}} = n $, where $ n $ is a non-negative integer.

Since square roots are always non-negative, $ n \geq 0 $. Squaring both sides, we get:

$$
120 - \sqrt{x} = n^2 \quad \Rightarrow \quad \sqrt{x} = 120 - n^2.
$$

Since the square root of $ x $ must be real, it must also be non-negative, so:

$$
120 - n^2 \geq 0 \quad \Rightarrow \quad n^2 \leq 120.
$$

Thus, $ n $ can be any **integer** such that $ n^2 \leq 120 $.

---

### Step 2: Find all valid non-negative integers $ n $ such that $ n^2 \leq 120 $

We know:

$$
\sqrt{120} \approx 10.95, \quad \text{so the largest integer } n \text{ such that } n^2 \leq 120 \text{ is } n = 10.
$$

Hence, $ n $ ranges from $ 0 $ to $ 10 $, inclusive. That gives us **11** possible values for $ n $.

---

### Step 3: For each such $ n $, compute $ x $

From the equation:

$$
x = (\sqrt{120 - \sqrt{x}})^2 = (120 - n^2)^2.
$$

So for each $ n $ in $ \{0, 1, 2, \ldots, 10\} $, we compute:

$$
x = (120 - n^2)^2.
$$

Now, each of these values gives a unique real $ x $, and all such $ x $ are in the range $ [0, 120^2 = 14400] $, satisfying all required conditions.

---

### Step 4: Ensure no duplicates among calculated $ x $

Note that for **each** $ n \in \{0, 1, 2, \ldots, 10\} $, we compute a unique value of $ x $, because:

- The square of a number is **unique for distinct inputs**
- $ 120 - n^2 $ is **unique** for each $ n $
- Therefore, $ x = (120 - n^2)^2 $ is **unique** for each $ n $

Hence, the mapping from $ n $ to $ x $ is injective (one-to-one), and thus the list of values for $ x $ contains exactly 11 **distinct real** values.

---

### Final Conclusion

There are **11** real values of $ x $ such that $ \sqrt{120 - \sqrt{x}} $ is an integer.

$$
\boxed{11}
$$

Solution:
Suppose that  $k = \sqrt{120 - \sqrt{x}}$ is an integer. Then $0\le k \le \sqrt{120}$, and because $k$ is an integer, we have $0\le k \le 10$. Thus there are 11 possible integer values of $k$. For each such $k$, the corresponding value of $x$ is $\left(120 - k^2\right)^2$. Because  $\left(120 -
k^2\right)^2$ is positive and decreasing for $0\le k \le 10$, the $\boxed{11}$ values of $x$ are distinct.


Explanation:
Okay, let's see. There are three concentric circles with radii 4, 6, and 7. The diagram shows regions labeled X, Y, and Z. I need to find the difference between the area of the region with the greatest area and the one with the smallest area.

First, I should figure out what each region represents. The Asymptote code draws three circles: the largest one with radius 7, then a medium one with radius 6, and the smallest with radius 4. The regions are filled with different colors: the largest circle is light gray, the middle one is gray, and the smallest is white. Then there are labels X, Y, Z at positions (2,0), (5,0), and (6.5,0).

Since the circles are concentric, they all share the same center. The regions X, Y, Z are probably the areas between the circles. Let me think. The outermost circle (radius 7) is light gray. Then the middle circle (radius 6) is gray. The innermost circle (radius 4) is white. So, maybe the regions are:

- X is the area between radius 4 and 6? Or between 6 and 7?

Wait, the labels are placed at (2,0), (5,0), (6.5,0). So X is near radius 2, which is between 0 and 4? Wait, but the inner circle is radius 4, so maybe X is the innermost region (radius 4)? But the innermost circle is white. But Y is at (5,0), which is between 4 and 6, and Z is at (6.5,0), which is between 6 and 7. Wait, but the labels are placed along the x-axis. Let me visualize:

- The center is at (0,0). The point (2,0) is on the x-axis at distance 2 from the center. Since the inner circle has radius 4, so (2,0) is inside the inner circle. But X is labeled there. Wait, but the inner circle is white. So maybe X is the innermost circle (radius 4), which is white? But the middle circle is gray, and the outermost is light gray. Wait, but the middle circle (radius 6) is gray, so the area between radius 4 and 6 would be the gray part minus the white part? But the Asymptote code says:

filldraw(circle((0,0),7), lightgray, black+linewidth(1));
filldraw(circle((0,0),6), gray, black+linewidth(1));
filldraw(circle((0,0),4), white, black+linewidth(1));

So, the largest circle is filled light gray, then the middle circle (radius 6) is filled gray. But when you draw a circle over another, the later ones cover the previous ones. So the area of the middle circle (radius 6) would be gray, but the inner circle (radius 4) is white. So the regions would be:

- The innermost region (radius 4) is white.

- The area between 4 and 6 is gray (since the middle circle is filled gray, but the inner circle is white. So the gray area is the middle circle minus the white inner circle? Wait, no. Wait, if you draw a circle of radius 6 over a larger circle of radius 7, then the middle circle (radius 6) is filled gray, meaning that the area inside radius 6 is gray? But the inner circle (radius 4) is filled white. So the white is over the gray? So the area between 4 and 6 is gray minus white? Wait, no. The order of drawing is important. The first circle drawn is radius 7, light gray. Then the middle circle (radius 6) is drawn on top, filled gray. Then the inner circle (radius 4) is drawn on top of that, filled white. Therefore, the regions are:

- The outermost area (radius >6) is light gray.

- The area between 4 and 6 is gray (since the middle circle is gray, but the inner circle (radius 4) is white, so the gray area is between 4 and 6? Wait, no. If you draw the middle circle (radius 6) over the large circle (radius 7), then the middle circle is filled gray. But then the innermost circle (radius 4) is drawn on top, filled white. So actually:

- The area inside radius 4 is white.

- The area between 4 and 6 is gray (since the middle circle is filled gray, but the inner circle is white, which covers the inner part).

- The area between 6 and 7 is light gray (since the largest circle is filled light gray, but the middle circle (radius 6) covers up to radius 6, so the area outside radius 6 is light gray).

So regions X, Y, Z are labeled along the x-axis. The label X is at (2,0), which is within radius 4, so that's the white region (inner circle). Y is at (5,0), which is between 4 and 6, so that's the gray area. Z is at (6.5,0), which is between 6 and 7, so that's the light gray area.

Wait, but the problem says: "three regions are labeled X, Y, or Z below." So the regions are:

X: inside radius 4 (white)

Y: between 4 and 6 (gray)

Z: between 6 and 7 (light gray)

But the problem says "Of these three regions, what is the difference between the area of the region with the greatest area and the area of the region with the smallest area?"

So I need to compute the area of X, Y, Z, then find the max - min.

But first, let me confirm the regions again. The Asymptote code labels X at (2,0), Y at (5,0), Z at (6.5,0). Given that, and the coloring:

- X is inside the innermost circle (radius 4): area is π*(4)^2.

- Y is between radius 4 and 6: area is π*(6)^2 - π*(4)^2.

- Z is between radius 6 and 7: area is π*(7)^2 - π*(6)^2.

Therefore, areas:

X: π*(4²) = 16π

Y: π*(6² - 4²) = π*(36 - 16) = 20π

Z: π*(7² - 6²) = π*(49 - 36) = 13π

So the areas are 16π, 20π, 13π. Therefore, the largest area is Y (20π), the smallest is Z (13π). The difference is 20π - 13π = 7π.

Wait, but hold on. Let me check again. Wait, the regions are X, Y, Z. If X is the innermost circle, Y is the middle ring, Z is the outer ring. Then their areas are:

X: π(4²) = 16π

Y: π(6² - 4²) = 36π - 16π = 20π

Z: π(7² - 6²) = 49π - 36π = 13π

So the regions have areas 16π, 20π, 13π. So the largest area is Y (20π), the smallest is Z (13π). The difference is 20π - 13π = 7π. So the answer is 7π.

But let me make sure that the regions are indeed as I thought. The Asymptote code labels X at (2,0), which is inside radius 4, Y at (5,0) which is between 4 and 6, Z at (6.5,0) which is between 6 and 7. So yes. So that seems correct.

But wait, the problem says "three regions are labeled X, Y, or Z below." So maybe the regions are X, Y, Z, but perhaps the problem isn't referring to the colored regions? Wait, but the Asymptote code has three regions: the inner circle (white), the middle ring (gray), and the outer ring (light gray). So those are three regions, each labeled X, Y, Z. So the areas are as calculated.

But maybe I misread the regions? Let me think again. The Asymptote code draws:

filldraw(circle((0,0),7), lightgray, black+linewidth(1));

filldraw(circle((0,0),6), gray, black+linewidth(1));

filldraw(circle((0,0),4), white, black+linewidth(1));

So the first circle is drawn, which is lightgray. Then on top of it, the circle with radius 6 is drawn in gray. Then on top of that, the circle with radius 4 is drawn in white. Therefore, the regions:

- The area inside radius 4 is white.

- The area between 4 and 6 is gray (because the gray circle is drawn on top of the lightgray, covering the inner part, but the inner part is white. Wait, actually, the order is:

First, draw the largest circle (radius 7) as lightgray. Then draw the middle circle (radius 6) as gray. Since the middle circle is drawn on top, the area inside radius 6 is now gray, but the area outside radius 6 is still lightgray. Then draw the inner circle (radius 4) as white. So the area inside radius 4 is white, the area between 4 and 6 is gray, and the area between 6 and 7 is lightgray. Therefore, regions X, Y, Z are:

- X is inside radius 4: white

- Y is between 4 and 6: gray

- Z is between 6 and 7: lightgray

Therefore, their areas are as I calculated before: X is 16π, Y is 20π, Z is 13π. Hence the difference between the largest (Y=20π) and smallest (Z=13π) is 7π.

Wait, but hold on. The problem says "three regions are labeled X, Y, or Z below." In the Asymptote code, the labels are placed at (2,0), (5,0), (6.5,0). So X is at (2,0), which is inside the inner circle (radius 4). Therefore, X is the innermost region. Y is at (5,0), which is between 4 and 6, so that's the middle annular region. Z is at (6.5,0), which is between 6 and 7, so that's the outer annular region. Therefore, the areas are as calculated.

But just to make sure, let me compute all three areas:

Area of X: π*(4)^2 = 16π

Area of Y: Area between 4 and 6: π*(6^2 - 4^2) = π*(36 - 16) = 20π

Area of Z: Area between 6 and 7: π*(7^2 - 6^2) = π*(49 - 36) = 13π

Thus, X=16π, Y=20π, Z=13π. Therefore, the largest area is Y (20π) and the smallest is Z (13π). So difference is 20π - 13π = 7π.

But wait, I should check if maybe the problem is referring to different regions? For example, maybe X is the area between 6 and 7? But no, because the label is at (2,0). Let me check again the Asymptote code labels:

label("$X$",(2,0));

label("$Y$",(5,0));

label("$Z$",(6.5,0));

So (2,0) is 2 units from center, which is inside radius 4. So X is inside the inner circle. (5,0) is between 4 and 6. (6.5, 0) is between 6 and 7. Therefore, regions are correct. Hence, the areas are correct.

Alternatively, perhaps the problem considers the entire circles? Wait, no. The regions are the areas between the circles. Since they are concentric, the regions are annular regions.

Alternatively, maybe the regions are different? Wait, the problem says "three regions are labeled X, Y, or Z below." So the Asymptote code shows three regions, which are the three annular regions. But perhaps the problem is considering the entire circles? Wait, but how?

Wait, if you think about the diagram: the largest circle (radius 7) is lightgray. The middle circle (radius 6) is gray, so the area inside radius 6 is gray, but the inner circle (radius 4) is white, which covers the inner part. Therefore, the regions:

- White region: inside 4.

- Gray region: between 4 and 6.

- Lightgray region: between 6 and 7.

Therefore, the three regions. Hence, their areas are as calculated. Therefore, the answer is 7π.

But let me check again. Maybe I made a mistake in the calculation:

Area of X: πr² where r=4: 16π. Correct.

Area of Y: π(6² - 4²) = π(36 - 16) = 20π. Correct.

Area of Z: π(7² - 6²) = π(49 - 36) = 13π. Correct.

So difference between largest (Y=20π) and smallest (Z=13π) is 7π.

But let me check if the problem says "difference between the area of the region with the greatest area and the area of the region with the smallest area." So that is 20π - 13π = 7π.

Alternatively, is there a possibility that the problem considers the regions differently? For example, maybe X is the area outside radius 7? But no, since the largest circle is drawn, and the label is at (2,0). Alternatively, maybe the regions are different?

Wait, let me think again. The Asymptote code draws three circles:

1. The largest circle (radius 7) filled lightgray.

2. Then a circle (radius 6) filled gray, so this covers the inner part of the lightgray circle.

3. Then a circle (radius 4) filled white, covering the inner part of the gray circle.

Therefore, if you imagine the diagram:

- The outermost area (radius >6) is lightgray.

- The middle area (between 4 and 6) is gray.

- The innermost area (radius <4) is white.

Therefore, the regions X, Y, Z are those three areas. So if the labels are placed at (2,0), (5,0), (6.5,0), then:

- X is white (innermost region)

- Y is gray (middle region)

- Z is lightgray (outer region)

Therefore, their areas are as calculated. Therefore, the answer is 7π.

But just to make sure that the problem isn't referring to some other regions. For example, maybe the regions are the three circles? But the three circles are overlapping. But the problem says "three regions are labeled X, Y, or Z." Given the Asymptote code, it's more likely that X, Y, Z are the three annular regions.

Alternatively, maybe the regions are the three circles? But that would be three circles, but they are concentric. However, the problem says "regions," which are areas between circles. So I think the answer is 7π.

Alternatively, maybe the problem is in the order of X, Y, Z? For example, maybe X is the outermost region? But according to the labels, X is at (2,0), which is near the center, so it's the innermost.

Alternatively, maybe the problem has a different configuration? For example, maybe X is between 6 and 7, Y between 4 and 6, Z inside 4? But the labels are at (2,0), (5,0), (6.5,0). If X is at (2,0), that's closer to center, so it's the innermost. So I think my original calculation is correct.

Therefore, I think the answer is 7π.

**Final Answer**
The difference between the areas is \boxed{7\pi}.

To determine the difference between the area of the region with the greatest area and the area of the region with the smallest area, we need to analyze the configuration of the three concentric circles and the regions they define.

---

### **Understanding the Regions**

We are given three concentric circles with radii:

- Innermost: **$ r = 4 $**
- Middle: **$ r = 6 $**
- Outermost: **$ r = 7 $**

From the Asymptote diagram, the three regions **X**, **Y**, and **Z** are labeled **at positions (2,0)**, **(5,0)**, and **(6.5,0)** along the x-axis. These positions suggest:

- **X** is within the innermost circle ($ r \leq 4 $)
- **Y** is between radii 4 and 6
- **Z** is between radii 6 and 7

From this, the **three distinct regions** are:

- **X**: The **interior region** of the smallest circle, with radius $ r = 4 $
- **Y**: The **annular ring** between radii $ r = 4 $ and $ r = 6 $
- **Z**: The **outer annular ring** between radii $ r = 6 $ and $ r = 7 $

---

### **Calculating the Areas**

The area of a circle is given by the formula:

$$
A = \pi r^2
$$

**Area of X (inner circle with radius 4):**

$$
A_X = \pi (4)^2 = 16\pi
$$

**Area of Y (annular region between radii 4 and 6):**

$$
A_Y = \pi (6^2 - 4^2) = \pi (36 - 16) = 20\pi
$$

**Area of Z (annular region between radii 6 and 7):**

$$
A_Z = \pi (7^2 - 6^2) = \pi (49 - 36) = 13\pi
$$

---

### **Finding the Difference Between the Largest and Smallest Areas**

Now compare the three areas:

- $ A_X = 16\pi $
- $ A_Y = 20\pi $
- $ A_Z = 13\pi $

The largest area is **$ A_Y = 20\pi $** and the smallest is **$ A_Z = 13\pi $**.

$$
\text{Difference} = A_Y - A_Z = 20\pi - 13\pi = 7\pi
$$

---

### **Final Answer**

$$
\boxed{7\pi}
$$

[Assistant]
Explanation:
Okay, let's see. There are three concentric circles with radii 4, 6, and 7. The diagram shows regions labeled X, Y, and Z. I need to find the difference between the area of the region with the greatest area and the one with the smallest area.

First, I should figure out what each region represents. The Asymptote code draws three circles: the largest one with radius 7, then a medium one with radius 6, and the smallest with radius 4. The regions are filled with different colors: the largest circle is light gray, the middle one is gray, and the smallest is white. Then there are labels X, Y, Z at positions (2,0), (5,0), and (6.5,0).

Since the circles are concentric, they all share the same center. The regions X, Y, Z are probably the areas between the circles. Let me think. The outermost circle (radius 7) is light gray. Then the middle circle (radius 6) is gray. The innermost circle (radius 4) is white. So, maybe the regions are:

- X is the area between radius 4 and 6? Or between 6 and 7?

Wait, the labels are placed at (2,0), (5,0), (6.5,0). So X is near radius 2, which is between 0 and 4? Wait, but the inner circle is radius 4, so maybe X is the innermost region (radius 4)? But the innermost circle is white. But Y is at (5,0), which is between 4 and 6, and Z is at (6.5,0), which is between 6 and 7. Wait, but the labels are placed along the x-axis. Let me visualize:

- The center is at (0,0). The point (2,0) is on the x-axis at distance 2 from the center. Since the inner circle has radius 4, so (2,0) is inside the inner circle. But X is labeled there. Wait, but the inner circle is white. So maybe X is the innermost circle (radius 4), which is white? But the middle circle is gray, and the outermost is light gray. Wait, but the middle circle (radius 6) is gray, so the area between radius 4 and 6 would be the gray part minus the white part? But the Asymptote code says:

filldraw(circle((0,0),7), lightgray, black+linewidth(1));
filldraw(circle((0,0),6), gray, black+linewidth(1));
filldraw(circle((0,0),4), white, black+linewidth(1));

So, the largest circle is filled light gray, then the middle circle (radius 6) is filled gray. But when you draw a circle over another, the later ones cover the previous ones. So the area of the middle circle (radius 6) would be gray, but the inner circle (radius 4) is white. So the regions would be:

- The innermost region (radius 4) is white.

- The area between 4 and 6 is gray (since the middle circle is filled gray, but the inner circle is white. So the gray area is the middle circle minus the white inner circle? Wait, no. Wait, if you draw a circle of radius 6 over a larger circle of radius 7, then the middle circle (radius 6) is filled gray, meaning that the area inside radius 6 is gray? But the inner circle (radius 4) is filled white. So the white is over the gray? So the area between 4 and 6 is gray minus white? Wait, no. The order of drawing is important. The first circle drawn is radius 7, light gray. Then the middle circle (radius 6) is drawn on top, filled gray. Then the inner circle (radius 4) is drawn on top of that, filled white. Therefore, the regions are:

- The outermost area (radius >6) is light gray.

- The area between 4 and 6 is gray (since the middle circle is gray, but the inner circle (radius 4) is white, so the gray area is between 4 and 6? Wait, no. If you draw the middle circle (radius 6) over the large circle (radius 7), then the middle circle is filled gray. But then the innermost circle (radius 4) is drawn on top, filled white. So actually:

- The area inside radius 4 is white.

- The area between 4 and 6 is gray (since the middle circle is filled gray, but the inner circle is white, which covers the inner part).

- The area between 6 and 7 is light gray (since the largest circle is filled light gray, but the middle circle (radius 6) covers up to radius 6, so the area outside radius 6 is light gray).

So regions X, Y, Z are labeled along the x-axis. The label X is at (2,0), which is within radius 4, so that's the white region (inner circle). Y is at (5,0), which is between 4 and 6, so that's the gray area. Z is at (6.5,0), which is between 6 and 7, so that's the light gray area.

Wait, but the problem says: "three regions are labeled X, Y, or Z below." So the regions are:

X: inside radius 4 (white)

Y: between 4 and 6 (gray)

Z: between 6 and 7 (light gray)

But the problem says "Of these three regions, what is the difference between the area of the region with the greatest area and the area of the region with the smallest area?"

So I need to compute the area of X, Y, Z, then find the max - min.

But first, let me confirm the regions again. The Asymptote code labels X at (2,0), Y at (5,0), Z at (6.5,0). Given that, and the coloring:

- X is inside the innermost circle (radius 4): area is π*(4)^2.

- Y is between radius 4 and 6: area is π*(6)^2 - π*(4)^2.

- Z is between radius 6 and 7: area is π*(7)^2 - π*(6)^2.

Therefore, areas:

X: π*(4²) = 16π

Y: π*(6² - 4²) = π*(36 - 16) = 20π

Z: π*(7² - 6²) = π*(49 - 36) = 13π

So the areas are 16π, 20π, 13π. Therefore, the largest area is Y (20π), the smallest is Z (13π). The difference is 20π - 13π = 7π.

Wait, but hold on. Let me check again. Wait, the regions are X, Y, Z. If X is the innermost circle, Y is the middle ring, Z is the outer ring. Then their areas are:

X: π(4²) = 16π

Y: π(6² - 4²) = 36π - 16π = 20π

Z: π(7² - 6²) = 49π - 36π = 13π

So the regions have areas 16π, 20π, 13π. So the largest area is Y (20π), the smallest is Z (13π). The difference is 20π - 13π = 7π. So the answer is 7π.

But let me make sure that the regions are indeed as I thought. The Asymptote code labels X at (2,0), which is inside radius 4, Y at (5,0) which is between 4 and 6, Z at (6.5,0) which is between 6 and 7. So yes. So that seems correct.

But wait, the problem says "three regions are labeled X, Y, or Z below." So maybe the regions are X, Y, Z, but perhaps the problem isn't referring to the colored regions? Wait, but the Asymptote code has three regions: the inner circle (white), the middle ring (gray), and the outer ring (light gray). So those are three regions, each labeled X, Y, Z. So the areas are as calculated.

But maybe I misread the regions? Let me think again. The Asymptote code draws:

filldraw(circle((0,0),7), lightgray, black+linewidth(1));

filldraw(circle((0,0),6), gray, black+linewidth(1));

filldraw(circle((0,0),4), white, black+linewidth(1));

So the first circle is drawn, which is lightgray. Then on top of it, the circle with radius 6 is drawn in gray. Then on top of that, the circle with radius 4 is drawn in white. Therefore, the regions:

- The area inside radius 4 is white.

- The area between 4 and 6 is gray (because the gray circle is drawn on top of the lightgray, covering the inner part, but the inner part is white. Wait, actually, the order is:

First, draw the largest circle (radius 7) as lightgray. Then draw the middle circle (radius 6) as gray. Since the middle circle is drawn on top, the area inside radius 6 is now gray, but the area outside radius 6 is still lightgray. Then draw the inner circle (radius 4) as white. So the area inside radius 4 is white, the area between 4 and 6 is gray, and the area between 6 and 7 is lightgray. Therefore, regions X, Y, Z are:

- X is inside radius 4: white

- Y is between 4 and 6: gray

- Z is between 6 and 7: lightgray

Therefore, their areas are as I calculated before: X is 16π, Y is 20π, Z is 13π. Hence the difference between the largest (Y=20π) and smallest (Z=13π) is 7π.

Wait, but hold on. The problem says "three regions are labeled X, Y, or Z below." In the Asymptote code, the labels are placed at (2,0), (5,0), (6.5,0). So X is at (2,0), which is inside the inner circle (radius 4). Therefore, X is the innermost region. Y is at (5,0), which is between 4 and 6, so that's the middle annular region. Z is at (6.5,0), which is between 6 and 7, so that's the outer annular region. Therefore, the areas are as calculated.

But just to make sure, let me compute all three areas:

Area of X: π*(4)^2 = 16π

Area of Y: Area between 4 and 6: π*(6^2 - 4^2) = π*(36 - 16) = 20π

Area of Z: Area between 6 and 7: π*(7^2 - 6^2) = π*(49 - 36) = 13π

Thus, X=16π, Y=20π, Z=13π. Therefore, the largest area is Y (20π) and the smallest is Z (13π). So difference is 20π - 13π = 7π.

But wait, I should check if maybe the problem is referring to different regions? For example, maybe X is the area between 6 and 7? But no, because the label is at (2,0). Let me check again the Asymptote code labels:

label("$X$",(2,0));

label("$Y$",(5,0));

label("$Z$",(6.5,0));

So (2,0) is 2 units from center, which is inside radius 4. So X is inside the inner circle. (5,0) is between 4 and 6. (6.5, 0) is between 6 and 7. Therefore, regions are correct. Hence, the areas are correct.

Alternatively, perhaps the problem considers the entire circles? Wait, no. The regions are the areas between the circles. Since they are concentric, the regions are annular regions.

Alternatively, maybe the regions are different? Wait, the problem says "three regions are labeled X, Y, or Z below." So the Asymptote code shows three regions, which are the three annular regions. But perhaps the problem is considering the entire circles? Wait, but how?

Wait, if you think about the diagram: the largest circle (radius 7) is lightgray. The middle circle (radius 6) is gray, so the area inside radius 6 is gray, but the inner circle (radius 4) is white, which covers the inner part. Therefore, the regions:

- White region: inside 4.

- Gray region: between 4 and 6.

- Lightgray region: between 6 and 7.

Therefore, the three regions. Hence, their areas are as calculated. Therefore, the answer is 7π.

But let me check again. Maybe I made a mistake in the calculation:

Area of X: πr² where r=4: 16π. Correct.

Area of Y: π(6² - 4²) = π(36 - 16) = 20π. Correct.

Area of Z: π(7² - 6²) = π(49 - 36) = 13π. Correct.

So difference between largest (Y=20π) and smallest (Z=13π) is 7π.

But let me check if the problem says "difference between the area of the region with the greatest area and the area of the region with the smallest area." So that is 20π - 13π = 7π.

Alternatively, is there a possibility that the problem considers the regions differently? For example, maybe X is the area outside radius 7? But no, since the largest circle is drawn, and the label is at (2,0). Alternatively, maybe the regions are different?

Wait, let me think again. The Asymptote code draws three circles:

1. The largest circle (radius 7) filled lightgray.

2. Then a circle (radius 6) filled gray, so this covers the inner part of the lightgray circle.

3. Then a circle (radius 4) filled white, covering the inner part of the gray circle.

Therefore, if you imagine the diagram:

- The outermost area (radius >6) is lightgray.

- The middle area (between 4 and 6) is gray.

- The innermost area (radius <4) is white.

Therefore, the regions X, Y, Z are those three areas. So if the labels are placed at (2,0), (5,0), (6.5,0), then:

- X is white (innermost region)

- Y is gray (middle region)

- Z is lightgray (outer region)

Therefore, their areas are as calculated. Therefore, the answer is 7π.

But just to make sure that the problem isn't referring to some other regions. For example, maybe the regions are the three circles? But the three circles are overlapping. But the problem says "three regions are labeled X, Y, or Z." Given the Asymptote code, it's more likely that X, Y, Z are the three annular regions.

Alternatively, maybe the regions are the three circles? But that would be three circles, but they are concentric. However, the problem says "regions," which are areas between circles. So I think the answer is 7π.

Alternatively, maybe the problem is in the order of X, Y, Z? For example, maybe X is the outermost region? But according to the labels, X is at (2,0), which is near the center, so it's the innermost.

Alternatively, maybe the problem has a different configuration? For example, maybe X is between 6 and 7, Y between 4 and 6, Z inside 4? But the labels are at (2,0), (5,0), (6.5,0). If X is at (2,0), that's closer to center, so it's the innermost. So I think my original calculation is correct.

Therefore, I think the answer is 7π.

**Final Answer**
The difference between the areas is \boxed{7\pi}.

To determine the difference between the area of the region with the greatest area and the area of the region with the smallest area, we need to analyze the configuration of the three concentric circles and the regions they define.

---

### **Understanding the Regions**

We are given three concentric circles with radii:

- Innermost: **$ r = 4 $**
- Middle: **$ r = 6 $**
- Outermost: **$ r = 7 $**

From the Asymptote diagram, the three regions **X**, **Y**, and **Z** are labeled **at positions (2,0)**, **(5,0)**, and **(6.5,0)** along the x-axis. These positions suggest:

- **X** is within the innermost circle ($ r \leq 4 $)
- **Y** is between radii 4 and 6
- **Z** is between radii 6 and 7

From this, the **three distinct regions** are:

- **X**: The **interior region** of the smallest circle, with radius $ r = 4 $
- **Y**: The **annular ring** between radii $ r = 4 $ and $ r = 6 $
- **Z**: The **outer annular ring** between radii $ r = 6 $ and $ r = 7 $

---

### **Calculating the Areas**

The area of a circle is given by the formula:

$$
A = \pi r^2
$$

**Area of X (inner circle with radius 4):**

$$
A_X = \pi (4)^2 = 16\pi
$$

**Area of Y (annular region between radii 4 and 6):**

$$
A_Y = \pi (6^2 - 4^2) = \pi (36 - 16) = 20\pi
$$

**Area of Z (annular region between radii 6 and 7):**

$$
A_Z = \pi (7^2 - 6^2) = \pi (49 - 36) = 13\pi
$$

---

### **Finding the Difference Between the Largest and Smallest Areas**

Now compare the three areas:

- $ A_X = 16\pi $
- $ A_Y = 20\pi $
- $ A_Z = 13\pi $

The largest area is **$ A_Y = 20\pi $** and the smallest is **$ A_Z = 13\pi $**.

$$
\text{Difference} = A_Y - A_Z = 20\pi - 13\pi = 7\pi
$$

---

### **Final Answer**

$$
\boxed{7\pi}
$$

Solution:
The area of the inside circle (region $X$) is $\pi\cdot 4^2=16\pi.$

Using a similar technique, the area of the middle ring (region $Y$) is $$\pi\cdot 6^2-\pi\cdot 4^2=36\pi-16\pi = 20\pi.$$ Also, the area of the outer ring (region $Z$) is $$\pi\cdot 7^2-\pi\cdot 6^2=49\pi - 36\pi = 13\pi.$$ Therefore, region $Y$ has the largest area and region $Z$ has the smallest area. The difference in their areas is $20\pi-13\pi = \boxed{7\pi}.$