Explanation:
Okay, so I need to figure out how many inches on the map correspond to 53.25 miles in real life. The problem says that 3 inches on the map equals 10 miles in real life. Hmm, let me think about how to approach this.

First, maybe I should find out how many inches correspond to one mile. If 3 inches is 10 miles, then dividing both numbers by 10 should give me the inches per mile. Let me write that down:

3 inches / 10 miles = 0.3 inches per mile. Wait, is that right? Let me check. If 3 inches is 10 miles, then 1 mile would be 3 inches divided by 10, which is indeed 0.3 inches. Yeah, that seems correct.

Alternatively, maybe I can set up a proportion. Let me try that method too. If 3 inches corresponds to 10 miles, then x inches corresponds to 53.25 miles. So the proportion would be:

3 inches / 10 miles = x inches / 53.25 miles

Cross-multiplying, that would be 10x = 3 * 53.25. Then solving for x would give me x = (3 * 53.25) / 10. Let me compute that.

First, 3 times 53.25. Let me calculate 53.25 * 3. Well, 50 * 3 is 150, 3.25 * 3 is 9.75, so total is 150 + 9.75 = 159.75. Then divide by 10, so 159.75 / 10 = 15.975 inches. Hmm, so that would be 15.975 inches. But the question says to express it as a fraction. So I need to convert 15.975 into a fraction.

Wait, let me check my steps again to make sure I didn't make a mistake. So the original ratio is 3 inches = 10 miles. So miles to inches is 3/10 inches per mile. Therefore, for 53.25 miles, multiply 53.25 by 3/10. Which is the same as (53.25 * 3)/10. Which is 159.75 / 10. Which is 15.975. So that's correct.

But converting 15.975 to a fraction. Let me think. 0.975 is the decimal part. Let me convert that. 0.975 can be written as 975/1000. Let me simplify that. Dividing numerator and denominator by 25: 975 ÷ 25 = 39, 1000 ÷ 25 = 40. So 39/40. Wait, 39 divided by 40 is 0.975. So 0.975 is 39/40. Therefore, 15.975 is 15 + 39/40, which is 15 39/40 inches. But the question says to express it as a fraction. So maybe they want an improper fraction?

So 15 39/40 is equal to (15*40 + 39)/40. Let me compute that. 15*40 is 600, plus 39 is 639. So 639/40. Let me check if that reduces. Let's see, 639 and 40. The factors of 40 are 2^3 * 5. Let me check if 639 is divisible by 2? No, it's odd. Divisible by 5? Ends with 9, so no. So 639/40 is the simplified fraction. So the answer is 639/40 inches.

Wait, but let me check if my decimal conversion was correct. 15.975. Let me verify 639 divided by 40. 40*15 = 600, 639 - 600 = 39. So yes, 639/40 is 15.975. So that's correct.

Alternatively, maybe there's another way to approach the problem by converting the miles to inches directly. Let me try that. Since 10 miles is 3 inches, then 1 mile is 3/10 inches. Therefore, 53.25 miles is 53.25 * 3/10 inches. Which is the same as before. So same result.

Alternatively, maybe I can use unit conversion. Let me think. The scale is 3 inches per 10 miles. So to convert miles to inches, multiply by (3 inches / 10 miles). Therefore, 53.25 miles * (3 inches / 10 miles) = (53.25 * 3)/10 inches. Which again is 159.75 / 10 = 15.975 inches. So same answer.

So, converting 15.975 to a fraction. Let me think again. 0.975. Let me write 15.975 as 15.975 = 15 + 0.975. As before. 0.975. Let me think of 0.975 as 975/1000. Then simplifying. Let me divide numerator and denominator by 5: 975 ÷ 5 = 195, 1000 ÷ 5 = 200. Then divide by 5 again: 195 ÷ 5 = 39, 200 ÷ 5 = 40. So 39/40. So yes, same as before. Therefore, 15.975 is 15 39/40, which is 639/40. So that's the fraction.

But let me check if 639 and 40 have any common factors. 639 divided by 3 is 213. 40 divided by 3 is not an integer. 639 divided by 2? No. Divided by 5? No. Divided by 7? Let me check 7*91 is 637, so 639-637=2, so no. Divided by 11? 11*58 is 638, so 639-638=1, so no. So 639/40 is reduced completely. So that's the answer.

Alternatively, maybe I can do this without converting decimals. Let me think. Since 53.25 is a decimal, maybe I can write it as a fraction. 53.25 is equal to 53 and 1/4, which is 213/4. Let me check: 53 * 4 = 212, plus 1 is 213. So 53.25 = 213/4. Then, using the original ratio, 3 inches / 10 miles = x inches / (213/4) miles. So cross-multiplying, 10x = 3*(213/4). Therefore, x = (3*213)/(4*10) = 639/40. Which is the same as before. So that's another way to do it, avoiding decimals. So that's good. So that gives me 639/40 inches. So that's the answer.

Therefore, the distance on the map is 639/40 inches. Let me check once more. If 10 miles is 3 inches, then 213/4 miles is 3*(213/4)/10 inches. Which is 639/40. Yep. So that's correct.

Alternatively, maybe I can think in terms of scaling factors. The scale is 3 inches per 10 miles. So the scale factor is 3/10 inches per mile. Therefore, multiplying 53.25 miles by 3/10 inches per mile gives the map distance. Which again is 53.25 * 3 / 10. Which is same as before.

So, all methods lead to 639/40 inches. Therefore, I think that is the correct answer. Let me just check if 639 divided by 40 is indeed 15.975. 40*15 = 600, 639 - 600 = 39. 39/40 = 0.975. So yes, 15.975. So that's correct.

Therefore, the answer is 639/40 inches. Let me check if that can be simplified further. As I thought before, 639 and 40 have no common factors besides 1. So yes, 639/40 is the simplest form.

Alternatively, maybe the problem expects a mixed number? But the question says "expressed as a fraction," so improper fraction is probably acceptable. But let me check if 639/40 can be simplified. Let me check GCD of 639 and 40. Prime factors of 40: 2^3 * 5. Prime factors of 639: Let's divide by 3: 6 + 3 + 9 = 18, which is divisible by 3. 639 ÷ 3 = 213. Then 213 ÷ 3 = 71. So 639 = 3*3*71. 71 is a prime number. So factors of 639 are 3^2 * 71. 40 is 2^3 * 5. No common factors. Therefore, 639/40 is reduced completely. So that's the answer.

Therefore, the distance on the map is 639/40 inches.

**Final Answer**
The distance between the buildings on the map is \boxed{\dfrac{639}{40}} inches.

To determine the distance between two buildings on a map, we are given a scale: **3 inches on the map corresponds to 10 miles in real life**. We are also told that the **real-life distance** between the buildings is **53.25 miles**, and we are to find the **map distance in inches**, expressed as a **fraction**.

---

### Step 1: Understand the Scale

The scale tells us that:

$$
3 \text{ inches} = 10 \text{ miles}
$$

This means that **1 mile** corresponds to:

$$
\frac{3}{10} \text{ inches}
$$

---

### Step 2: Apply the Scale to the Given Distance

We are given a real-life distance of **53.25 miles**. To find the corresponding map distance in inches, we multiply the real-life distance by the scale factor:

$$
\text{Map distance} = 53.25 \times \frac{3}{10}
$$

---

### Step 3: Perform the Multiplication

First, convert 53.25 to a fraction:

$$
53.25 = \frac{213}{4}
$$

Now multiply:

$$
\frac{213}{4} \times \frac{3}{10} = \frac{213 \times 3}{4 \times 10} = \frac{639}{40}
$$

---

### Step 4: Simplify the Fraction

Check if the numerator and denominator have any common factors. The prime factorization of 639 is:

$$
639 = 3 \times 3 \times 71
$$

The prime factorization of 40 is:

$$
40 = 2^3 \times 5
$$

There are **no common factors** between 639 and 40, so the fraction is already in its simplest form.

---

### Final Answer

$$
\boxed{\dfrac{639}{40}}
$$

This is the distance between the buildings on the map, expressed as a fraction.

[Assistant]
Explanation:
Okay, so I need to figure out how many inches on the map correspond to 53.25 miles in real life. The problem says that 3 inches on the map equals 10 miles in real life. Hmm, let me think about how to approach this.

First, maybe I should find out how many inches correspond to one mile. If 3 inches is 10 miles, then dividing both numbers by 10 should give me the inches per mile. Let me write that down:

3 inches / 10 miles = 0.3 inches per mile. Wait, is that right? Let me check. If 3 inches is 10 miles, then 1 mile would be 3 inches divided by 10, which is indeed 0.3 inches. Yeah, that seems correct.

Alternatively, maybe I can set up a proportion. Let me try that method too. If 3 inches corresponds to 10 miles, then x inches corresponds to 53.25 miles. So the proportion would be:

3 inches / 10 miles = x inches / 53.25 miles

Cross-multiplying, that would be 10x = 3 * 53.25. Then solving for x would give me x = (3 * 53.25) / 10. Let me compute that.

First, 3 times 53.25. Let me calculate 53.25 * 3. Well, 50 * 3 is 150, 3.25 * 3 is 9.75, so total is 150 + 9.75 = 159.75. Then divide by 10, so 159.75 / 10 = 15.975 inches. Hmm, so that would be 15.975 inches. But the question says to express it as a fraction. So I need to convert 15.975 into a fraction.

Wait, let me check my steps again to make sure I didn't make a mistake. So the original ratio is 3 inches = 10 miles. So miles to inches is 3/10 inches per mile. Therefore, for 53.25 miles, multiply 53.25 by 3/10. Which is the same as (53.25 * 3)/10. Which is 159.75 / 10. Which is 15.975. So that's correct.

But converting 15.975 to a fraction. Let me think. 0.975 is the decimal part. Let me convert that. 0.975 can be written as 975/1000. Let me simplify that. Dividing numerator and denominator by 25: 975 ÷ 25 = 39, 1000 ÷ 25 = 40. So 39/40. Wait, 39 divided by 40 is 0.975. So 0.975 is 39/40. Therefore, 15.975 is 15 + 39/40, which is 15 39/40 inches. But the question says to express it as a fraction. So maybe they want an improper fraction?

So 15 39/40 is equal to (15*40 + 39)/40. Let me compute that. 15*40 is 600, plus 39 is 639. So 639/40. Let me check if that reduces. Let's see, 639 and 40. The factors of 40 are 2^3 * 5. Let me check if 639 is divisible by 2? No, it's odd. Divisible by 5? Ends with 9, so no. So 639/40 is the simplified fraction. So the answer is 639/40 inches.

Wait, but let me check if my decimal conversion was correct. 15.975. Let me verify 639 divided by 40. 40*15 = 600, 639 - 600 = 39. So yes, 639/40 is 15.975. So that's correct.

Alternatively, maybe there's another way to approach the problem by converting the miles to inches directly. Let me try that. Since 10 miles is 3 inches, then 1 mile is 3/10 inches. Therefore, 53.25 miles is 53.25 * 3/10 inches. Which is the same as before. So same result.

Alternatively, maybe I can use unit conversion. Let me think. The scale is 3 inches per 10 miles. So to convert miles to inches, multiply by (3 inches / 10 miles). Therefore, 53.25 miles * (3 inches / 10 miles) = (53.25 * 3)/10 inches. Which again is 159.75 / 10 = 15.975 inches. So same answer.

So, converting 15.975 to a fraction. Let me think again. 0.975. Let me write 15.975 as 15.975 = 15 + 0.975. As before. 0.975. Let me think of 0.975 as 975/1000. Then simplifying. Let me divide numerator and denominator by 5: 975 ÷ 5 = 195, 1000 ÷ 5 = 200. Then divide by 5 again: 195 ÷ 5 = 39, 200 ÷ 5 = 40. So 39/40. So yes, same as before. Therefore, 15.975 is 15 39/40, which is 639/40. So that's the fraction.

But let me check if 639 and 40 have any common factors. 639 divided by 3 is 213. 40 divided by 3 is not an integer. 639 divided by 2? No. Divided by 5? No. Divided by 7? Let me check 7*91 is 637, so 639-637=2, so no. Divided by 11? 11*58 is 638, so 639-638=1, so no. So 639/40 is reduced completely. So that's the answer.

Alternatively, maybe I can do this without converting decimals. Let me think. Since 53.25 is a decimal, maybe I can write it as a fraction. 53.25 is equal to 53 and 1/4, which is 213/4. Let me check: 53 * 4 = 212, plus 1 is 213. So 53.25 = 213/4. Then, using the original ratio, 3 inches / 10 miles = x inches / (213/4) miles. So cross-multiplying, 10x = 3*(213/4). Therefore, x = (3*213)/(4*10) = 639/40. Which is the same as before. So that's another way to do it, avoiding decimals. So that's good. So that gives me 639/40 inches. So that's the answer.

Therefore, the distance on the map is 639/40 inches. Let me check once more. If 10 miles is 3 inches, then 213/4 miles is 3*(213/4)/10 inches. Which is 639/40. Yep. So that's correct.

Alternatively, maybe I can think in terms of scaling factors. The scale is 3 inches per 10 miles. So the scale factor is 3/10 inches per mile. Therefore, multiplying 53.25 miles by 3/10 inches per mile gives the map distance. Which again is 53.25 * 3 / 10. Which is same as before.

So, all methods lead to 639/40 inches. Therefore, I think that is the correct answer. Let me just check if 639 divided by 40 is indeed 15.975. 40*15 = 600, 639 - 600 = 39. 39/40 = 0.975. So yes, 15.975. So that's correct.

Therefore, the answer is 639/40 inches. Let me check if that can be simplified further. As I thought before, 639 and 40 have no common factors besides 1. So yes, 639/40 is the simplest form.

Alternatively, maybe the problem expects a mixed number? But the question says "expressed as a fraction," so improper fraction is probably acceptable. But let me check if 639/40 can be simplified. Let me check GCD of 639 and 40. Prime factors of 40: 2^3 * 5. Prime factors of 639: Let's divide by 3: 6 + 3 + 9 = 18, which is divisible by 3. 639 ÷ 3 = 213. Then 213 ÷ 3 = 71. So 639 = 3*3*71. 71 is a prime number. So factors of 639 are 3^2 * 71. 40 is 2^3 * 5. No common factors. Therefore, 639/40 is reduced completely. So that's the answer.

Therefore, the distance on the map is 639/40 inches.

**Final Answer**
The distance between the buildings on the map is \boxed{\dfrac{639}{40}} inches.

To determine the distance between two buildings on a map, we are given a scale: **3 inches on the map corresponds to 10 miles in real life**. We are also told that the **real-life distance** between the buildings is **53.25 miles**, and we are to find the **map distance in inches**, expressed as a **fraction**.

---

### Step 1: Understand the Scale

The scale tells us that:

$$
3 \text{ inches} = 10 \text{ miles}
$$

This means that **1 mile** corresponds to:

$$
\frac{3}{10} \text{ inches}
$$

---

### Step 2: Apply the Scale to the Given Distance

We are given a real-life distance of **53.25 miles**. To find the corresponding map distance in inches, we multiply the real-life distance by the scale factor:

$$
\text{Map distance} = 53.25 \times \frac{3}{10}
$$

---

### Step 3: Perform the Multiplication

First, convert 53.25 to a fraction:

$$
53.25 = \frac{213}{4}
$$

Now multiply:

$$
\frac{213}{4} \times \frac{3}{10} = \frac{213 \times 3}{4 \times 10} = \frac{639}{40}
$$

---

### Step 4: Simplify the Fraction

Check if the numerator and denominator have any common factors. The prime factorization of 639 is:

$$
639 = 3 \times 3 \times 71
$$

The prime factorization of 40 is:

$$
40 = 2^3 \times 5
$$

There are **no common factors** between 639 and 40, so the fraction is already in its simplest form.

---

### Final Answer

$$
\boxed{\dfrac{639}{40}}
$$

This is the distance between the buildings on the map, expressed as a fraction.

Solution: If we let the distance between the buildings on the map be $d$, then $\frac{d}{53.25} = \frac{3}{10}$. Cross-multiplying and solving for $d$, we obtain $10d = 159.75 \Rightarrow d=15.975$ inches. Expressed as a fraction, $d = 15\frac{975}{1000} = \boxed{15\frac{39}{40}},$ or $\boxed{\frac{639}{40}}$ inches.


Explanation:
Okay, let's try to figure out how many students showed up to the archery class this week. The problem says that between 30 and 50 students come each week, and usually they can split into groups of equal size. But this week, Betty noticed they couldn't split into groups of equal size. Then Wilma noticed that even if she and Betty joined the students, they still couldn't split into equal groups. So we need to find the number of students that satisfies these conditions.

First, let me restate the problem in my own words. Let's denote the number of students as N. We know that 30 < N < 50. Normally, N can be divided into groups of equal size, which means that N is a composite number, right? Because if it's prime, you can't split into groups of equal size unless you have groups of size 1 or N. But since they usually break into multiple groups, maybe N is not prime? Wait, actually, the problem says "multiple groups of equal size," so maybe N is composite? But maybe not necessarily. Wait, if N is prime, you can't split into multiple groups of equal size unless you have groups of size 1. But maybe the problem implies that they split into more than one group? So if N is prime, then you can't split into groups of equal size unless you have groups of size 1. But maybe the problem is considering that splitting into groups of size 1 is not considered? Hmm, the problem says "multiple groups," so maybe they need at least two groups. Therefore, if N is prime, you can't split into multiple groups of equal size. But if N is composite, you can split into multiple groups. However, the problem says "usually the students break up into groups of equal size," so maybe N is composite. But this week, Betty noticed that she couldn't break them into groups of equal size. So this week, N is not composite? Wait, but N is between 30 and 50. So maybe N is a prime number? But wait, if N is prime, then you can't split into groups of equal size unless you have groups of size 1. But maybe the problem is considering that splitting into groups of size 1 is not allowed? So maybe N is a prime number? But then Wilma noticed that if she and Betty joined the students, meaning N + 2, they still couldn't split into groups of equal size. So N + 2 is also a prime number? But then N and N + 2 are both primes. So that would mean N is a prime number such that N + 2 is also prime. So twin primes? But N is between 30 and 50. Let me check the primes between 30 and 50. Let me list them:

Primes between 30 and 50: 31, 37, 41, 43, 47. So these are primes. Then N + 2 would be 33, 39, 43, 45, 49. Now, check if these are primes. 33 is 3*11, not prime. 39 is 3*13, not prime. 43 is prime. 45 is 9*5, not prime. 49 is 7*7, not prime. So only N=41 would make N + 2=43, which is prime. So if N=41, then N is prime, and N + 2 is also prime. Therefore, if N=41, then Betty couldn't split into groups of equal size (since it's prime), and adding 2 people (Betty and Wilma) would make 43, which is also prime, so they still can't split into groups. So that would satisfy the conditions. But wait, but the problem says "usually the students break up into groups of equal size." So if N is prime, then usually they can't split into groups. But the problem says "usually" they can. So maybe N is not prime? Wait, this is conflicting. Let me think again.

Wait, the problem says "usually the students break up into groups of equal size." So that suggests that normally, N is such that it can be divided into groups of equal size. So N is composite. But this week, Betty noticed that she couldn't break them into groups of equal size. So this week, N is not composite? But if N is prime, then it's not composite. But then Wilma noticed that if she and Betty joined the students, meaning N + 2, they still couldn't break into groups of equal size. So N + 2 is also prime? But if N is prime, then N + 2 is prime only in some cases. But if N is composite, then N + 2 could be prime or composite. Wait, but if N is composite, then maybe N is not prime, but N + 2 is also not prime? But the problem says that even after adding 2, they still couldn't split into groups. So N and N + 2 are both not composite? Wait, but if N is composite, then it can be split into groups. But this week, Betty noticed that she couldn't split them. So maybe N is not composite? But that contradicts the "usually" statement. Wait, maybe the problem is that "usually" they can split into groups, but this week they couldn't. So maybe N is such that it's not a composite number? But if N is composite, then it can be split into groups. So maybe N is a prime number? But then "usually" they can split into groups. So maybe "usually" they can split into groups, meaning that N is composite, but this week N is prime? But how can that be? If N is between 30 and 50, and usually they can split into groups, meaning that N is composite. But this week, N is prime? But how can N be prime? Unless there's some other condition. Wait, maybe the problem is that "usually" they can split into groups, but this week they couldn't. So maybe N is a number that is not a composite number? But if N is prime, then it's not composite. But if N is 1, but N is between 30 and 50. So maybe N is prime? But then the problem says "usually" they can split into groups. So maybe "usually" they can split into groups, meaning that N is composite, but this week N is prime? But how can that be? Unless the number of students varies each week? Wait, the problem says "Each week, between 30 and 50 students show up..." So each week, the number of students is between 30 and 50. But this week, the number is such that they can't split into groups. So maybe the number is prime? But then, if N is prime, then usually they can split into groups? That would mean that other weeks, the number of students is composite? But the problem says "usually" they break up into groups. So maybe "usually" implies that the number is composite, but this week it's prime. But then, the problem says that Wilma noticed that if she and Betty joined the students, they still couldn't split into groups. So N + 2 is also prime. Therefore, N and N + 2 are both primes. So twin primes. So between 30 and 50, the primes are 31, 37, 41, 43, 47. So twin primes would be (31, 33) but 33 is not prime. (37, 39) nope. (41, 43) yes, both primes. (43, 45) nope. (47, 49) nope. So only 41 and 43 are twin primes. Therefore, if N is 41, then N + 2 is 43, which is prime. Therefore, if N is 41, then this week they couldn't split into groups (since 41 is prime), and adding 2 people (Betty and Wilma) makes 43, which is also prime, so they still can't split. Therefore, N=41. But wait, but the problem says "usually the students break up into groups of equal size." So if N is 41, which is prime, then usually they can split into groups? That would mean that other weeks, the number of students is composite? But the problem says "Each week, between 30 and 50 students show up..." So maybe the number of students varies each week? So sometimes it's composite, sometimes prime? But the problem says "usually" they break up into groups. So maybe "usually" means that most of the time, the number is composite? But the problem is about this week. So maybe the answer is 41? But let me check if there are other possibilities. Wait, maybe N is not prime, but it's a number that is not divisible by any number except 1 and itself? Wait, no, that's prime. Wait, maybe N is a composite number, but it's not possible to split into groups? Wait, but if N is composite, then it can be split into groups. For example, if N is 32, which is composite, you can split into groups of 2, 4, 8, etc. So if N is composite, you can split into groups. Therefore, if Betty noticed that she couldn't split into groups, that would mean that N is prime. Therefore, N is prime. Then Wilma noticed that adding 2 people (Betty and Wilma) still couldn't split into groups, meaning N + 2 is also prime. Therefore, N and N + 2 are both primes. So twin primes. As we saw, only 41 and 43. Therefore, N=41. But let me check if there are other numbers between 30 and 50 where N is prime and N + 2 is also prime. For example, 31 and 33. 33 is not prime. 37 and 39, 39 is not prime. 41 and 43, both primes. 43 and 45, 45 not prime. 47 and 49, 49 not prime. So only 41. Therefore, N=41. But let me check if there's another way. Maybe N is not prime, but it's a number that is not divisible by any number except 1 and itself? Wait, that's prime. So maybe N is a prime. Therefore, answer is 41. But let me check if there's another possibility. Suppose that N is composite, but for some reason, Betty couldn't split into groups. Wait, but if N is composite, you can split into groups. Unless the number of groups is specified? Wait, the problem says "break up into multiple groups of equal size." So if N is composite, you can split into groups. For example, if N is 32, you can split into 2 groups of 16, 4 groups of 8, etc. So if N is composite, you can split into groups. Therefore, if Betty couldn't split into groups, N must be prime. Therefore, N is prime. Then N + 2 is also prime. Therefore, N=41. Therefore, the answer is 41. But let me check if 41 is between 30 and 50. Yes, 41 is between 30 and 50. So that seems to fit. But let me check if there's another possible number. For example, suppose N is 49. Wait, 49 is 7 squared. So it's composite. So if N=49, then Betty could split into groups of 7. But the problem says that Betty noticed she couldn't split into groups. Therefore, N=49 is not possible. Similarly, N=47 is prime. If N=47, then N + 2=49, which is composite. Therefore, if N=47, then adding 2 people would make 49, which is composite. Therefore, Wilma noticed that they could split into groups. But the problem says Wilma noticed that they still couldn't split into groups. Therefore, N=47 is not possible. Similarly, N=43. Then N + 2=45, which is composite. Therefore, adding 2 people would make 45, which is composite. Therefore, they could split into groups. Therefore, Wilma would not have noticed that they couldn't split. Therefore, N=43 is not possible. N=37. N + 2=39, which is composite. So same thing. N=31. N + 2=33, composite. Therefore, only N=41 gives N + 2=43, which is prime. Therefore, answer is 41. Therefore, the number of students is 41. Wait, but let me check if there's another way. Maybe the problem is not about primes? Maybe there's another condition. For example, maybe the number of students is such that it's not divisible by any number between 2 and N-1? But that's prime. So same thing. Alternatively, maybe the number of students is such that it's not divisible by any number except 1 and itself. Which is prime. So same thing. Therefore, answer is 41. Therefore, I think the answer is 41. But let me check again. Let me check if 41 is between 30 and 50. Yes. If N=41, then Betty couldn't split into groups (since it's prime). Then adding 2 people (Betty and Wilma) makes 43, which is also prime, so they still can't split. Therefore, that fits. Therefore, the answer is 41. But let me check if there's another number. For example, suppose N is 25. But 25 is less than 30. Not in the range. How about N= 35? 35 is composite. So Betty could split into groups. So not applicable. N= 33. Composite. N= 37. Prime. N + 2=39, composite. So Wilma would have noticed that they could split. Therefore, not applicable. N= 39. Composite. N= 41. Prime. N + 2=43. Prime. N= 43. Prime. N + 2=45. Composite. N= 47. Prime. N + 2=49. Composite. So only N=41. Therefore, answer is 41. Therefore, I think the answer is 41. So I think the answer is \boxed{41}.

**Final Answer**
\boxed{41}

To determine how many students showed up to the archery class this week, we analyze the conditions given:

---

### **Understanding the Problem**

- The number of students, $ N $, is between **30 and 50**.
- **Usually**, the students can be split into **multiple groups of equal size**.
- **This week**, Betty noticed that **they could not** be split into groups of equal size.
- **Wilma noticed** that even if **Betty and Wilma joined the students**, the total number of people (i.e., $ N + 2 $) **still could not** be split into groups of equal size.

---

### **Key Observations**

1. **Splitting into groups of equal size** implies that the number must be **composite** (i.e., not prime), because a prime number can only be divided into groups of size 1 or itself. If the number is prime, and the group size is more than 1, it's not possible to split into equal groups.

2. Therefore, **this week**, the number of students $ N $ must be **prime**, since they couldn't be split into equal groups.

3. **After adding 2 people** (Betty and Wilma), the total becomes $ N + 2 $, and **still** couldn't be split into equal groups. So $ N + 2 $ must also be **prime**.

4. Thus, we are looking for a **prime number $ N $** such that $ N + 2 $ is also **prime**. These are known as **twin primes**.

---

### **Step-by-Step Search for Twin Primes Between 30 and 50**

Let’s list the prime numbers between 30 and 50:

- 31, 37, 41, 43, 47

Now, check which of these have a **twin prime** (i.e., $ N + 2 $ is also prime):

- $ 31 + 2 = 33 $ → Not prime
- $ 37 + 2 = 39 $ → Not prime
- $ 41 + 2 = 43 $ → **Prime**
- $ 43 + 2 = 45 $ → Not prime
- $ 47 + 2 = 49 $ → Not prime

Only **41** and **43** are twin primes in this range.

---

### **Conclusion**

- $ N = 41 $ is **prime**, so the students couldn't be split into equal groups.
- $ N + 2 = 43 $ is also **prime**, so even with Betty and Wilma joining, the total number still couldn't be split into equal groups.

This satisfies **all the conditions** of the problem.

---

### **Final Answer**

$$
\boxed{41}
$$

[Assistant]
Explanation:
Okay, let's try to figure out how many students showed up to the archery class this week. The problem says that between 30 and 50 students come each week, and usually they can split into groups of equal size. But this week, Betty noticed they couldn't split into groups of equal size. Then Wilma noticed that even if she and Betty joined the students, they still couldn't split into equal groups. So we need to find the number of students that satisfies these conditions.

First, let me restate the problem in my own words. Let's denote the number of students as N. We know that 30 < N < 50. Normally, N can be divided into groups of equal size, which means that N is a composite number, right? Because if it's prime, you can't split into groups of equal size unless you have groups of size 1 or N. But since they usually break into multiple groups, maybe N is not prime? Wait, actually, the problem says "multiple groups of equal size," so maybe N is composite? But maybe not necessarily. Wait, if N is prime, you can't split into multiple groups of equal size unless you have groups of size 1. But maybe the problem implies that they split into more than one group? So if N is prime, then you can't split into groups of equal size unless you have groups of size 1. But maybe the problem is considering that splitting into groups of size 1 is not considered? Hmm, the problem says "multiple groups," so maybe they need at least two groups. Therefore, if N is prime, you can't split into multiple groups of equal size. But if N is composite, you can split into multiple groups. However, the problem says "usually the students break up into groups of equal size," so maybe N is composite. But this week, Betty noticed that she couldn't break them into groups of equal size. So this week, N is not composite? Wait, but N is between 30 and 50. So maybe N is a prime number? But wait, if N is prime, then you can't split into groups of equal size unless you have groups of size 1. But maybe the problem is considering that splitting into groups of size 1 is not allowed? So maybe N is a prime number? But then Wilma noticed that if she and Betty joined the students, meaning N + 2, they still couldn't split into groups of equal size. So N + 2 is also a prime number? But then N and N + 2 are both primes. So that would mean N is a prime number such that N + 2 is also prime. So twin primes? But N is between 30 and 50. Let me check the primes between 30 and 50. Let me list them:

Primes between 30 and 50: 31, 37, 41, 43, 47. So these are primes. Then N + 2 would be 33, 39, 43, 45, 49. Now, check if these are primes. 33 is 3*11, not prime. 39 is 3*13, not prime. 43 is prime. 45 is 9*5, not prime. 49 is 7*7, not prime. So only N=41 would make N + 2=43, which is prime. So if N=41, then N is prime, and N + 2 is also prime. Therefore, if N=41, then Betty couldn't split into groups of equal size (since it's prime), and adding 2 people (Betty and Wilma) would make 43, which is also prime, so they still can't split into groups. So that would satisfy the conditions. But wait, but the problem says "usually the students break up into groups of equal size." So if N is prime, then usually they can't split into groups. But the problem says "usually" they can. So maybe N is not prime? Wait, this is conflicting. Let me think again.

Wait, the problem says "usually the students break up into groups of equal size." So that suggests that normally, N is such that it can be divided into groups of equal size. So N is composite. But this week, Betty noticed that she couldn't break them into groups of equal size. So this week, N is not composite? But if N is prime, then it's not composite. But then Wilma noticed that if she and Betty joined the students, meaning N + 2, they still couldn't break into groups of equal size. So N + 2 is also prime? But if N is prime, then N + 2 is prime only in some cases. But if N is composite, then N + 2 could be prime or composite. Wait, but if N is composite, then maybe N is not prime, but N + 2 is also not prime? But the problem says that even after adding 2, they still couldn't split into groups. So N and N + 2 are both not composite? Wait, but if N is composite, then it can be split into groups. But this week, Betty noticed that she couldn't split them. So maybe N is not composite? But that contradicts the "usually" statement. Wait, maybe the problem is that "usually" they can split into groups, but this week they couldn't. So maybe N is such that it's not a composite number? But if N is composite, then it can be split into groups. So maybe N is a prime number? But then "usually" they can split into groups. So maybe "usually" they can split into groups, meaning that N is composite, but this week N is prime? But how can that be? If N is between 30 and 50, and usually they can split into groups, meaning that N is composite. But this week, N is prime? But how can N be prime? Unless there's some other condition. Wait, maybe the problem is that "usually" they can split into groups, but this week they couldn't. So maybe N is a number that is not a composite number? But if N is prime, then it's not composite. But if N is 1, but N is between 30 and 50. So maybe N is prime? But then the problem says "usually" they can split into groups. So maybe "usually" they can split into groups, meaning that N is composite, but this week N is prime? But how can that be? Unless the number of students varies each week? Wait, the problem says "Each week, between 30 and 50 students show up..." So each week, the number of students is between 30 and 50. But this week, the number is such that they can't split into groups. So maybe the number is prime? But then, if N is prime, then usually they can split into groups? That would mean that other weeks, the number of students is composite? But the problem says "usually" they break up into groups. So maybe "usually" implies that the number is composite, but this week it's prime. But then, the problem says that Wilma noticed that if she and Betty joined the students, they still couldn't split into groups. So N + 2 is also prime. Therefore, N and N + 2 are both primes. So twin primes. So between 30 and 50, the primes are 31, 37, 41, 43, 47. So twin primes would be (31, 33) but 33 is not prime. (37, 39) nope. (41, 43) yes, both primes. (43, 45) nope. (47, 49) nope. So only 41 and 43 are twin primes. Therefore, if N is 41, then N + 2 is 43, which is prime. Therefore, if N is 41, then this week they couldn't split into groups (since 41 is prime), and adding 2 people (Betty and Wilma) makes 43, which is also prime, so they still can't split. Therefore, N=41. But wait, but the problem says "usually the students break up into groups of equal size." So if N is 41, which is prime, then usually they can split into groups? That would mean that other weeks, the number of students is composite? But the problem says "Each week, between 30 and 50 students show up..." So maybe the number of students varies each week? So sometimes it's composite, sometimes prime? But the problem says "usually" they break up into groups. So maybe "usually" means that most of the time, the number is composite? But the problem is about this week. So maybe the answer is 41? But let me check if there are other possibilities. Wait, maybe N is not prime, but it's a number that is not divisible by any number except 1 and itself? Wait, no, that's prime. Wait, maybe N is a composite number, but it's not possible to split into groups? Wait, but if N is composite, then it can be split into groups. For example, if N is 32, which is composite, you can split into groups of 2, 4, 8, etc. So if N is composite, you can split into groups. Therefore, if Betty noticed that she couldn't split into groups, that would mean that N is prime. Therefore, N is prime. Then Wilma noticed that adding 2 people (Betty and Wilma) still couldn't split into groups, meaning N + 2 is also prime. Therefore, N and N + 2 are both primes. So twin primes. As we saw, only 41 and 43. Therefore, N=41. But let me check if there are other numbers between 30 and 50 where N is prime and N + 2 is also prime. For example, 31 and 33. 33 is not prime. 37 and 39, 39 is not prime. 41 and 43, both primes. 43 and 45, 45 not prime. 47 and 49, 49 not prime. So only 41. Therefore, N=41. But let me check if there's another way. Maybe N is not prime, but it's a number that is not divisible by any number except 1 and itself? Wait, that's prime. So maybe N is a prime. Therefore, answer is 41. But let me check if there's another possibility. Suppose that N is composite, but for some reason, Betty couldn't split into groups. Wait, but if N is composite, you can split into groups. Unless the number of groups is specified? Wait, the problem says "break up into multiple groups of equal size." So if N is composite, you can split into groups. For example, if N is 32, you can split into 2 groups of 16, 4 groups of 8, etc. So if N is composite, you can split into groups. Therefore, if Betty couldn't split into groups, N must be prime. Therefore, N is prime. Then N + 2 is also prime. Therefore, N=41. Therefore, the answer is 41. But let me check if 41 is between 30 and 50. Yes, 41 is between 30 and 50. So that seems to fit. But let me check if there's another possible number. For example, suppose N is 49. Wait, 49 is 7 squared. So it's composite. So if N=49, then Betty could split into groups of 7. But the problem says that Betty noticed she couldn't split into groups. Therefore, N=49 is not possible. Similarly, N=47 is prime. If N=47, then N + 2=49, which is composite. Therefore, if N=47, then adding 2 people would make 49, which is composite. Therefore, Wilma noticed that they could split into groups. But the problem says Wilma noticed that they still couldn't split into groups. Therefore, N=47 is not possible. Similarly, N=43. Then N + 2=45, which is composite. Therefore, adding 2 people would make 45, which is composite. Therefore, they could split into groups. Therefore, Wilma would not have noticed that they couldn't split. Therefore, N=43 is not possible. N=37. N + 2=39, which is composite. So same thing. N=31. N + 2=33, composite. Therefore, only N=41 gives N + 2=43, which is prime. Therefore, answer is 41. Therefore, the number of students is 41. Wait, but let me check if there's another way. Maybe the problem is not about primes? Maybe there's another condition. For example, maybe the number of students is such that it's not divisible by any number between 2 and N-1? But that's prime. So same thing. Alternatively, maybe the number of students is such that it's not divisible by any number except 1 and itself. Which is prime. So same thing. Therefore, answer is 41. Therefore, I think the answer is 41. But let me check again. Let me check if 41 is between 30 and 50. Yes. If N=41, then Betty couldn't split into groups (since it's prime). Then adding 2 people (Betty and Wilma) makes 43, which is also prime, so they still can't split. Therefore, that fits. Therefore, the answer is 41. But let me check if there's another number. For example, suppose N is 25. But 25 is less than 30. Not in the range. How about N= 35? 35 is composite. So Betty could split into groups. So not applicable. N= 33. Composite. N= 37. Prime. N + 2=39, composite. So Wilma would have noticed that they could split. Therefore, not applicable. N= 39. Composite. N= 41. Prime. N + 2=43. Prime. N= 43. Prime. N + 2=45. Composite. N= 47. Prime. N + 2=49. Composite. So only N=41. Therefore, answer is 41. Therefore, I think the answer is 41. So I think the answer is \boxed{41}.

**Final Answer**
\boxed{41}

To determine how many students showed up to the archery class this week, we analyze the conditions given:

---

### **Understanding the Problem**

- The number of students, $ N $, is between **30 and 50**.
- **Usually**, the students can be split into **multiple groups of equal size**.
- **This week**, Betty noticed that **they could not** be split into groups of equal size.
- **Wilma noticed** that even if **Betty and Wilma joined the students**, the total number of people (i.e., $ N + 2 $) **still could not** be split into groups of equal size.

---

### **Key Observations**

1. **Splitting into groups of equal size** implies that the number must be **composite** (i.e., not prime), because a prime number can only be divided into groups of size 1 or itself. If the number is prime, and the group size is more than 1, it's not possible to split into equal groups.

2. Therefore, **this week**, the number of students $ N $ must be **prime**, since they couldn't be split into equal groups.

3. **After adding 2 people** (Betty and Wilma), the total becomes $ N + 2 $, and **still** couldn't be split into equal groups. So $ N + 2 $ must also be **prime**.

4. Thus, we are looking for a **prime number $ N $** such that $ N + 2 $ is also **prime**. These are known as **twin primes**.

---

### **Step-by-Step Search for Twin Primes Between 30 and 50**

Let’s list the prime numbers between 30 and 50:

- 31, 37, 41, 43, 47

Now, check which of these have a **twin prime** (i.e., $ N + 2 $ is also prime):

- $ 31 + 2 = 33 $ → Not prime
- $ 37 + 2 = 39 $ → Not prime
- $ 41 + 2 = 43 $ → **Prime**
- $ 43 + 2 = 45 $ → Not prime
- $ 47 + 2 = 49 $ → Not prime

Only **41** and **43** are twin primes in this range.

---

### **Conclusion**

- $ N = 41 $ is **prime**, so the students couldn't be split into equal groups.
- $ N + 2 = 43 $ is also **prime**, so even with Betty and Wilma joining, the total number still couldn't be split into equal groups.

This satisfies **all the conditions** of the problem.

---

### **Final Answer**

$$
\boxed{41}
$$

Solution: Let $n$ be the number of students at class.  Betty noticed that $n$ has no divisors between 1 and itself, so $n$ is prime.  Wilma noticed that $n + 2$ is prime.  This means we are looking for the smaller of two primes that differ by 2 that are between 30 and 50.  The primes in that range are 31, 37, 41, 43, and 47.  Since 41 and 43 differ by 2, $n = \boxed{41}$.


Explanation:
Okay, so I need to solve this system of logarithmic equations:

First equation: log base 6 of (x - y) plus log base 6 of (x + y) equals 2.

Second equation: log base y of 5x equals 2.

And I need to find the value of x. Alright, let me start by recalling some logarithm properties.

For the first equation, since both terms have the same base, 6, I can use the logarithm property that says log_b(a) + log_b(c) = log_b(a*c). So combining those two logs into one:

log_6[(x - y)(x + y)] = 2.

Then, converting that logarithmic equation to its exponential form. Remember that log_b(A) = C is equivalent to b^C = A. So here, that would be:

6^2 = (x - y)(x + y).

Calculating 6 squared is 36, so:

(x - y)(x + y) = 36.

But (x - y)(x + y) is a difference of squares, so that simplifies to x² - y² = 36. Let me write that down:

Equation (1): x² - y² = 36.

Now moving on to the second equation: log base y of 5x equals 2. Again, converting that to exponential form. log_y(5x) = 2 means that y² = 5x. So:

Equation (2): y² = 5x.

So now I have two equations:

1) x² - y² = 36

2) y² = 5x

Hmm, maybe I can substitute equation (2) into equation (1). Let me try that. Since equation (2) gives me y² in terms of x, I can replace y² in equation (1) with 5x. Let me do that:

x² - (5x) = 36.

So that simplifies to x² - 5x - 36 = 0.

Now, solving this quadratic equation for x. Let me check if I can factor it. Let me think of two numbers that multiply to -36 and add up to -5. Let me see... factors of 36 are 1 & 36, 2 & 18, 3 & 12, 4 & 9, 6 & 6. Since the product is negative, one has to be positive and the other negative. Let me try 4 and -9: 4 * (-9) = -36, and 4 + (-9) = -5. Yes! Perfect. So the quadratic factors as:

(x + 4)(x - 9) = 0.

Therefore, the solutions are x = -4 or x = 9.

But wait, I need to check if these solutions are valid in the original equations, especially considering the domains of the logarithms.

First, let's recall that logarithms have certain domain restrictions. For log_6(x - y) and log_6(x + y) to be defined, the arguments must be positive. So:

x - y > 0 and x + y > 0.

Also, for log_y(5x) to be defined, the base y must be positive and not equal to 1, and the argument 5x must be positive. So:

y > 0, y ≠ 1, and 5x > 0 ⇒ x > 0.

So, let's check each solution.

First, x = -4. If x is -4, then from equation (2): y² = 5x = 5*(-4) = -20. But y² can't be negative. So this is impossible. Therefore, x = -4 is rejected.

So the only possible solution is x = 9. Let me check if this works.

If x = 9, then from equation (2): y² = 5*9 = 45 ⇒ y = √45 or y = -√45. But since y must be positive (from the domain of log_y(5x)), we take y = √45. Simplify √45: √(9*5) = 3√5. So y = 3√5.

Now check the domain conditions for the first equation. Let's check x - y and x + y.

x - y = 9 - 3√5. Let me approximate √5 ≈ 2.236, so 3√5 ≈ 6.708. Then 9 - 6.708 ≈ 2.292, which is positive. Good.

x + y = 9 + 3√5 ≈ 9 + 6.708 ≈ 15.708, which is also positive. So both arguments are positive. Also, check that y ≠ 1. Since y = 3√5 ≈ 6.708, which is not 1. Also, 5x = 5*9 = 45 > 0. So all domain conditions are satisfied.

Therefore, x = 9 is a valid solution.

But just to be thorough, let me check if substituting x = 9 and y = 3√5 into the original equations works.

First equation: log_6(x - y) + log_6(x + y) = log_6[(x - y)(x + y)] = log_6(x² - y²). From equation (1), x² - y² = 36, so log_6(36). Since 6² = 36, log_6(36) = 2. Which matches the first equation. Good.

Second equation: log_y(5x) = log_{3√5}(45). Let me check if this equals 2.

By definition, log_{3√5}(45) = 2 means that (3√5)^2 = 45. Let me compute (3√5)^2: 3²*(√5)^2 = 9*5 = 45. Yes! So that's correct. Therefore, both equations are satisfied.

Therefore, x = 9 is the correct solution.

Wait, but let me just make sure there are no other possible solutions. For example, when solving the quadratic equation, we had x = -4 and x = 9. We rejected x = -4 because y² would be negative. But is there a possibility that maybe y could be negative? Wait, no. Because in the second equation, log_y(5x) requires that y > 0 and y ≠ 1. So even if we had y negative, it would not be allowed. Therefore, only x = 9 is valid.

Therefore, the answer is x = 9.

But let me just recap all steps to make sure I didn't skip anything.

Original equations:

1) log_6(x - y) + log_6(x + y) = 2

2) log_y(5x) = 2

First equation combined to x² - y² = 36.

Second equation converted to y² = 5x.

Substituted into first equation: x² - 5x = 36 ⇒ x² -5x -36 = 0 ⇒ (x - 9)(x + 4) = 0 ⇒ x = 9 or x = -4. Discarded x = -4 because y² would be negative. Then checked x = 9, found y = 3√5, checked domain conditions, and verified both equations. All good.

Therefore, x = 9 is the solution.

I think that's solid. I don't see any mistakes here. Maybe I can check if there's another way to approach the problem, but I think this is straightforward.

Alternatively, maybe solving for y from equation (2) and substituting into equation (1). Let me try that approach as a check.

From equation (2): y² = 5x ⇒ y = √(5x). Since y must be positive, we take the positive root.

Then, substitute into equation (1): x² - y² = 36 ⇒ x² - 5x = 36 ⇒ same quadratic equation. So same result. So same conclusion. So that's consistent.

Alternatively, maybe solving for x in terms of y from equation (2). From equation (2): 5x = y² ⇒ x = y² / 5. Then substitute into equation (1):

x² - y² = 36 ⇒ (y² / 5)^2 - y² = 36 ⇒ y^4 / 25 - y² - 36 = 0. Multiply both sides by 25: y^4 - 25y² - 900 = 0. Let me set z = y², so equation becomes z² -25z -900 = 0. Solve for z:

z = [25 ± √(625 + 3600)] / 2 = [25 ± √4225]/2 = [25 ± 65]/2. So z = (25 + 65)/2 = 90/2 = 45 or z = (25 - 65)/2 = (-40)/2 = -20. Since z = y² can't be negative, so z = 45. Therefore, y² = 45 ⇒ y = 3√5. Then x = y² / 5 = 45 / 5 = 9. Same result. So that's another way to get x = 9. So that's consistent.

Therefore, regardless of the method, x = 9 is the solution.

Therefore, I can be confident that the answer is 9.

**Final Answer**
The value of $ x $ is \boxed{9}.

To solve the system of equations:

$$
\log_6 (x - y) + \log_6 (x + y) = 2 \quad \text{(1)}
$$
$$
\log_y (5x) = 2 \quad \text{(2)}
$$

we proceed step by step using logarithmic identities and algebraic manipulation.

---

### Step 1: Simplify the First Equation

Using the logarithmic identity:

$$
\log_b a + \log_b c = \log_b (ac)
$$

we combine the two logarithms in equation (1):

$$
\log_6 (x - y) + \log_6 (x + y) = \log_6 [(x - y)(x + y)] = \log_6 (x^2 - y^2)
$$

So the equation becomes:

$$
\log_6 (x^2 - y^2) = 2
$$

Convert this to exponential form:

$$
x^2 - y^2 = 6^2 = 36
$$

This gives us our first equation:

$$
x^2 - y^2 = 36 \quad \text{(1')}
$$

---

### Step 2: Simplify the Second Equation

From equation (2):

$$
\log_y (5x) = 2
$$

Convert to exponential form:

$$
y^2 = 5x
$$

This gives us our second equation:

$$
y^2 = 5x \quad \text{(2')}
$$

---

### Step 3: Substitute Equation (2') into Equation (1')

Substitute $ y^2 = 5x $ into $ x^2 - y^2 = 36 $:

$$
x^2 - 5x = 36
$$

Rewriting:

$$
x^2 - 5x - 36 = 0
$$

Solve this quadratic equation. Factor:

$$
(x - 9)(x + 4) = 0
$$

So the solutions are:

$$
x = 9 \quad \text{or} \quad x = -4
$$

---

### Step 4: Check for Validity of Solutions

We must ensure that the logarithms are defined, which imposes the following conditions:

- $ x - y > 0 $, $ x + y > 0 $ (arguments of log base 6 must be positive)
- $ y > 0 $, $ y \ne 1 $ (base of log must be positive and not 1)
- $ 5x > 0 $ (argument of log base y must be positive)

#### Case 1: $ x = -4 $

From equation (2'):

$$
y^2 = 5x = 5(-4) = -20
$$

This is **not possible**, since $ y^2 $ cannot be negative. So $ x = -4 $ is **invalid**.

#### Case 2: $ x = 9 $

From equation (2'):

$$
y^2 = 5x = 5(9) = 45 \Rightarrow y = \sqrt{45} = 3\sqrt{5}
$$

Check domain conditions:

- $ x - y = 9 - 3\sqrt{5} \approx 9 - 6.708 = 2.292 > 0 $
- $ x + y = 9 + 3\sqrt{5} \approx 15.708 > 0 $
- $ y = 3\sqrt{5} > 0 $, and $ y \ne 1 $
- $ 5x = 45 > 0 $

All conditions are satisfied.

---

### Step 5: Verify the Solution in Original Equations

#### First Equation:

$$
\log_6 (x - y) + \log_6 (x + y) = \log_6 [(x - y)(x + y)] = \log_6 (x^2 - y^2)
$$

From earlier, $ x^2 - y^2 = 36 $, so:

$$
\log_6 (36) = 2 \quad \text{✓}
$$

#### Second Equation:

$$
\log_y (5x) = \log_{3\sqrt{5}} (45)
$$

Check if $ (3\sqrt{5})^2 = 45 $:

$$
(3\sqrt{5})^2 = 9 \cdot 5 = 45 \quad \text{✓}
$$

So the equation is satisfied.

---

### Final Answer

$$
\boxed{9}
$$

[Assistant]
Explanation:
Okay, so I need to solve this system of logarithmic equations:

First equation: log base 6 of (x - y) plus log base 6 of (x + y) equals 2.

Second equation: log base y of 5x equals 2.

And I need to find the value of x. Alright, let me start by recalling some logarithm properties.

For the first equation, since both terms have the same base, 6, I can use the logarithm property that says log_b(a) + log_b(c) = log_b(a*c). So combining those two logs into one:

log_6[(x - y)(x + y)] = 2.

Then, converting that logarithmic equation to its exponential form. Remember that log_b(A) = C is equivalent to b^C = A. So here, that would be:

6^2 = (x - y)(x + y).

Calculating 6 squared is 36, so:

(x - y)(x + y) = 36.

But (x - y)(x + y) is a difference of squares, so that simplifies to x² - y² = 36. Let me write that down:

Equation (1): x² - y² = 36.

Now moving on to the second equation: log base y of 5x equals 2. Again, converting that to exponential form. log_y(5x) = 2 means that y² = 5x. So:

Equation (2): y² = 5x.

So now I have two equations:

1) x² - y² = 36

2) y² = 5x

Hmm, maybe I can substitute equation (2) into equation (1). Let me try that. Since equation (2) gives me y² in terms of x, I can replace y² in equation (1) with 5x. Let me do that:

x² - (5x) = 36.

So that simplifies to x² - 5x - 36 = 0.

Now, solving this quadratic equation for x. Let me check if I can factor it. Let me think of two numbers that multiply to -36 and add up to -5. Let me see... factors of 36 are 1 & 36, 2 & 18, 3 & 12, 4 & 9, 6 & 6. Since the product is negative, one has to be positive and the other negative. Let me try 4 and -9: 4 * (-9) = -36, and 4 + (-9) = -5. Yes! Perfect. So the quadratic factors as:

(x + 4)(x - 9) = 0.

Therefore, the solutions are x = -4 or x = 9.

But wait, I need to check if these solutions are valid in the original equations, especially considering the domains of the logarithms.

First, let's recall that logarithms have certain domain restrictions. For log_6(x - y) and log_6(x + y) to be defined, the arguments must be positive. So:

x - y > 0 and x + y > 0.

Also, for log_y(5x) to be defined, the base y must be positive and not equal to 1, and the argument 5x must be positive. So:

y > 0, y ≠ 1, and 5x > 0 ⇒ x > 0.

So, let's check each solution.

First, x = -4. If x is -4, then from equation (2): y² = 5x = 5*(-4) = -20. But y² can't be negative. So this is impossible. Therefore, x = -4 is rejected.

So the only possible solution is x = 9. Let me check if this works.

If x = 9, then from equation (2): y² = 5*9 = 45 ⇒ y = √45 or y = -√45. But since y must be positive (from the domain of log_y(5x)), we take y = √45. Simplify √45: √(9*5) = 3√5. So y = 3√5.

Now check the domain conditions for the first equation. Let's check x - y and x + y.

x - y = 9 - 3√5. Let me approximate √5 ≈ 2.236, so 3√5 ≈ 6.708. Then 9 - 6.708 ≈ 2.292, which is positive. Good.

x + y = 9 + 3√5 ≈ 9 + 6.708 ≈ 15.708, which is also positive. So both arguments are positive. Also, check that y ≠ 1. Since y = 3√5 ≈ 6.708, which is not 1. Also, 5x = 5*9 = 45 > 0. So all domain conditions are satisfied.

Therefore, x = 9 is a valid solution.

But just to be thorough, let me check if substituting x = 9 and y = 3√5 into the original equations works.

First equation: log_6(x - y) + log_6(x + y) = log_6[(x - y)(x + y)] = log_6(x² - y²). From equation (1), x² - y² = 36, so log_6(36). Since 6² = 36, log_6(36) = 2. Which matches the first equation. Good.

Second equation: log_y(5x) = log_{3√5}(45). Let me check if this equals 2.

By definition, log_{3√5}(45) = 2 means that (3√5)^2 = 45. Let me compute (3√5)^2: 3²*(√5)^2 = 9*5 = 45. Yes! So that's correct. Therefore, both equations are satisfied.

Therefore, x = 9 is the correct solution.

Wait, but let me just make sure there are no other possible solutions. For example, when solving the quadratic equation, we had x = -4 and x = 9. We rejected x = -4 because y² would be negative. But is there a possibility that maybe y could be negative? Wait, no. Because in the second equation, log_y(5x) requires that y > 0 and y ≠ 1. So even if we had y negative, it would not be allowed. Therefore, only x = 9 is valid.

Therefore, the answer is x = 9.

But let me just recap all steps to make sure I didn't skip anything.

Original equations:

1) log_6(x - y) + log_6(x + y) = 2

2) log_y(5x) = 2

First equation combined to x² - y² = 36.

Second equation converted to y² = 5x.

Substituted into first equation: x² - 5x = 36 ⇒ x² -5x -36 = 0 ⇒ (x - 9)(x + 4) = 0 ⇒ x = 9 or x = -4. Discarded x = -4 because y² would be negative. Then checked x = 9, found y = 3√5, checked domain conditions, and verified both equations. All good.

Therefore, x = 9 is the solution.

I think that's solid. I don't see any mistakes here. Maybe I can check if there's another way to approach the problem, but I think this is straightforward.

Alternatively, maybe solving for y from equation (2) and substituting into equation (1). Let me try that approach as a check.

From equation (2): y² = 5x ⇒ y = √(5x). Since y must be positive, we take the positive root.

Then, substitute into equation (1): x² - y² = 36 ⇒ x² - 5x = 36 ⇒ same quadratic equation. So same result. So same conclusion. So that's consistent.

Alternatively, maybe solving for x in terms of y from equation (2). From equation (2): 5x = y² ⇒ x = y² / 5. Then substitute into equation (1):

x² - y² = 36 ⇒ (y² / 5)^2 - y² = 36 ⇒ y^4 / 25 - y² - 36 = 0. Multiply both sides by 25: y^4 - 25y² - 900 = 0. Let me set z = y², so equation becomes z² -25z -900 = 0. Solve for z:

z = [25 ± √(625 + 3600)] / 2 = [25 ± √4225]/2 = [25 ± 65]/2. So z = (25 + 65)/2 = 90/2 = 45 or z = (25 - 65)/2 = (-40)/2 = -20. Since z = y² can't be negative, so z = 45. Therefore, y² = 45 ⇒ y = 3√5. Then x = y² / 5 = 45 / 5 = 9. Same result. So that's another way to get x = 9. So that's consistent.

Therefore, regardless of the method, x = 9 is the solution.

Therefore, I can be confident that the answer is 9.

**Final Answer**
The value of $ x $ is \boxed{9}.

To solve the system of equations:

$$
\log_6 (x - y) + \log_6 (x + y) = 2 \quad \text{(1)}
$$
$$
\log_y (5x) = 2 \quad \text{(2)}
$$

we proceed step by step using logarithmic identities and algebraic manipulation.

---

### Step 1: Simplify the First Equation

Using the logarithmic identity:

$$
\log_b a + \log_b c = \log_b (ac)
$$

we combine the two logarithms in equation (1):

$$
\log_6 (x - y) + \log_6 (x + y) = \log_6 [(x - y)(x + y)] = \log_6 (x^2 - y^2)
$$

So the equation becomes:

$$
\log_6 (x^2 - y^2) = 2
$$

Convert this to exponential form:

$$
x^2 - y^2 = 6^2 = 36
$$

This gives us our first equation:

$$
x^2 - y^2 = 36 \quad \text{(1')}
$$

---

### Step 2: Simplify the Second Equation

From equation (2):

$$
\log_y (5x) = 2
$$

Convert to exponential form:

$$
y^2 = 5x
$$

This gives us our second equation:

$$
y^2 = 5x \quad \text{(2')}
$$

---

### Step 3: Substitute Equation (2') into Equation (1')

Substitute $ y^2 = 5x $ into $ x^2 - y^2 = 36 $:

$$
x^2 - 5x = 36
$$

Rewriting:

$$
x^2 - 5x - 36 = 0
$$

Solve this quadratic equation. Factor:

$$
(x - 9)(x + 4) = 0
$$

So the solutions are:

$$
x = 9 \quad \text{or} \quad x = -4
$$

---

### Step 4: Check for Validity of Solutions

We must ensure that the logarithms are defined, which imposes the following conditions:

- $ x - y > 0 $, $ x + y > 0 $ (arguments of log base 6 must be positive)
- $ y > 0 $, $ y \ne 1 $ (base of log must be positive and not 1)
- $ 5x > 0 $ (argument of log base y must be positive)

#### Case 1: $ x = -4 $

From equation (2'):

$$
y^2 = 5x = 5(-4) = -20
$$

This is **not possible**, since $ y^2 $ cannot be negative. So $ x = -4 $ is **invalid**.

#### Case 2: $ x = 9 $

From equation (2'):

$$
y^2 = 5x = 5(9) = 45 \Rightarrow y = \sqrt{45} = 3\sqrt{5}
$$

Check domain conditions:

- $ x - y = 9 - 3\sqrt{5} \approx 9 - 6.708 = 2.292 > 0 $
- $ x + y = 9 + 3\sqrt{5} \approx 15.708 > 0 $
- $ y = 3\sqrt{5} > 0 $, and $ y \ne 1 $
- $ 5x = 45 > 0 $

All conditions are satisfied.

---

### Step 5: Verify the Solution in Original Equations

#### First Equation:

$$
\log_6 (x - y) + \log_6 (x + y) = \log_6 [(x - y)(x + y)] = \log_6 (x^2 - y^2)
$$

From earlier, $ x^2 - y^2 = 36 $, so:

$$
\log_6 (36) = 2 \quad \text{✓}
$$

#### Second Equation:

$$
\log_y (5x) = \log_{3\sqrt{5}} (45)
$$

Check if $ (3\sqrt{5})^2 = 45 $:

$$
(3\sqrt{5})^2 = 9 \cdot 5 = 45 \quad \text{✓}
$$

So the equation is satisfied.

---

### Final Answer

$$
\boxed{9}
$$

Solution: Working on the first equation, we have from the difference of squares factorization that $\log_6 (x-y) + \log_6 (x+y) = \log_6 (x^2-y^2) = 2$, so $x^2 - y^2 = 6^2 = 36$. Using the change of base formula, the second equation becomes $$\frac{\log(5x)}{\log y} = 2 \Longrightarrow \log(5x) = 2\log y = \log y^2.$$Substituting that $y^2 = x^2 - 36$, it follows that $\log (x^2 - 36) = \log y^2 = 2\log y = \log 5x$. Since the logarithm is a one-to-one function, it follows that $x^2 - 36 = 5x$, so $x^2 - 5x - 36 = (x - 9)(x + 4) = 0$. Thus, $x = 9, - 4$, but the second does not work. Thus, our answer is $x = \boxed{9}$.


Explanation:
Okay, so I need to evaluate the ceiling of the square root of 20, and then square that result. Let me break this down step by step. First, let me recall what the ceiling function does. The ceiling of a number is the smallest integer that is greater than or equal to that number. So, if I have a number like 4.3, the ceiling would be 5, right? Because 5 is the next integer higher than 4.3. Similarly, if the number is already an integer, like 5, the ceiling is just 5.

Alright, so the first part is √20. Let me figure out what √20 is approximately. I know that √16 is 4 and √25 is 5. So √20 must be somewhere between 4 and 5. Let me try to get a better approximation. Let me think... 4.4 squared is 19.36, because 4.4 * 4.4. Let me check that. 4 * 4 is 16, 4 * 0.4 is 1.6, 0.4 * 4 is another 1.6, and 0.4 * 0.4 is 0.16. Adding those up: 16 + 1.6 + 1.6 + 0.16 = 16 + 3.2 + 0.16 = 19.36. Yeah, that's right. So 4.4 squared is 19.36.

Now, 4.5 squared is 20.25. Let me confirm that. 4.5 * 4.5. Well, 4 * 4 is 16, 4 * 0.5 is 2, 0.5 * 4 is another 2, and 0.5 * 0.5 is 0.25. Adding those up: 16 + 2 + 2 + 0.25 = 20.25. Correct. So √20 is between 4.4 and 4.5 because 4.4² = 19.36 and 4.5² = 20.25. Since 20 is between 19.36 and 20.25, √20 is between 4.4 and 4.5.

But I need to find the ceiling of √20. The ceiling function rounds up to the nearest integer. Since √20 is between 4.4 and 4.5, which are both between 4 and 5, the ceiling of √20 would be 5, because 5 is the smallest integer greater than √20. Wait, but let me make sure that √20 isn't actually exactly an integer. But 20 isn't a perfect square. The perfect squares around 20 are 16 (4²), 25 (5²). So 20 isn't a perfect square, so √20 is irrational. Therefore, it's not an integer. Therefore, the ceiling of √20 is indeed 5.

So, √20 ≈ 4.472... something. Let me check with a calculator method. Let me try to compute √20 more accurately. Let me use the Newton-Raphson method or something. Let me take an initial guess of 4.47. Let me square that. 4.47 * 4.47. Let me compute 4 * 4.47 = 17.88, 0.47 * 4.47. Wait, maybe another way. Let me do 4.47 squared.

First, 4.4 squared is 19.36. Then, 4.47 is 4.4 + 0.07. So, (a + b)^2 = a² + 2ab + b². So, a = 4.4, b = 0.07. Therefore, (4.4 + 0.07)^2 = 4.4² + 2*4.4*0.07 + 0.07².

Calculating each term:

4.4² = 19.36

2*4.4*0.07 = 8.8 * 0.07 = 0.616

0.07² = 0.0049

Adding them up: 19.36 + 0.616 = 19.976 + 0.0049 = 19.9809.

So, 4.47 squared is approximately 19.9809. That's very close to 20. So, 4.47² ≈ 19.9809. Therefore, √20 is a bit more than 4.47. Let me try 4.472. Let me square that.

Again, using the same method. Let me write 4.472 as 4.47 + 0.002. So, (a + b)^2 where a = 4.47, b = 0.002.

So, a² = 19.9809 (from before), 2ab = 2 * 4.47 * 0.002 = 8.94 * 0.002 = 0.01788, and b² = (0.002)^2 = 0.000004.

Adding them up: 19.9809 + 0.01788 = 19.99878 + 0.000004 ≈ 19.998784.

So, 4.472 squared is approximately 19.998784, which is still less than 20. So, the difference is 20 - 19.998784 = 0.001216.

So, to get a better approximation, let me try 4.472 + x, where x is a small number. Let me use linear approximation. Let me denote f(x) = (4.472 + x)^2. We know that f(0) = 19.998784. We want to find x such that f(x) = 20.

The derivative f’(x) = 2*(4.472 + x). At x = 0, f’(0) = 2*4.472 = 8.944.

So, using linear approximation: f(x) ≈ f(0) + f’(0)*x.

Set this equal to 20:

19.998784 + 8.944*x ≈ 20

So, 8.944*x ≈ 20 - 19.998784 = 0.001216

Therefore, x ≈ 0.001216 / 8.944 ≈ Let me compute that.

Dividing 0.001216 by 8.944. Let me convert this to decimal.

First, note that 8.944 * 0.0001 = 0.0008944.

So, 0.0008944 is 0.0001 * 8.944.

Our numerator is 0.001216. Let me see how many times 0.0008944 fits into 0.001216.

Dividing 0.001216 / 0.0008944 ≈ Let me compute 0.001216 / 0.0008944.

Multiply numerator and denominator by 1,000,000 to eliminate decimals:

1216 / 894.4 ≈ Let me compute that.

Divide numerator and denominator by 16: 1216 / 16 = 76; 894.4 / 16 = 55.9.

So, 76 / 55.9 ≈ 1.36.

Therefore, x ≈ 0.0001 * 1.36 ≈ 0.000136.

Therefore, √20 ≈ 4.472 + 0.000136 ≈ 4.472136.

So, approximately 4.4721. Therefore, √20 ≈ 4.4721.

So, the value of √20 is approximately 4.4721, which is between 4.47 and 4.48. Therefore, the ceiling of √20 is 5, since it's greater than 4.47 and less than 5.

Wait, but let me check if it's actually less than 4.5. Since 4.4721 is less than 4.5, right? Yes, because 4.4721 is 4 and 0.4721, which is less than 0.5. Therefore, the ceiling is 5.

Therefore, the ceiling of √20 is 5.

Therefore, the problem asks for the ceiling of √20 squared, which is (⌈√20⌉)^2 = 5² = 25.

But let me just make sure I didn't make any mistakes here. Let me recap.

First, √20 is between 4 and 5. Then, since it's not an integer, the ceiling is the next integer, which is 5. Then, squaring 5 gives 25.

Alternatively, maybe I can check with another approach. Let me think.

Alternatively, maybe I can find the exact value of the ceiling. Let me think. Since √20 is between 4.47 and 4.48, as we saw, so definitely less than 4.5. Therefore, the ceiling is 5.

Alternatively, if I didn't remember the exact value of √20, I could note that 4.47² ≈ 19.98, which is less than 20, and 4.48² is (4.47 + 0.01)^2 = 4.47² + 2*4.47*0.01 + 0.01² ≈ 19.98 + 0.0894 + 0.0001 ≈ 20.0695. So, 4.48² ≈ 20.0695, which is greater than 20. Therefore, √20 is between 4.47 and 4.48. Therefore, the ceiling is 5.

Therefore, the answer is 5 squared, which is 25.

Wait, but let me check if there's any chance that the ceiling could be 4? No, because √20 is greater than 4. So, the ceiling of a number greater than 4 but less than 5 is 5. So, yes, that's correct.

Alternatively, if I didn't know the approximate value, I could think of the definition of ceiling function. The ceiling of x is the least integer greater than or equal to x. Since √20 is not an integer, the ceiling is the next integer. Since √20 is between 4 and 5, the next integer is 5.

Therefore, regardless of the exact decimal value, as long as √20 is not an integer, which it isn't, the ceiling is 5.

Therefore, squaring that gives 25.

So, I think that's solid.

But just to make sure, let me check with another method. Let me think about the square numbers.

We know that 4² = 16, 5² = 25. Since 20 is between 16 and 25, √20 is between 4 and 5. Therefore, the ceiling is 5.

Therefore, the answer is 5² = 25.

Therefore, I think the answer is 25.

**Final Answer**
\boxed{25}

To evaluate the expression $\lceil{\sqrt{20}}\rceil^2$, we proceed step by step:

---

### Step 1: Understand the Ceiling Function

The **ceiling function**, denoted $\lceil x \rceil$, gives the **smallest integer greater than or equal to** $x$. For example:
- $\lceil 4.2 \rceil = 5$
- $\lceil 5 \rceil = 5$
- $\lceil 4.999 \rceil = 5$

---

### Step 2: Estimate $\sqrt{20}$

We know that:
- $4^2 = 16$
- $5^2 = 25$

So, $\sqrt{20}$ lies between 4 and 5. To narrow it down further:
- $4.4^2 = 19.36$
- $4.5^2 = 20.25$

Since $19.36 < 20 < 20.25$, it follows that:
$$
4.4 < \sqrt{20} < 4.5
$$

This tells us that $\sqrt{20}$ is **not an integer**, and lies strictly between 4 and 5.

---

### Step 3: Apply the Ceiling Function

Since $\sqrt{20}$ is not an integer and lies between 4 and 5, the **smallest integer greater than or equal to** $\sqrt{20}$ is:
$$
\lceil \sqrt{20} \rceil = 5
$$

---

### Step 4: Square the Result

Now, we square the result:
$$
\lceil \sqrt{20} \rceil^2 = 5^2 = 25
$$

---

### Final Answer

$$
\boxed{25}
$$

[Assistant]
Explanation:
Okay, so I need to evaluate the ceiling of the square root of 20, and then square that result. Let me break this down step by step. First, let me recall what the ceiling function does. The ceiling of a number is the smallest integer that is greater than or equal to that number. So, if I have a number like 4.3, the ceiling would be 5, right? Because 5 is the next integer higher than 4.3. Similarly, if the number is already an integer, like 5, the ceiling is just 5.

Alright, so the first part is √20. Let me figure out what √20 is approximately. I know that √16 is 4 and √25 is 5. So √20 must be somewhere between 4 and 5. Let me try to get a better approximation. Let me think... 4.4 squared is 19.36, because 4.4 * 4.4. Let me check that. 4 * 4 is 16, 4 * 0.4 is 1.6, 0.4 * 4 is another 1.6, and 0.4 * 0.4 is 0.16. Adding those up: 16 + 1.6 + 1.6 + 0.16 = 16 + 3.2 + 0.16 = 19.36. Yeah, that's right. So 4.4 squared is 19.36.

Now, 4.5 squared is 20.25. Let me confirm that. 4.5 * 4.5. Well, 4 * 4 is 16, 4 * 0.5 is 2, 0.5 * 4 is another 2, and 0.5 * 0.5 is 0.25. Adding those up: 16 + 2 + 2 + 0.25 = 20.25. Correct. So √20 is between 4.4 and 4.5 because 4.4² = 19.36 and 4.5² = 20.25. Since 20 is between 19.36 and 20.25, √20 is between 4.4 and 4.5.

But I need to find the ceiling of √20. The ceiling function rounds up to the nearest integer. Since √20 is between 4.4 and 4.5, which are both between 4 and 5, the ceiling of √20 would be 5, because 5 is the smallest integer greater than √20. Wait, but let me make sure that √20 isn't actually exactly an integer. But 20 isn't a perfect square. The perfect squares around 20 are 16 (4²), 25 (5²). So 20 isn't a perfect square, so √20 is irrational. Therefore, it's not an integer. Therefore, the ceiling of √20 is indeed 5.

So, √20 ≈ 4.472... something. Let me check with a calculator method. Let me try to compute √20 more accurately. Let me use the Newton-Raphson method or something. Let me take an initial guess of 4.47. Let me square that. 4.47 * 4.47. Let me compute 4 * 4.47 = 17.88, 0.47 * 4.47. Wait, maybe another way. Let me do 4.47 squared.

First, 4.4 squared is 19.36. Then, 4.47 is 4.4 + 0.07. So, (a + b)^2 = a² + 2ab + b². So, a = 4.4, b = 0.07. Therefore, (4.4 + 0.07)^2 = 4.4² + 2*4.4*0.07 + 0.07².

Calculating each term:

4.4² = 19.36

2*4.4*0.07 = 8.8 * 0.07 = 0.616

0.07² = 0.0049

Adding them up: 19.36 + 0.616 = 19.976 + 0.0049 = 19.9809.

So, 4.47 squared is approximately 19.9809. That's very close to 20. So, 4.47² ≈ 19.9809. Therefore, √20 is a bit more than 4.47. Let me try 4.472. Let me square that.

Again, using the same method. Let me write 4.472 as 4.47 + 0.002. So, (a + b)^2 where a = 4.47, b = 0.002.

So, a² = 19.9809 (from before), 2ab = 2 * 4.47 * 0.002 = 8.94 * 0.002 = 0.01788, and b² = (0.002)^2 = 0.000004.

Adding them up: 19.9809 + 0.01788 = 19.99878 + 0.000004 ≈ 19.998784.

So, 4.472 squared is approximately 19.998784, which is still less than 20. So, the difference is 20 - 19.998784 = 0.001216.

So, to get a better approximation, let me try 4.472 + x, where x is a small number. Let me use linear approximation. Let me denote f(x) = (4.472 + x)^2. We know that f(0) = 19.998784. We want to find x such that f(x) = 20.

The derivative f’(x) = 2*(4.472 + x). At x = 0, f’(0) = 2*4.472 = 8.944.

So, using linear approximation: f(x) ≈ f(0) + f’(0)*x.

Set this equal to 20:

19.998784 + 8.944*x ≈ 20

So, 8.944*x ≈ 20 - 19.998784 = 0.001216

Therefore, x ≈ 0.001216 / 8.944 ≈ Let me compute that.

Dividing 0.001216 by 8.944. Let me convert this to decimal.

First, note that 8.944 * 0.0001 = 0.0008944.

So, 0.0008944 is 0.0001 * 8.944.

Our numerator is 0.001216. Let me see how many times 0.0008944 fits into 0.001216.

Dividing 0.001216 / 0.0008944 ≈ Let me compute 0.001216 / 0.0008944.

Multiply numerator and denominator by 1,000,000 to eliminate decimals:

1216 / 894.4 ≈ Let me compute that.

Divide numerator and denominator by 16: 1216 / 16 = 76; 894.4 / 16 = 55.9.

So, 76 / 55.9 ≈ 1.36.

Therefore, x ≈ 0.0001 * 1.36 ≈ 0.000136.

Therefore, √20 ≈ 4.472 + 0.000136 ≈ 4.472136.

So, approximately 4.4721. Therefore, √20 ≈ 4.4721.

So, the value of √20 is approximately 4.4721, which is between 4.47 and 4.48. Therefore, the ceiling of √20 is 5, since it's greater than 4.47 and less than 5.

Wait, but let me check if it's actually less than 4.5. Since 4.4721 is less than 4.5, right? Yes, because 4.4721 is 4 and 0.4721, which is less than 0.5. Therefore, the ceiling is 5.

Therefore, the ceiling of √20 is 5.

Therefore, the problem asks for the ceiling of √20 squared, which is (⌈√20⌉)^2 = 5² = 25.

But let me just make sure I didn't make any mistakes here. Let me recap.

First, √20 is between 4 and 5. Then, since it's not an integer, the ceiling is the next integer, which is 5. Then, squaring 5 gives 25.

Alternatively, maybe I can check with another approach. Let me think.

Alternatively, maybe I can find the exact value of the ceiling. Let me think. Since √20 is between 4.47 and 4.48, as we saw, so definitely less than 4.5. Therefore, the ceiling is 5.

Alternatively, if I didn't remember the exact value of √20, I could note that 4.47² ≈ 19.98, which is less than 20, and 4.48² is (4.47 + 0.01)^2 = 4.47² + 2*4.47*0.01 + 0.01² ≈ 19.98 + 0.0894 + 0.0001 ≈ 20.0695. So, 4.48² ≈ 20.0695, which is greater than 20. Therefore, √20 is between 4.47 and 4.48. Therefore, the ceiling is 5.

Therefore, the answer is 5 squared, which is 25.

Wait, but let me check if there's any chance that the ceiling could be 4? No, because √20 is greater than 4. So, the ceiling of a number greater than 4 but less than 5 is 5. So, yes, that's correct.

Alternatively, if I didn't know the approximate value, I could think of the definition of ceiling function. The ceiling of x is the least integer greater than or equal to x. Since √20 is not an integer, the ceiling is the next integer. Since √20 is between 4 and 5, the next integer is 5.

Therefore, regardless of the exact decimal value, as long as √20 is not an integer, which it isn't, the ceiling is 5.

Therefore, squaring that gives 25.

So, I think that's solid.

But just to make sure, let me check with another method. Let me think about the square numbers.

We know that 4² = 16, 5² = 25. Since 20 is between 16 and 25, √20 is between 4 and 5. Therefore, the ceiling is 5.

Therefore, the answer is 5² = 25.

Therefore, I think the answer is 25.

**Final Answer**
\boxed{25}

To evaluate the expression $\lceil{\sqrt{20}}\rceil^2$, we proceed step by step:

---

### Step 1: Understand the Ceiling Function

The **ceiling function**, denoted $\lceil x \rceil$, gives the **smallest integer greater than or equal to** $x$. For example:
- $\lceil 4.2 \rceil = 5$
- $\lceil 5 \rceil = 5$
- $\lceil 4.999 \rceil = 5$

---

### Step 2: Estimate $\sqrt{20}$

We know that:
- $4^2 = 16$
- $5^2 = 25$

So, $\sqrt{20}$ lies between 4 and 5. To narrow it down further:
- $4.4^2 = 19.36$
- $4.5^2 = 20.25$

Since $19.36 < 20 < 20.25$, it follows that:
$$
4.4 < \sqrt{20} < 4.5
$$

This tells us that $\sqrt{20}$ is **not an integer**, and lies strictly between 4 and 5.

---

### Step 3: Apply the Ceiling Function

Since $\sqrt{20}$ is not an integer and lies between 4 and 5, the **smallest integer greater than or equal to** $\sqrt{20}$ is:
$$
\lceil \sqrt{20} \rceil = 5
$$

---

### Step 4: Square the Result

Now, we square the result:
$$
\lceil \sqrt{20} \rceil^2 = 5^2 = 25
$$

---

### Final Answer

$$
\boxed{25}
$$

Solution: Since $\sqrt{16}<\sqrt{20}<\sqrt{25}$, or, equivalently, $4<\sqrt{20}<5$, the smallest integer greater than or equal to $\sqrt{20}$ must be $5$. Thus, $\lceil{\sqrt{20}}\rceil^2=5^2=\boxed{25}$.


Explanation:
Okay, so I need to find the class average for this test. Let me think about how to approach this. The class has 30 students total. Some scored 80, some 90, and some 100. The average is the sum of all the scores divided by the number of students, right? So I need to figure out the total points scored by all the students and then divide that by 30.

Let me break it down. First, there are 20 students who scored 80. So if I multiply 20 by 80, that should give me the total points from those students. Let me do that calculation: 20 times 80. Hmm, 20 times 80 is 1600. Okay, so 1600 points from the first group.

Next, there are 8 students who scored 90. So I need to calculate 8 times 90. Let me do that. 8 times 90... Well, 8 times 9 is 72, so adding a zero makes it 720. So that's 720 points from the second group.

Then there are 2 students who scored 100. Multiplying 2 by 100 is straightforward. That's 200 points.

Now, to get the total points for the whole class, I need to add up those three amounts: 1600, 720, and 200. Let me add them step by step. First, 1600 plus 720. Let me think: 1600 + 700 is 2300, and then plus 20 more is 2320. Then add the 200. So 2320 + 200 is 2520. So the total points scored by all students is 2520.

Now, the average is total points divided by number of students. There are 30 students. So I need to divide 2520 by 30. Let me do that division. Hmm, 2520 divided by 30. Well, 30 times 80 is 2400. So subtract 2400 from 2520, which leaves 120. Then 30 goes into 120 exactly 4 times. So 80 + 4 is 84. Therefore, the average is 84.

Wait, let me check that again to make sure I didn't make a mistake. So total points: 20*80=1600, 8*90=720, 2*100=200. Adding them up: 1600+720 is 2320, plus 200 is 2520. Divided by 30. 2520 divided by 30. Let me think of another way. Dividing both numerator and denominator by 10 first: 2520/10=252, 30/10=3. So now it's 252 divided by 3. 3 goes into 252 how many times? 3*80=240, so 252-240=12, and 3 goes into 12 four times. So 80+4=84. Yep, same result. So that seems right.

Alternatively, maybe I can check with another method. Let me think. If all 30 students scored 80, the total would be 30*80=2400. But some students scored higher. The 8 students who scored 90 are 10 points above 80, so that's 8*10=80 extra points. The 2 students who scored 100 are 20 points above 80, so that's 2*20=40 extra points. So total extra points are 80+40=120. Therefore, total points would be 2400 + 120 = 2520, which matches what I had before. Then average is 2520/30=84. Yep, same answer. So that seems to confirm it.

Alternatively, maybe I can think of the average as a weighted average. The average would be (20/30)*80 + (8/30)*90 + (2/30)*100. Let me compute that.

First, 20/30 is 2/3. 2/3 of 80 is (2*80)/3 = 160/3 ≈ 53.333...

Then 8/30 is 4/15. 4/15 of 90 is (4*90)/15 = 360/15 = 24.

Then 2/30 is 1/15. 1/15 of 100 is 100/15 ≈ 6.666...

Adding those together: 53.333... + 24 + 6.666... Let me add 53.333 and 6.666 first. That's 60. Then 60 +24 is 84. So that also gives 84. So that's another way to get the same answer.

So all methods lead to 84. Therefore, I think the class average is 84.

But just to make sure I didn't make any arithmetic errors. Let me verify once again:

Total points:

20 students * 80 = 1600

8 students * 90 = 720

2 students * 100 = 200

Total: 1600 + 720 = 2320; 2320 + 200 = 2520. Correct.

Number of students: 20 + 8 + 2 = 30. Correct.

Average: 2520 / 30. Let me divide 2520 by 30:

Divide numerator and denominator by 10: 252 / 3 = 84. Correct.

Yes, that's straightforward. So I think that's solid.

Alternatively, if I use decimal division: 2520 divided by 30. 30 goes into 2520 how many times? Well, 30*80=2400, as before. 2520-2400=120. 30*4=120. So total is 84. Yep.

So all methods confirm that the average is 84. Therefore, I can be confident that the answer is 84.

**Final Answer**
The class average on the test was \boxed{84}.

To find the **class average** (mean) on the test, we need to calculate the **total score** of all students and then divide it by the **total number of students**.

---

### Step 1: Calculate the total score

We are given the following:

- 20 students scored **80**
- 8 students scored **90**
- 2 students scored **100**

We compute the total score by multiplying the number of students by their respective scores:

$$
\text{Total score} = (20 \times 80) + (8 \times 90) + (2 \times 100)
$$

$$
= 1600 + 720 + 200 = 2520
$$

---

### Step 2: Calculate the class average

The class has **30 students**, so the average is:

$$
\text{Average} = \frac{\text{Total score}}{\text{Number of students}} = \frac{2520}{30}
$$

$$
= 84
$$

---

### Final Answer

$$
\boxed{84}
$$

[Assistant]
Explanation:
Okay, so I need to find the class average for this test. Let me think about how to approach this. The class has 30 students total. Some scored 80, some 90, and some 100. The average is the sum of all the scores divided by the number of students, right? So I need to figure out the total points scored by all the students and then divide that by 30.

Let me break it down. First, there are 20 students who scored 80. So if I multiply 20 by 80, that should give me the total points from those students. Let me do that calculation: 20 times 80. Hmm, 20 times 80 is 1600. Okay, so 1600 points from the first group.

Next, there are 8 students who scored 90. So I need to calculate 8 times 90. Let me do that. 8 times 90... Well, 8 times 9 is 72, so adding a zero makes it 720. So that's 720 points from the second group.

Then there are 2 students who scored 100. Multiplying 2 by 100 is straightforward. That's 200 points.

Now, to get the total points for the whole class, I need to add up those three amounts: 1600, 720, and 200. Let me add them step by step. First, 1600 plus 720. Let me think: 1600 + 700 is 2300, and then plus 20 more is 2320. Then add the 200. So 2320 + 200 is 2520. So the total points scored by all students is 2520.

Now, the average is total points divided by number of students. There are 30 students. So I need to divide 2520 by 30. Let me do that division. Hmm, 2520 divided by 30. Well, 30 times 80 is 2400. So subtract 2400 from 2520, which leaves 120. Then 30 goes into 120 exactly 4 times. So 80 + 4 is 84. Therefore, the average is 84.

Wait, let me check that again to make sure I didn't make a mistake. So total points: 20*80=1600, 8*90=720, 2*100=200. Adding them up: 1600+720 is 2320, plus 200 is 2520. Divided by 30. 2520 divided by 30. Let me think of another way. Dividing both numerator and denominator by 10 first: 2520/10=252, 30/10=3. So now it's 252 divided by 3. 3 goes into 252 how many times? 3*80=240, so 252-240=12, and 3 goes into 12 four times. So 80+4=84. Yep, same result. So that seems right.

Alternatively, maybe I can check with another method. Let me think. If all 30 students scored 80, the total would be 30*80=2400. But some students scored higher. The 8 students who scored 90 are 10 points above 80, so that's 8*10=80 extra points. The 2 students who scored 100 are 20 points above 80, so that's 2*20=40 extra points. So total extra points are 80+40=120. Therefore, total points would be 2400 + 120 = 2520, which matches what I had before. Then average is 2520/30=84. Yep, same answer. So that seems to confirm it.

Alternatively, maybe I can think of the average as a weighted average. The average would be (20/30)*80 + (8/30)*90 + (2/30)*100. Let me compute that.

First, 20/30 is 2/3. 2/3 of 80 is (2*80)/3 = 160/3 ≈ 53.333...

Then 8/30 is 4/15. 4/15 of 90 is (4*90)/15 = 360/15 = 24.

Then 2/30 is 1/15. 1/15 of 100 is 100/15 ≈ 6.666...

Adding those together: 53.333... + 24 + 6.666... Let me add 53.333 and 6.666 first. That's 60. Then 60 +24 is 84. So that also gives 84. So that's another way to get the same answer.

So all methods lead to 84. Therefore, I think the class average is 84.

But just to make sure I didn't make any arithmetic errors. Let me verify once again:

Total points:

20 students * 80 = 1600

8 students * 90 = 720

2 students * 100 = 200

Total: 1600 + 720 = 2320; 2320 + 200 = 2520. Correct.

Number of students: 20 + 8 + 2 = 30. Correct.

Average: 2520 / 30. Let me divide 2520 by 30:

Divide numerator and denominator by 10: 252 / 3 = 84. Correct.

Yes, that's straightforward. So I think that's solid.

Alternatively, if I use decimal division: 2520 divided by 30. 30 goes into 2520 how many times? Well, 30*80=2400, as before. 2520-2400=120. 30*4=120. So total is 84. Yep.

So all methods confirm that the average is 84. Therefore, I can be confident that the answer is 84.

**Final Answer**
The class average on the test was \boxed{84}.

To find the **class average** (mean) on the test, we need to calculate the **total score** of all students and then divide it by the **total number of students**.

---

### Step 1: Calculate the total score

We are given the following:

- 20 students scored **80**
- 8 students scored **90**
- 2 students scored **100**

We compute the total score by multiplying the number of students by their respective scores:

$$
\text{Total score} = (20 \times 80) + (8 \times 90) + (2 \times 100)
$$

$$
= 1600 + 720 + 200 = 2520
$$

---

### Step 2: Calculate the class average

The class has **30 students**, so the average is:

$$
\text{Average} = \frac{\text{Total score}}{\text{Number of students}} = \frac{2520}{30}
$$

$$
= 84
$$

---

### Final Answer

$$
\boxed{84}
$$

Solution: From the given information, the total amount of marks obtained by the class is   $$20(80)+8(90)+2(100)=2520.$$Therefore, the class average is  $$\frac{2520}{30} = \boxed{84}.$$