Explanation:
Okay, so I need to solve this problem about a two-digit number. Let me read it again:

"The sum of the digits of a two-digit number is 13. The difference between the number and the number with its digits reversed is 27. What is the sum of the original number and the number with its digits reversed?"

Alright, let me break this down. First, there's a two-digit number. Let me think about how to represent that. Usually, for two-digit numbers, we can let the tens digit be one variable and the units digit be another. Let me denote the original number as having a tens digit of 'a' and a units digit of 'b'. So, the original number would be 10a + b. Then, when we reverse the digits, the number becomes 10b + a.

The problem says the sum of the digits is 13. So, that would be a + b = 13. Got that. Then, the difference between the original number and the reversed number is 27. So, that would be (original number) - (reversed number) = 27. Let me write that as an equation: (10a + b) - (10b + a) = 27. Let me simplify that equation.

First, expanding the left side: 10a + b - 10b - a. Combining like terms: (10a - a) + (b - 10b) = 9a - 9b. So, 9a - 9b = 27. I can factor out a 9: 9(a - b) = 27. Dividing both sides by 9: a - b = 3. So, that gives me another equation: a - b = 3.

So now I have two equations:

1. a + b = 13
2. a - b = 3

I can solve this system of equations. Let me add the two equations together. If I add equation 1 and equation 2:

(a + b) + (a - b) = 13 + 3

Simplify left side: a + b + a - b = 2a

Right side: 16

So, 2a = 16 => a = 8.

Then, substitute back into equation 1: 8 + b = 13 => b = 13 - 8 = 5.

So, the original number has a tens digit of 8 and a units digit of 5. Therefore, the original number is 10*8 + 5 = 85. The reversed number would be 10*5 + 8 = 58.

Now, the problem asks for the sum of the original number and the reversed number. So, 85 + 58. Let me compute that. 85 + 58. Let me do 80 + 50 = 130, and 5 + 8 = 13, so total is 130 + 13 = 143. So, the answer should be 143.

Wait, let me check if everything makes sense. Let me verify the original conditions.

First, the sum of the digits: 8 + 5 = 13. That's correct. Then, the difference between the original number and reversed number: 85 - 58 = 27. Let me check that. 85 - 58. 85 - 50 = 35, then subtract 8 more: 35 - 8 = 27. Yes, that's correct. So, the difference is indeed 27. Then, the sum is 85 + 58 = 143. That seems right.

But let me just make sure I didn't make any mistakes in solving the equations. Let me go through the equations again.

Original equations:

a + b = 13

a - b = 3

Adding them gives 2a = 16 => a = 8. Then b = 13 - 8 = 5. Correct. So, digits are 8 and 5. Original number is 85, reversed is 58. Sum is 143. Seems good.

Alternatively, maybe there's another way to approach this problem? Let me think. Suppose I didn't use variables. Let me try to think of two-digit numbers where digits add up to 13. Let me list possible two-digit numbers where digits add to 13.

Possible pairs (a, b) where a + b = 13. Since a and b are digits, a can be from 1 to 9 (since it's a two-digit number, the tens digit can't be 0), and b from 0 to 9.

So, possible pairs:

a=4, b=9 (sum 13)

a=5, b=8

a=6, b=7

a=7, b=6

a=8, b=5

a=9, b=4

So, the numbers are 49, 58, 67, 76, 85, 94.

Now, check which of these numbers have a difference of 27 when subtracted by their reversed numbers.

Let me check each one:

First, 49 reversed is 94. 49 - 94 = -45. The absolute difference is 45, which is not 27.

Next, 58 reversed is 85. 58 - 85 = -27. The difference is 27. So, that's the one. So, original number is 58? Wait, but earlier I found 85. Wait, but here, if original number is 58, then reversed is 85, and 58 - 85 = -27. But the problem says "the difference between the number and the number with its digits reversed is 27." So, difference is 27. Depending on which way you subtract, it could be positive or negative. But since difference is mentioned as 27, maybe they mean absolute difference? Or maybe they specify which one is larger.

Wait, the problem says "the difference between the number and the number with its digits reversed is 27." So, if the original number is larger, then original - reversed = 27. If reversed is larger, then reversed - original = 27. But in the problem statement, they might be referring to the absolute difference, but maybe not. Let me check.

In my previous calculation, when I set up the equation (original - reversed) = 27, I got a solution of original number 85, reversed 58, so 85 - 58 = 27. So that works. However, if the original number was 58, then reversed is 85, and 58 - 85 = -27. But if the problem says the difference is 27, then maybe they take absolute value? But in that case, both 85 and 58 would satisfy the condition. However, in my equations, I assumed original - reversed = 27, which gives 85 - 58 = 27. But if the problem says "the difference between the number and the reversed number is 27," then depending on which is larger, it could be either way. However, since the problem says "the difference between the number and the reversed number is 27," and doesn't specify which is larger, but in my equations, I assumed original - reversed = 27. However, if the original number is smaller, then reversed - original = 27. So, maybe there are two possibilities? Wait, but in the list above, when I checked 58, reversed is 85, and 85 - 58 = 27. So, if the original number is 58, then reversed - original = 27. So, depending on which one is considered the original number. Wait, but the problem says "the difference between the number and the number with its digits reversed is 27." So, if the original number is 58, then the difference between 58 and 85 is 27. But if the original number is 85, then the difference between 85 and 58 is 27. So, both numbers satisfy the condition. However, in my equations, I assumed original - reversed = 27, which gives me original number as 85. But if I had set reversed - original = 27, then I would have gotten original number as 58. However, since the problem says "the difference between the number and the number with its digits reversed is 27," it's possible that they mean the absolute difference, so both numbers are valid. However, in that case, the sum of the original number and reversed number would be the same in both cases. Let me check.

If original number is 85, reversed is 58, sum is 85 + 58 = 143.

If original number is 58, reversed is 85, sum is 58 + 85 = 143. Same result. So regardless, the answer is 143. Therefore, even if there are two numbers, their sum is same. Therefore, the answer is 143.

But let me check if there are other numbers in the list that could satisfy the difference of 27. Let me check the rest.

Original number 67, reversed is 76. 67 - 76 = -9. Absolute difference 9. Not 27.

Original number 76, reversed is 67. 76 - 67 = 9. Not 27.

Original number 94, reversed is 49. 94 - 49 = 45. Not 27.

Original number 49, reversed is 94. 49 - 94 = -45. Absolute difference 45. Not 27.

So only 58 and 85 satisfy the difference of 27. Therefore, the answer is 143. So, seems correct.

Alternatively, maybe I can solve it using algebra again, just to confirm.

We had:

a + b = 13

and

original - reversed = 27 => 10a + b - (10b + a) = 27 => 9a - 9b = 27 => a - b = 3.

So solving:

a + b = 13

a - b = 3

Adding equations:

2a = 16 => a = 8, then b = 5. So original number is 85, reversed is 58. Sum is 143. Correct.

Alternatively, if I had set reversed - original = 27, then:

(10b + a) - (10a + b) = 27 => 9b - 9a = 27 => b - a = 3. Then, with a + b = 13, so:

a + b = 13

b - a = 3

Adding equations:

2b = 16 => b = 8, then a = 13 - 8 = 5. So original number is 58, reversed is 85. Sum is same 58 + 85 = 143. So same answer.

Therefore, regardless of which way the difference is taken, the sum is same. Therefore, answer is 143.

Therefore, I think the answer is 143. Let me check once again.

Original number: 85. Sum of digits: 8 + 5 = 13. Correct. Difference between original and reversed: 85 - 58 = 27. Correct. Sum of original and reversed: 85 + 58 = 143. Correct.

Alternatively, if I think of the two numbers as x and y, where x + y = ?

But since x and y are original and reversed numbers, and we know that x - y = 27 or y - x = 27. But regardless, x + y can be found if we know x - y and x + y. Wait, but we can also find x + y from the equations.

Wait, let me think. If I have x = original number, y = reversed number.

We know that x - y = 27 or y - x = 27. Also, we know that x + y = ?

But we also know that the sum of digits is 13. Let me think if there's another way to find x + y.

Alternatively, since x = 10a + b, y = 10b + a. Then x + y = 11a + 11b = 11(a + b). Since a + b = 13, then x + y = 11 * 13 = 143. Oh! That's a quicker way. So, regardless of the values of a and b, as long as their sum is 13, the sum of the original number and reversed number is 11*(a + b) = 11*13 = 143. Therefore, the answer is 143. That's a much more straightforward way. So, even without solving for a and b, we can find the sum directly. That's clever.

So, the key insight here is that x + y = 11(a + b). Since a + b is given as 13, then x + y = 11*13 = 143. Therefore, the answer is 143. That's a good way to think about it. So, even if I didn't solve for a and b, I could have found the answer directly. But since I did solve for a and b, and verified, it's good.

Therefore, I think the answer is definitely 143. I don't see any mistakes in my reasoning. Let me just check once again with the numbers.

Original number: 85, reversed: 58. 85 + 58 = 143. Correct. Sum of digits: 8 + 5 = 13. Correct. Difference: 85 - 58 = 27. Correct. All conditions satisfied. So, answer is 143.

**Final Answer**
\boxed{143}

To solve the problem, we begin by representing the two-digit number algebraically. Let the tens digit be $ a $ and the units digit be $ b $. Then the original number can be written as:

$$
10a + b
$$

When the digits are reversed, the number becomes:

$$
10b + a
$$

---

### Step 1: Use the sum of digits condition

We are told that the sum of the digits is 13:

$$
a + b = 13 \tag{1}
$$

---

### Step 2: Use the difference condition

We are also told that the difference between the original number and the reversed number is 27:

$$
(10a + b) - (10b + a) = 27
$$

Simplify the left-hand side:

$$
10a + b - 10b - a = 9a - 9b = 27
$$

Factor out 9:

$$
9(a - b) = 27 \Rightarrow a - b = 3 \tag{2}
$$

---

### Step 3: Solve the system of equations

We now solve the system:

$$
\begin{cases}
a + b = 13 \\
a - b = 3
\end{cases}
$$

Add the two equations:

$$
2a = 16 \Rightarrow a = 8
$$

Substitute $ a = 8 $ into equation (1):

$$
8 + b = 13 \Rightarrow b = 5
$$

So, the original number is:

$$
10a + b = 10(8) + 5 = 85
$$

The reversed number is:

$$
10b + a = 10(5) + 8 = 58
$$

---

### Step 4: Find the sum of the original and reversed numbers

$$
85 + 58 = 143
$$

---

### Step 5: Alternative approach (more efficient)

We can also observe that the sum of the original number and the reversed number is:

$$
(10a + b) + (10b + a) = 11a + 11b = 11(a + b)
$$

Since $ a + b = 13 $, we have:

$$
11 \times 13 = 143
$$

This confirms the result without solving for $ a $ and $ b $ individually.

---

### Final Answer

$$
\boxed{143}
$$

[Assistant]
Explanation:
Okay, so I need to solve this problem about a two-digit number. Let me read it again:

"The sum of the digits of a two-digit number is 13. The difference between the number and the number with its digits reversed is 27. What is the sum of the original number and the number with its digits reversed?"

Alright, let me break this down. First, there's a two-digit number. Let me think about how to represent that. Usually, for two-digit numbers, we can let the tens digit be one variable and the units digit be another. Let me denote the original number as having a tens digit of 'a' and a units digit of 'b'. So, the original number would be 10a + b. Then, when we reverse the digits, the number becomes 10b + a.

The problem says the sum of the digits is 13. So, that would be a + b = 13. Got that. Then, the difference between the original number and the reversed number is 27. So, that would be (original number) - (reversed number) = 27. Let me write that as an equation: (10a + b) - (10b + a) = 27. Let me simplify that equation.

First, expanding the left side: 10a + b - 10b - a. Combining like terms: (10a - a) + (b - 10b) = 9a - 9b. So, 9a - 9b = 27. I can factor out a 9: 9(a - b) = 27. Dividing both sides by 9: a - b = 3. So, that gives me another equation: a - b = 3.

So now I have two equations:

1. a + b = 13
2. a - b = 3

I can solve this system of equations. Let me add the two equations together. If I add equation 1 and equation 2:

(a + b) + (a - b) = 13 + 3

Simplify left side: a + b + a - b = 2a

Right side: 16

So, 2a = 16 => a = 8.

Then, substitute back into equation 1: 8 + b = 13 => b = 13 - 8 = 5.

So, the original number has a tens digit of 8 and a units digit of 5. Therefore, the original number is 10*8 + 5 = 85. The reversed number would be 10*5 + 8 = 58.

Now, the problem asks for the sum of the original number and the reversed number. So, 85 + 58. Let me compute that. 85 + 58. Let me do 80 + 50 = 130, and 5 + 8 = 13, so total is 130 + 13 = 143. So, the answer should be 143.

Wait, let me check if everything makes sense. Let me verify the original conditions.

First, the sum of the digits: 8 + 5 = 13. That's correct. Then, the difference between the original number and reversed number: 85 - 58 = 27. Let me check that. 85 - 58. 85 - 50 = 35, then subtract 8 more: 35 - 8 = 27. Yes, that's correct. So, the difference is indeed 27. Then, the sum is 85 + 58 = 143. That seems right.

But let me just make sure I didn't make any mistakes in solving the equations. Let me go through the equations again.

Original equations:

a + b = 13

a - b = 3

Adding them gives 2a = 16 => a = 8. Then b = 13 - 8 = 5. Correct. So, digits are 8 and 5. Original number is 85, reversed is 58. Sum is 143. Seems good.

Alternatively, maybe there's another way to approach this problem? Let me think. Suppose I didn't use variables. Let me try to think of two-digit numbers where digits add up to 13. Let me list possible two-digit numbers where digits add to 13.

Possible pairs (a, b) where a + b = 13. Since a and b are digits, a can be from 1 to 9 (since it's a two-digit number, the tens digit can't be 0), and b from 0 to 9.

So, possible pairs:

a=4, b=9 (sum 13)

a=5, b=8

a=6, b=7

a=7, b=6

a=8, b=5

a=9, b=4

So, the numbers are 49, 58, 67, 76, 85, 94.

Now, check which of these numbers have a difference of 27 when subtracted by their reversed numbers.

Let me check each one:

First, 49 reversed is 94. 49 - 94 = -45. The absolute difference is 45, which is not 27.

Next, 58 reversed is 85. 58 - 85 = -27. The difference is 27. So, that's the one. So, original number is 58? Wait, but earlier I found 85. Wait, but here, if original number is 58, then reversed is 85, and 58 - 85 = -27. But the problem says "the difference between the number and the number with its digits reversed is 27." So, difference is 27. Depending on which way you subtract, it could be positive or negative. But since difference is mentioned as 27, maybe they mean absolute difference? Or maybe they specify which one is larger.

Wait, the problem says "the difference between the number and the number with its digits reversed is 27." So, if the original number is larger, then original - reversed = 27. If reversed is larger, then reversed - original = 27. But in the problem statement, they might be referring to the absolute difference, but maybe not. Let me check.

In my previous calculation, when I set up the equation (original - reversed) = 27, I got a solution of original number 85, reversed 58, so 85 - 58 = 27. So that works. However, if the original number was 58, then reversed is 85, and 58 - 85 = -27. But if the problem says the difference is 27, then maybe they take absolute value? But in that case, both 85 and 58 would satisfy the condition. However, in my equations, I assumed original - reversed = 27, which gives 85 - 58 = 27. But if the problem says "the difference between the number and the reversed number is 27," then depending on which is larger, it could be either way. However, since the problem says "the difference between the number and the reversed number is 27," and doesn't specify which is larger, but in my equations, I assumed original - reversed = 27. However, if the original number is smaller, then reversed - original = 27. So, maybe there are two possibilities? Wait, but in the list above, when I checked 58, reversed is 85, and 85 - 58 = 27. So, if the original number is 58, then reversed - original = 27. So, depending on which one is considered the original number. Wait, but the problem says "the difference between the number and the number with its digits reversed is 27." So, if the original number is 58, then the difference between 58 and 85 is 27. But if the original number is 85, then the difference between 85 and 58 is 27. So, both numbers satisfy the condition. However, in my equations, I assumed original - reversed = 27, which gives me original number as 85. But if I had set reversed - original = 27, then I would have gotten original number as 58. However, since the problem says "the difference between the number and the number with its digits reversed is 27," it's possible that they mean the absolute difference, so both numbers are valid. However, in that case, the sum of the original number and reversed number would be the same in both cases. Let me check.

If original number is 85, reversed is 58, sum is 85 + 58 = 143.

If original number is 58, reversed is 85, sum is 58 + 85 = 143. Same result. So regardless, the answer is 143. Therefore, even if there are two numbers, their sum is same. Therefore, the answer is 143.

But let me check if there are other numbers in the list that could satisfy the difference of 27. Let me check the rest.

Original number 67, reversed is 76. 67 - 76 = -9. Absolute difference 9. Not 27.

Original number 76, reversed is 67. 76 - 67 = 9. Not 27.

Original number 94, reversed is 49. 94 - 49 = 45. Not 27.

Original number 49, reversed is 94. 49 - 94 = -45. Absolute difference 45. Not 27.

So only 58 and 85 satisfy the difference of 27. Therefore, the answer is 143. So, seems correct.

Alternatively, maybe I can solve it using algebra again, just to confirm.

We had:

a + b = 13

and

original - reversed = 27 => 10a + b - (10b + a) = 27 => 9a - 9b = 27 => a - b = 3.

So solving:

a + b = 13

a - b = 3

Adding equations:

2a = 16 => a = 8, then b = 5. So original number is 85, reversed is 58. Sum is 143. Correct.

Alternatively, if I had set reversed - original = 27, then:

(10b + a) - (10a + b) = 27 => 9b - 9a = 27 => b - a = 3. Then, with a + b = 13, so:

a + b = 13

b - a = 3

Adding equations:

2b = 16 => b = 8, then a = 13 - 8 = 5. So original number is 58, reversed is 85. Sum is same 58 + 85 = 143. So same answer.

Therefore, regardless of which way the difference is taken, the sum is same. Therefore, answer is 143.

Therefore, I think the answer is 143. Let me check once again.

Original number: 85. Sum of digits: 8 + 5 = 13. Correct. Difference between original and reversed: 85 - 58 = 27. Correct. Sum of original and reversed: 85 + 58 = 143. Correct.

Alternatively, if I think of the two numbers as x and y, where x + y = ?

But since x and y are original and reversed numbers, and we know that x - y = 27 or y - x = 27. But regardless, x + y can be found if we know x - y and x + y. Wait, but we can also find x + y from the equations.

Wait, let me think. If I have x = original number, y = reversed number.

We know that x - y = 27 or y - x = 27. Also, we know that x + y = ?

But we also know that the sum of digits is 13. Let me think if there's another way to find x + y.

Alternatively, since x = 10a + b, y = 10b + a. Then x + y = 11a + 11b = 11(a + b). Since a + b = 13, then x + y = 11 * 13 = 143. Oh! That's a quicker way. So, regardless of the values of a and b, as long as their sum is 13, the sum of the original number and reversed number is 11*(a + b) = 11*13 = 143. Therefore, the answer is 143. That's a much more straightforward way. So, even without solving for a and b, we can find the sum directly. That's clever.

So, the key insight here is that x + y = 11(a + b). Since a + b is given as 13, then x + y = 11*13 = 143. Therefore, the answer is 143. That's a good way to think about it. So, even if I didn't solve for a and b, I could have found the answer directly. But since I did solve for a and b, and verified, it's good.

Therefore, I think the answer is definitely 143. I don't see any mistakes in my reasoning. Let me just check once again with the numbers.

Original number: 85, reversed: 58. 85 + 58 = 143. Correct. Sum of digits: 8 + 5 = 13. Correct. Difference: 85 - 58 = 27. Correct. All conditions satisfied. So, answer is 143.

**Final Answer**
\boxed{143}

To solve the problem, we begin by representing the two-digit number algebraically. Let the tens digit be $ a $ and the units digit be $ b $. Then the original number can be written as:

$$
10a + b
$$

When the digits are reversed, the number becomes:

$$
10b + a
$$

---

### Step 1: Use the sum of digits condition

We are told that the sum of the digits is 13:

$$
a + b = 13 \tag{1}
$$

---

### Step 2: Use the difference condition

We are also told that the difference between the original number and the reversed number is 27:

$$
(10a + b) - (10b + a) = 27
$$

Simplify the left-hand side:

$$
10a + b - 10b - a = 9a - 9b = 27
$$

Factor out 9:

$$
9(a - b) = 27 \Rightarrow a - b = 3 \tag{2}
$$

---

### Step 3: Solve the system of equations

We now solve the system:

$$
\begin{cases}
a + b = 13 \\
a - b = 3
\end{cases}
$$

Add the two equations:

$$
2a = 16 \Rightarrow a = 8
$$

Substitute $ a = 8 $ into equation (1):

$$
8 + b = 13 \Rightarrow b = 5
$$

So, the original number is:

$$
10a + b = 10(8) + 5 = 85
$$

The reversed number is:

$$
10b + a = 10(5) + 8 = 58
$$

---

### Step 4: Find the sum of the original and reversed numbers

$$
85 + 58 = 143
$$

---

### Step 5: Alternative approach (more efficient)

We can also observe that the sum of the original number and the reversed number is:

$$
(10a + b) + (10b + a) = 11a + 11b = 11(a + b)
$$

Since $ a + b = 13 $, we have:

$$
11 \times 13 = 143
$$

This confirms the result without solving for $ a $ and $ b $ individually.

---

### Final Answer

$$
\boxed{143}
$$

Solution:
The two digit number can be represented as $10x + y,$ where $x$ and $y$ are digits, with $x \neq 0.$ We are given that the sum of the digits is $13,$ so $x + y = 13.$ If we reverse the digits of this number, we have $10y + x.$ We are given that the difference is $27,$ but we don't know if the original number or if the number with its digits reversed is greater. We can show this as such: $$|(10x + y) - (10y + x)| = 27.$$ However, it doesn't matter which of the two numbers is greater, since we wish to find their sum. So, without loss of generality, we will let the first number be the larger of the two. This means that $x > y,$ so we can get rid of the absolute values in our last equation to obtain $9x - 9y = 27,$ equivalent to $x - y = 3.$

We now have two equations in two variables: $x + y = 13$ and $x - y = 3.$ Adding the two, we obtain $2x = 16,$ so $x = 8.$ Subtracting, we obtain $2y = 10,$ so $y = 5.$ Thus, the original number is $85,$ and our answer is $85 + 58 = \boxed{143}.$

OR

As before, the two digit number can be expressed as $10x + y,$ and the number with its digits reversed is $10y + x.$ We want to find the sum of these two numbers, which is $$(10x + y) + (10y + x) = 11x + 11y = 11(x + y).$$ We are given that the sum of the digits is $13,$ so $x + y = 13.$ Since all we want is $11(x + y),$ we can substitute for $x + y$ to obtain our answer of $11\cdot 13 = \boxed{143}.$


Explanation:
Okay, so I need to figure out how many representatives the 8th grade should have on the Student Council, given that there are 7th and 8th grades with enrollments of 520 and 650 respectively, and the total number of representatives is 18. The goal is to have a fair representation. Hmm, fair representation... I think that means the number of representatives each grade gets should be proportional to their enrollment. Let me check if that makes sense.

First, let me recall what proportional representation means. If something is proportional, then the ratio of the representatives should be the same as the ratio of their enrollments. So, if the 7th grade has 520 students and the 8th has 650, then the ratio of their enrollments is 520:650. Maybe I can simplify that ratio first?

Let me divide both numbers by 130. 520 divided by 130 is 4, and 650 divided by 130 is 5. So the simplified ratio is 4:5. That means for every 4 students in 7th grade, there are 5 students in 8th grade. Therefore, the ratio of representatives should also be 4:5.

But wait, the total number of representatives is 18. So if the ratio is 4:5, then the total parts are 4 + 5 = 9 parts. So each part is equal to 18 divided by 9, which is 2. Therefore, the 7th grade should have 4 parts times 2, which is 8 representatives, and the 8th grade should have 5 parts times 2, which is 10 representatives.

Let me check if that adds up. 8 + 10 is 18, which matches the total number of representatives. So that seems to work. But let me make sure I didn't skip any steps or make any wrong assumptions.

Alternatively, maybe I should approach it using fractions. The total enrollment is 520 + 650. Let me calculate that. 520 + 650 is 1170 students total. So the 7th grade is 520/1170 of the total enrollment, and the 8th grade is 650/1170. Then, the number of representatives should be proportional to these fractions.

So, for the 7th grade, the number of representatives would be (520/1170)*18, and for the 8th grade, it would be (650/1170)*18. Let me compute these.

First, simplifying 520/1170. Let me divide numerator and denominator by 130 again. 520 ÷ 130 = 4, 1170 ÷ 130 = 9. So that's 4/9. Therefore, 4/9 of 18 is (4*18)/9 = (72)/9 = 8. Similarly, 650/1170 simplifies to 5/9, so 5/9 of 18 is (5*18)/9 = 90/9 = 10. So that gives the same result as before: 8 and 10.

Therefore, the 8th grade should have 10 representatives.

But let me think again if there's another way to approach this. Maybe using the concept of proportionality. If the number of representatives is proportional to the number of students, then the ratio of representatives should be the same as the ratio of students.

So, let me denote the number of representatives for 7th grade as R7 and for 8th grade as R8. Then, R7/R8 = 520/650. Simplifying that ratio, as before, 520/650 = 4/5. Therefore, R7/R8 = 4/5. Also, R7 + R8 = 18.

So, if R7/R8 = 4/5, then R7 = (4/5) R8. Substituting into the total: (4/5) R8 + R8 = 18. Combining terms: (4/5 + 5/5) R8 = 18 => (9/5) R8 = 18. Therefore, R8 = 18 * (5/9) = 10.

Same answer again. So that seems consistent.

Alternatively, maybe I can think in terms of weighted averages or something else? Let me see. But I think the methods above are sufficient.

Wait, but maybe there's a catch here. For example, sometimes when you have to assign whole numbers of representatives, you might have to round, but in this case, the numbers worked out evenly. 520 and 650 are multiples of 130, and 18 is a multiple of 9, so the division comes out even.

But just to make sure, let me check if there's another interpretation of "fair representation." Sometimes, "fair representation" can refer to something else, like equal representation per student, but that would mean each student gets one representative, which is not feasible here. But since there are only 18 representatives, it's about proportionality.

Alternatively, maybe using some other method like the Hamilton method or Jefferson method for apportionment? But since the numbers are small and the proportion works out evenly, I don't think that's necessary.

But just for thoroughness, let me think. The Hamilton method involves calculating the standard divisor, which is total population divided by number of representatives. Then each state gets a quota, which is population divided by standard divisor, and then you allocate the representatives based on the integer part and then assign the remaining based on the fractional parts.

Let me try that.

Total enrollment is 1170. Number of representatives is 18. So standard divisor is 1170 / 18 = 65.

Then, the quota for 7th grade is 520 / 65 = 8.

Quota for 8th grade is 650 / 65 = 10.

So exactly 8 and 10, so no need for any further allocation. So that's the same result.

Therefore, regardless of the method, the answer is 10.

Alternatively, if the numbers didn't divide evenly, we might have to use some other method, but in this case, it's straightforward.

So, I think the answer is 10.

But let me check once more.

Total students: 520 + 650 = 1170.

Total representatives: 18.

So, the proportion of representatives for 8th grade is (650 / 1170) * 18.

Calculating 650 divided by 1170: Let me compute that decimal.

Divide numerator and denominator by 10: 65 / 117.

Divide numerator and denominator by 13: 65 ÷13=5, 117 ÷13=9. So 5/9 ≈ 0.5555...

Multiply by 18: 0.5555... * 18 = 10. Exactly.

So that's 10.

Therefore, the answer is 10.

I think that's solid. All methods lead to the same answer. So I can be confident that the 8th grade should have 10 representatives.

**Final Answer**
The 8th grade should have \boxed{10} representatives.

To determine how many representatives the 8th grade should have on the Student Council, we need to ensure **fair representation**, which means the number of representatives should be **proportional to the number of students** in each grade.

---

### Step 1: Understand the Given Data

- **7th grade enrollment**: 520 students
- **8th grade enrollment**: 650 students
- **Total representatives**: 18
- **Goal**: Distribute the 18 representatives in proportion to the enrollments.

---

### Step 2: Determine the Ratio of Enrollments

We start by finding the ratio of the enrollments:

$$
\frac{520}{650} = \frac{4}{5}
$$

This means for every 4 students in the 7th grade, there are 5 students in the 8th grade.

---

### Step 3: Use the Ratio to Allocate Representatives

Let the number of representatives for the 7th grade be $ R_7 $, and for the 8th grade be $ R_8 $. Then:

$$
\frac{R_7}{R_8} = \frac{4}{5}
$$

Also, the total number of representatives is:

$$
R_7 + R_8 = 18
$$

From the ratio, we can express $ R_7 $ in terms of $ R_8 $:

$$
R_7 = \frac{4}{5} R_8
$$

Substitute into the total:

$$
\frac{4}{5} R_8 + R_8 = 18
$$

$$
\left( \frac{4}{5} + 1 \right) R_8 = 18
$$

$$
\frac{9}{5} R_8 = 18
$$

$$
R_8 = 18 \times \frac{5}{9} = 10
$$

---

### Step 4: Verify the Solution

- 8th grade representatives: 10
- 7th grade representatives: $ 18 - 10 = 8 $
- Total: $ 10 + 8 = 18 $, which matches the total number of representatives.

Also, the ratio of representatives is:

$$
\frac{8}{10} = \frac{4}{5}
$$

Which matches the ratio of enrollments, confirming the fairness of the distribution.

---

### Final Answer

$$
\boxed{10}
$$

[Assistant]
Explanation:
Okay, so I need to figure out how many representatives the 8th grade should have on the Student Council, given that there are 7th and 8th grades with enrollments of 520 and 650 respectively, and the total number of representatives is 18. The goal is to have a fair representation. Hmm, fair representation... I think that means the number of representatives each grade gets should be proportional to their enrollment. Let me check if that makes sense.

First, let me recall what proportional representation means. If something is proportional, then the ratio of the representatives should be the same as the ratio of their enrollments. So, if the 7th grade has 520 students and the 8th has 650, then the ratio of their enrollments is 520:650. Maybe I can simplify that ratio first?

Let me divide both numbers by 130. 520 divided by 130 is 4, and 650 divided by 130 is 5. So the simplified ratio is 4:5. That means for every 4 students in 7th grade, there are 5 students in 8th grade. Therefore, the ratio of representatives should also be 4:5.

But wait, the total number of representatives is 18. So if the ratio is 4:5, then the total parts are 4 + 5 = 9 parts. So each part is equal to 18 divided by 9, which is 2. Therefore, the 7th grade should have 4 parts times 2, which is 8 representatives, and the 8th grade should have 5 parts times 2, which is 10 representatives.

Let me check if that adds up. 8 + 10 is 18, which matches the total number of representatives. So that seems to work. But let me make sure I didn't skip any steps or make any wrong assumptions.

Alternatively, maybe I should approach it using fractions. The total enrollment is 520 + 650. Let me calculate that. 520 + 650 is 1170 students total. So the 7th grade is 520/1170 of the total enrollment, and the 8th grade is 650/1170. Then, the number of representatives should be proportional to these fractions.

So, for the 7th grade, the number of representatives would be (520/1170)*18, and for the 8th grade, it would be (650/1170)*18. Let me compute these.

First, simplifying 520/1170. Let me divide numerator and denominator by 130 again. 520 ÷ 130 = 4, 1170 ÷ 130 = 9. So that's 4/9. Therefore, 4/9 of 18 is (4*18)/9 = (72)/9 = 8. Similarly, 650/1170 simplifies to 5/9, so 5/9 of 18 is (5*18)/9 = 90/9 = 10. So that gives the same result as before: 8 and 10.

Therefore, the 8th grade should have 10 representatives.

But let me think again if there's another way to approach this. Maybe using the concept of proportionality. If the number of representatives is proportional to the number of students, then the ratio of representatives should be the same as the ratio of students.

So, let me denote the number of representatives for 7th grade as R7 and for 8th grade as R8. Then, R7/R8 = 520/650. Simplifying that ratio, as before, 520/650 = 4/5. Therefore, R7/R8 = 4/5. Also, R7 + R8 = 18.

So, if R7/R8 = 4/5, then R7 = (4/5) R8. Substituting into the total: (4/5) R8 + R8 = 18. Combining terms: (4/5 + 5/5) R8 = 18 => (9/5) R8 = 18. Therefore, R8 = 18 * (5/9) = 10.

Same answer again. So that seems consistent.

Alternatively, maybe I can think in terms of weighted averages or something else? Let me see. But I think the methods above are sufficient.

Wait, but maybe there's a catch here. For example, sometimes when you have to assign whole numbers of representatives, you might have to round, but in this case, the numbers worked out evenly. 520 and 650 are multiples of 130, and 18 is a multiple of 9, so the division comes out even.

But just to make sure, let me check if there's another interpretation of "fair representation." Sometimes, "fair representation" can refer to something else, like equal representation per student, but that would mean each student gets one representative, which is not feasible here. But since there are only 18 representatives, it's about proportionality.

Alternatively, maybe using some other method like the Hamilton method or Jefferson method for apportionment? But since the numbers are small and the proportion works out evenly, I don't think that's necessary.

But just for thoroughness, let me think. The Hamilton method involves calculating the standard divisor, which is total population divided by number of representatives. Then each state gets a quota, which is population divided by standard divisor, and then you allocate the representatives based on the integer part and then assign the remaining based on the fractional parts.

Let me try that.

Total enrollment is 1170. Number of representatives is 18. So standard divisor is 1170 / 18 = 65.

Then, the quota for 7th grade is 520 / 65 = 8.

Quota for 8th grade is 650 / 65 = 10.

So exactly 8 and 10, so no need for any further allocation. So that's the same result.

Therefore, regardless of the method, the answer is 10.

Alternatively, if the numbers didn't divide evenly, we might have to use some other method, but in this case, it's straightforward.

So, I think the answer is 10.

But let me check once more.

Total students: 520 + 650 = 1170.

Total representatives: 18.

So, the proportion of representatives for 8th grade is (650 / 1170) * 18.

Calculating 650 divided by 1170: Let me compute that decimal.

Divide numerator and denominator by 10: 65 / 117.

Divide numerator and denominator by 13: 65 ÷13=5, 117 ÷13=9. So 5/9 ≈ 0.5555...

Multiply by 18: 0.5555... * 18 = 10. Exactly.

So that's 10.

Therefore, the answer is 10.

I think that's solid. All methods lead to the same answer. So I can be confident that the 8th grade should have 10 representatives.

**Final Answer**
The 8th grade should have \boxed{10} representatives.

To determine how many representatives the 8th grade should have on the Student Council, we need to ensure **fair representation**, which means the number of representatives should be **proportional to the number of students** in each grade.

---

### Step 1: Understand the Given Data

- **7th grade enrollment**: 520 students
- **8th grade enrollment**: 650 students
- **Total representatives**: 18
- **Goal**: Distribute the 18 representatives in proportion to the enrollments.

---

### Step 2: Determine the Ratio of Enrollments

We start by finding the ratio of the enrollments:

$$
\frac{520}{650} = \frac{4}{5}
$$

This means for every 4 students in the 7th grade, there are 5 students in the 8th grade.

---

### Step 3: Use the Ratio to Allocate Representatives

Let the number of representatives for the 7th grade be $ R_7 $, and for the 8th grade be $ R_8 $. Then:

$$
\frac{R_7}{R_8} = \frac{4}{5}
$$

Also, the total number of representatives is:

$$
R_7 + R_8 = 18
$$

From the ratio, we can express $ R_7 $ in terms of $ R_8 $:

$$
R_7 = \frac{4}{5} R_8
$$

Substitute into the total:

$$
\frac{4}{5} R_8 + R_8 = 18
$$

$$
\left( \frac{4}{5} + 1 \right) R_8 = 18
$$

$$
\frac{9}{5} R_8 = 18
$$

$$
R_8 = 18 \times \frac{5}{9} = 10
$$

---

### Step 4: Verify the Solution

- 8th grade representatives: 10
- 7th grade representatives: $ 18 - 10 = 8 $
- Total: $ 10 + 8 = 18 $, which matches the total number of representatives.

Also, the ratio of representatives is:

$$
\frac{8}{10} = \frac{4}{5}
$$

Which matches the ratio of enrollments, confirming the fairness of the distribution.

---

### Final Answer

$$
\boxed{10}
$$

Solution: The 8th grade has $\frac{650}{520+650} = \frac{650}{1170} = \frac{65}{117}$ of the total students. To simplify this fraction further, we notice that $65 = 5 \cdot 13$. Since $117$ is not divisible by $5$, we test if it's divisible by $13$ and find that $117 = 9 \cdot 13$. Thus to have fair representation, the 8th grade should have $\frac{65}{117} \times 18 = \frac{5}{9} \times 18 = \boxed{10}$ of the $18$ representatives.


Explanation:
Okay, so I need to figure out how many ways 5 students can be selected from a group of 6 students. Hmm, let me think. This seems like a combination problem because the order in which we select the students doesn't matter, right? Like, if I pick student A first and then student B, it's the same as picking student B first and then student A. So combinations are used when the order doesn't matter, and permutations when it does. Since the question is about selecting students without any mention of order, combinations must be the way to go.

Alright, so the formula for combinations is C(n, k) = n! / (k! * (n - k)!), where n is the total number of items, and k is the number of items to choose. In this case, n is 6 and k is 5. Let me plug those numbers into the formula.

So, C(6, 5) = 6! / (5! * (6 - 5)!). Let me compute that step by step. First, 6! is 6 factorial, which is 6 × 5 × 4 × 3 × 2 × 1. But wait, maybe I can simplify this before multiplying everything out. Let me see. The denominator is 5! times (6 - 5)! which is 5! times 1!. So, 6! divided by 5! times 1!.

I remember that 6! is equal to 6 × 5!, so maybe I can cancel out the 5! in the numerator and denominator. Let me write that down:

C(6, 5) = (6 × 5!) / (5! × 1!)

Oh, right! The 5! in the numerator and denominator cancels out. So that simplifies to 6 / 1! which is 6 / 1 = 6. So, does that mean there are 6 ways to select 5 students from 6?

Wait, let me verify that. If there are 6 students and I need to choose 5, that's the same as leaving out 1 student. So, how many ways can I leave out 1 student? Well, there are 6 students, so there are 6 different ways to leave out one student. Therefore, the number of ways to choose 5 students is equal to the number of ways to leave out 1 student, which is 6. That makes sense. So, the answer is 6.

But let me check with the combination formula again. If I use the formula C(n, k) = C(n, n - k), which is a property of combinations. So, C(6, 5) should be equal to C(6, 1). And C(6, 1) is 6, which matches the previous result. So that's another way to think about it. Choosing 5 out of 6 is the same as choosing 1 out of 6 to exclude. Therefore, the answer is 6.

Alternatively, if I didn't remember that property, I could compute it directly. Let's do that again. C(6, 5) = 6! / (5! * 1!) = (720) / (120 * 1) = 720 / 120 = 6. Yep, same answer.

Is there another way to think about this? Maybe by listing all possible combinations? Let's say the students are labeled A, B, C, D, E, F. If I need to choose 5, then the possible selections are all groups that exclude one student. So:

- Exclude A: B, C, D, E, F
- Exclude B: A, C, D, E, F
- Exclude C: A, B, D, E, F
- Exclude D: A, B, C, E, F
- Exclude E: A, B, C, D, F
- Exclude F: A, B, C, D, E

That's 6 different groups, each missing one student. So, again, 6 ways. So that's concrete evidence that the answer is 6.

Wait, but just to make sure I didn't make a mistake in the combination formula. Let me recall the formula again. The combination formula is n! divided by (k! times (n - k)!). So, for n=6 and k=5, that's 6! / (5! * 1!) which is indeed 720 / (120 * 1) = 6. So that's correct.

Alternatively, if I use the multiplicative formula for combinations. The formula can also be written as C(n, k) = (n * (n - 1) * ... * (n - k + 1)) / k! So, for C(6, 5), that would be (6 * 5) / 5! Wait, no. Wait, let me think. Wait, the multiplicative formula is:

C(n, k) = n! / (k! (n - k)!) which can also be written as [n × (n - 1) × ... × (n - k + 1)] / k!

So for C(6, 5), that would be [6 × 5 × 4 × 3 × 2] / 5! Wait, but that's 6 × 5 × 4 × 3 × 2 divided by 5! Let me compute that. The numerator is 6 × 5 × 4 × 3 × 2 = 720. The denominator is 5! = 120. So 720 / 120 = 6. Same answer. So that works too.

Alternatively, if I think of it as 6 × 5 × 4 × 3 × 2 divided by 5 × 4 × 3 × 2 × 1. Wait, that's the same as 6 / 1 = 6. So yeah, that's another way to see it. The numerator is 6 × 5! and the denominator is 5! × 1!, so cancels out to 6.

So all these different methods lead me to the same answer, 6. Therefore, I think that's correct.

But just to make sure, let me think of another example. Suppose there are 3 students and I want to choose 2. How many ways? Using the same logic, it should be 3. Let's check with the formula: C(3, 2) = 3! / (2! * 1!) = 6 / 2 = 3. Which is correct. Alternatively, leaving out 1 student: 3 ways. So that works. Similarly, if I have 4 students and choose 3, that's 4 ways. Which is the same as leaving out 1 student. So yeah, the pattern holds.

Therefore, applying the same logic, with 6 students choosing 5, it's 6 ways. So I think that's solid.

Another way to think about it is using Pascal's triangle. The 6th row (if we start counting from row 0) would be 1, 6, 15, 20, 15, 6, 1. Wait, so the 6th row (row index 5 if starting from 0) would be... Wait, maybe I need to check. Let me recall that the nth row corresponds to the coefficients for (a + b)^n. So, row 0 is 1. Row 1 is 1, 1. Row 2 is 1, 2, 1. Row 3 is 1, 3, 3, 1. Row 4 is 1, 4, 6, 4, 1. Row 5 is 1, 5, 10, 10, 5, 1. Row 6 is 1, 6, 15, 20, 15, 6, 1. So, if we are looking for C(6, 5), that's the entry in row 6, position 5 (if starting from 0). Which is 6. So that's consistent. So, again, the answer is 6.

Therefore, all these different methods confirm that the number of ways is 6. I don't see any contradictions here. So I think the answer is 6.

**Final Answer**
The number of ways is \boxed{6}.

To determine how many ways 5 students can be selected from a group of 6, we need to consider the nature of the selection: **order does not matter**. This is a classic **combinations** problem.

---

### Step 1: Understand the Problem

We are selecting **5 students** from a group of **6**, and the order in which we select them is **not important**. This means that the group {A, B, C, D, E} is the same as {B, A, C, D, E}, and so on. Therefore, we use the **combination formula**:

$$
C(n, k) = \frac{n!}{k!(n - k)!}
$$

Where:
- $ n $ is the total number of items (students),
- $ k $ is the number of items to choose (students to select),
- $ ! $ denotes factorial.

---

### Step 2: Apply the Formula

Given:
- $ n = 6 $
- $ k = 5 $

$$
C(6, 5) = \frac{6!}{5!(6 - 5)!} = \frac{6!}{5! \cdot 1!}
$$

Now simplify:

$$
6! = 6 \times 5! \quad \text{so} \quad \frac{6!}{5!} = 6
$$

$$
C(6, 5) = \frac{6}{1} = 6
$$

---

### Step 3: Interpret the Result

This means there are **6 different ways** to choose 5 students from a group of 6. Another way to think about it is: **each selection corresponds to leaving out one student**, and since there are 6 students, there are 6 such selections.

---

### Final Answer

$$
\boxed{6}
$$

[Assistant]
Explanation:
Okay, so I need to figure out how many ways 5 students can be selected from a group of 6 students. Hmm, let me think. This seems like a combination problem because the order in which we select the students doesn't matter, right? Like, if I pick student A first and then student B, it's the same as picking student B first and then student A. So combinations are used when the order doesn't matter, and permutations when it does. Since the question is about selecting students without any mention of order, combinations must be the way to go.

Alright, so the formula for combinations is C(n, k) = n! / (k! * (n - k)!), where n is the total number of items, and k is the number of items to choose. In this case, n is 6 and k is 5. Let me plug those numbers into the formula.

So, C(6, 5) = 6! / (5! * (6 - 5)!). Let me compute that step by step. First, 6! is 6 factorial, which is 6 × 5 × 4 × 3 × 2 × 1. But wait, maybe I can simplify this before multiplying everything out. Let me see. The denominator is 5! times (6 - 5)! which is 5! times 1!. So, 6! divided by 5! times 1!.

I remember that 6! is equal to 6 × 5!, so maybe I can cancel out the 5! in the numerator and denominator. Let me write that down:

C(6, 5) = (6 × 5!) / (5! × 1!)

Oh, right! The 5! in the numerator and denominator cancels out. So that simplifies to 6 / 1! which is 6 / 1 = 6. So, does that mean there are 6 ways to select 5 students from 6?

Wait, let me verify that. If there are 6 students and I need to choose 5, that's the same as leaving out 1 student. So, how many ways can I leave out 1 student? Well, there are 6 students, so there are 6 different ways to leave out one student. Therefore, the number of ways to choose 5 students is equal to the number of ways to leave out 1 student, which is 6. That makes sense. So, the answer is 6.

But let me check with the combination formula again. If I use the formula C(n, k) = C(n, n - k), which is a property of combinations. So, C(6, 5) should be equal to C(6, 1). And C(6, 1) is 6, which matches the previous result. So that's another way to think about it. Choosing 5 out of 6 is the same as choosing 1 out of 6 to exclude. Therefore, the answer is 6.

Alternatively, if I didn't remember that property, I could compute it directly. Let's do that again. C(6, 5) = 6! / (5! * 1!) = (720) / (120 * 1) = 720 / 120 = 6. Yep, same answer.

Is there another way to think about this? Maybe by listing all possible combinations? Let's say the students are labeled A, B, C, D, E, F. If I need to choose 5, then the possible selections are all groups that exclude one student. So:

- Exclude A: B, C, D, E, F
- Exclude B: A, C, D, E, F
- Exclude C: A, B, D, E, F
- Exclude D: A, B, C, E, F
- Exclude E: A, B, C, D, F
- Exclude F: A, B, C, D, E

That's 6 different groups, each missing one student. So, again, 6 ways. So that's concrete evidence that the answer is 6.

Wait, but just to make sure I didn't make a mistake in the combination formula. Let me recall the formula again. The combination formula is n! divided by (k! times (n - k)!). So, for n=6 and k=5, that's 6! / (5! * 1!) which is indeed 720 / (120 * 1) = 6. So that's correct.

Alternatively, if I use the multiplicative formula for combinations. The formula can also be written as C(n, k) = (n * (n - 1) * ... * (n - k + 1)) / k! So, for C(6, 5), that would be (6 * 5) / 5! Wait, no. Wait, let me think. Wait, the multiplicative formula is:

C(n, k) = n! / (k! (n - k)!) which can also be written as [n × (n - 1) × ... × (n - k + 1)] / k!

So for C(6, 5), that would be [6 × 5 × 4 × 3 × 2] / 5! Wait, but that's 6 × 5 × 4 × 3 × 2 divided by 5! Let me compute that. The numerator is 6 × 5 × 4 × 3 × 2 = 720. The denominator is 5! = 120. So 720 / 120 = 6. Same answer. So that works too.

Alternatively, if I think of it as 6 × 5 × 4 × 3 × 2 divided by 5 × 4 × 3 × 2 × 1. Wait, that's the same as 6 / 1 = 6. So yeah, that's another way to see it. The numerator is 6 × 5! and the denominator is 5! × 1!, so cancels out to 6.

So all these different methods lead me to the same answer, 6. Therefore, I think that's correct.

But just to make sure, let me think of another example. Suppose there are 3 students and I want to choose 2. How many ways? Using the same logic, it should be 3. Let's check with the formula: C(3, 2) = 3! / (2! * 1!) = 6 / 2 = 3. Which is correct. Alternatively, leaving out 1 student: 3 ways. So that works. Similarly, if I have 4 students and choose 3, that's 4 ways. Which is the same as leaving out 1 student. So yeah, the pattern holds.

Therefore, applying the same logic, with 6 students choosing 5, it's 6 ways. So I think that's solid.

Another way to think about it is using Pascal's triangle. The 6th row (if we start counting from row 0) would be 1, 6, 15, 20, 15, 6, 1. Wait, so the 6th row (row index 5 if starting from 0) would be... Wait, maybe I need to check. Let me recall that the nth row corresponds to the coefficients for (a + b)^n. So, row 0 is 1. Row 1 is 1, 1. Row 2 is 1, 2, 1. Row 3 is 1, 3, 3, 1. Row 4 is 1, 4, 6, 4, 1. Row 5 is 1, 5, 10, 10, 5, 1. Row 6 is 1, 6, 15, 20, 15, 6, 1. So, if we are looking for C(6, 5), that's the entry in row 6, position 5 (if starting from 0). Which is 6. So that's consistent. So, again, the answer is 6.

Therefore, all these different methods confirm that the number of ways is 6. I don't see any contradictions here. So I think the answer is 6.

**Final Answer**
The number of ways is \boxed{6}.

To determine how many ways 5 students can be selected from a group of 6, we need to consider the nature of the selection: **order does not matter**. This is a classic **combinations** problem.

---

### Step 1: Understand the Problem

We are selecting **5 students** from a group of **6**, and the order in which we select them is **not important**. This means that the group {A, B, C, D, E} is the same as {B, A, C, D, E}, and so on. Therefore, we use the **combination formula**:

$$
C(n, k) = \frac{n!}{k!(n - k)!}
$$

Where:
- $ n $ is the total number of items (students),
- $ k $ is the number of items to choose (students to select),
- $ ! $ denotes factorial.

---

### Step 2: Apply the Formula

Given:
- $ n = 6 $
- $ k = 5 $

$$
C(6, 5) = \frac{6!}{5!(6 - 5)!} = \frac{6!}{5! \cdot 1!}
$$

Now simplify:

$$
6! = 6 \times 5! \quad \text{so} \quad \frac{6!}{5!} = 6
$$

$$
C(6, 5) = \frac{6}{1} = 6
$$

---

### Step 3: Interpret the Result

This means there are **6 different ways** to choose 5 students from a group of 6. Another way to think about it is: **each selection corresponds to leaving out one student**, and since there are 6 students, there are 6 such selections.

---

### Final Answer

$$
\boxed{6}
$$

Solution: We can choose 5 students out of a group of 6 students  without regard to order in $\binom{6}{5} = \boxed{6}$ ways.


Explanation:
Okay, so Linda, Sherry, June, and Connie sold cookies and earned some money. Now they want to convert all their coins into dollars, but they need to know how much money in cents is left after they convert as many coins as possible into bills. Hmm, let me think.

First, I need to figure out the total amount of money they have. Let me add up each person's earnings. Let me write down the amounts:

- Linda: $27.47
- Sherry: $35.23
- June: $37.37
- Connie: $26.52

So adding these up. Let me do it step by step. Let me add Linda and Sherry first. 27.47 + 35.23. Let me add the dollars first: 27 + 35 = 62. Then the cents: 47 cents + 23 cents = 70 cents. So that's $62.70.

Now add June's amount: $37.37. So adding $62.70 + $37.37. Again, dollars: 62 + 37 = 99. Cents: 70 + 37 = 107 cents. Wait, 107 cents is equal to $1.07. So that would make the total dollars 99 + 1 = 100, and then 0.07. So total is $100.07?

Wait, let me check that again. Maybe I should add them all together properly.

Alternatively, maybe convert all amounts to cents first to avoid decimal confusion. Let me try that.

Linda: $27.47 is 27 dollars and 47 cents. So 27*100 + 47 = 2700 + 47 = 2747 cents.

Sherry: $35.23 is 35*100 + 23 = 3500 + 23 = 3523 cents.

June: $37.37 is 37*100 + 37 = 3700 + 37 = 3737 cents.

Connie: $26.52 is 26*100 + 52 = 2600 + 52 = 2652 cents.

Now total cents is 2747 + 3523 + 3737 + 2652. Let me add them step by step.

First, 2747 + 3523. Let me compute that. 2747 + 3523. Let me add 2700 + 3500 = 6200. Then 47 + 23 = 70. So total is 6200 + 70 = 6270 cents.

Next, add 3737. So 6270 + 3737. Let me do 6270 + 3000 = 9270, then +737 = 9270 + 700 = 9970, +37 = 10007 cents.

Then add 2652. So 10007 + 2652. Let me compute that. 10007 + 2000 = 12007, then +652 = 12007 + 600 = 12607, +52 = 12659 cents.

So total money is 12,659 cents. Wait, let me check that again because adding step by step can have errors.

Wait, first step: Linda + Sherry: 2747 + 3523. Let me do 2747 + 3523:

2747 + 3523:

Let me add 2747 + 3523:

Start with 2747 + 3500 = 6247, then +23 = 6270. That's correct.

Then adding June's 3737: 6270 + 3737.

Let me do 6270 + 3737:

Breakdown: 6270 + 3000 = 9270, 9270 + 700 = 9970, 9970 + 37 = 10007. Correct.

Then adding Connie's 2652: 10007 + 2652.

Let me do 10007 + 2000 = 12007, 12007 + 600 = 12607, 12607 + 52 = 12659. Yes, that's 12,659 cents total.

So total money is 12,659 cents. Now, they want to convert as many coins into bills as possible. So the question is, how much money in cents is left after converting as many coins as possible into bills?

Wait, but when you convert coins into bills, you can only convert whole bills. So, for example, if you have a certain number of cents, you can convert as many dollars as possible, and the leftover cents would be the remainder.

But wait, but the problem says they pooled their money together and went to the bank to convert the coins into dollars. So maybe they have a mix of coins and bills? Wait, but the problem says they "converted the coins into dollars." Wait, maybe they had all their money in coins, and they want to convert all coins into bills, and find out how much is left in cents? But if they convert all coins into bills, then there should be no cents left. Wait, that doesn't make sense. Maybe there's a misunderstanding here.

Wait, let me re-read the problem:

"After the sale, they pooled their money together and went to the bank to convert the coins into dollars. How much money, in cents, was left after they converted as many of the coins into bills as they could?"

Wait, so maybe they had some coins and some bills, and they wanted to convert the coins into bills. But the problem says they "pooled their money together" which might mean that they had all their money in coins? Or maybe they had some coins and some bills, but they wanted to convert the coins into bills. But the problem says "convert the coins into dollars", so maybe they had coins, and they wanted to exchange them for bills, and the question is how much money in cents was left after converting as many coins as possible into bills. But if they converted all coins into bills, then there would be no coins left. But the question says "how much money, in cents, was left after they converted as many of the coins into bills as they could?"

Wait, maybe they had some coins and some bills, and they wanted to convert the coins into bills, but the total amount of money they had was in dollars and cents. Wait, but the problem says they "pooled their money together" which is the total amount of money they earned, which is in dollars and cents. But maybe they had that money in coins? Or maybe they had some coins and some bills, but the problem is not clear. Wait, but the problem says "convert the coins into dollars", so maybe they had coins, and they wanted to convert them into dollars, and the question is how much money in cents was left after converting as many coins as possible into bills. But if they converted all coins into bills, then there would be no coins left. But the problem says "how much money, in cents, was left after they converted as many of the coins into bills as they could?"

Wait, maybe the problem is that they had all their money in coins, and they wanted to convert them into bills, but they couldn't convert all of them because they couldn't make exact bills? Wait, but if you have a certain amount of cents, you can convert as many dollars as possible, and the leftover cents would be the remainder. For example, if you have 123 cents, you can convert 1 dollar (100 cents) and have 23 cents left. So maybe the question is, if they had all their money in cents (i.e., all coins), then converting as many as possible into dollars, how much is left in cents?

But the problem says they "pooled their money together" which is the total amount they earned, which is in dollars and cents. But maybe they had that amount in coins? For example, if they had $27.47, that would be 27 dollars and 47 cents, but if they had that in coins, then they would have 27*100 + 47 = 2747 cents. But if they had all their money in coins, then the total is 12,659 cents as calculated before. Then converting as many coins into bills as possible would mean converting as many dollars as possible, and the leftover cents would be the remainder when dividing total cents by 100. Wait, but that would be 12,659 divided by 100, which is 126 dollars and 59 cents. So the leftover cents would be 59 cents. But that seems too straightforward. But maybe that's the answer? But why would the problem mention converting coins into dollars? Maybe there's a trick here.

Wait, but maybe they had some coins and some bills, and they wanted to convert the coins into bills, but the problem is not clear. However, given that they earned money, which is in dollars and cents, but the problem says they converted the coins into dollars. Maybe they had all their money in coins, so the total amount is 12,659 cents. Then converting as many coins into bills as possible would mean converting 12,659 cents into dollars. Since 1 dollar is 100 cents, the number of dollars is 12,659 divided by 100, which is 126 dollars with a remainder of 59 cents. Therefore, the leftover cents would be 59 cents. Therefore, the answer is 59 cents. But that seems too easy. However, maybe there's more to it?

Wait, but maybe the problem is not about converting all the cents into dollars, but converting coins into bills, meaning that they have coins of different denominations, and they want to exchange them for bills, but the question is how much is left in cents after exchanging as much as possible. But if they have all their money in coins, then converting into bills would just be exchanging all coins for bills, leaving no coins. But the problem says "how much money, in cents, was left after they converted as many of the coins into bills as they could?" So maybe they had some coins and some bills, and they wanted to convert the coins into bills, but the problem is not clear. However, given the problem statement, maybe the answer is 59 cents. But let me check again.

Alternatively, maybe the problem is that they had the total amount of money, which is in dollars and cents, and they wanted to convert the coins (i.e., the cents part) into dollars. Wait, but if they have, for example, $27.47, that is 27 dollars and 47 cents. If they wanted to convert the coins (the 47 cents) into dollars, but you can't convert 47 cents into a dollar. So maybe the question is about converting the cents part into dollars? But that doesn't make sense. Alternatively, maybe they had all their money in coins, and they wanted to convert as much as possible into bills, so the leftover is the cents part. Which would be 59 cents. But maybe the problem is more complicated?

Wait, let me check the problem again:

"Linda, Sherry, June, and Connie walked around their neighborhoods selling girl scout cookies. Linda earned $27.47, Sherry earned $35.23, June earned $37.37, and Connie earned $26.52. After the sale, they pooled their money together and went to the bank to convert the coins into dollars. How much money, in cents, was left after they converted as many of the coins into bills as they could?"

So they pooled their money together, which is the total amount of money they earned, which is in dollars and cents. Then they went to the bank to convert the coins into dollars. So maybe they had some coins and some bills, and they wanted to convert the coins into bills. But the problem is, how much money in cents was left after converting as many coins as possible into bills. But if they had some coins, converting them into bills would leave them with bills, but the question is about cents left. Wait, maybe they had all their money in coins, so they had 12,659 cents. Then converting as many coins into bills as possible would mean converting 12,659 cents into 126 dollars and 59 cents. Therefore, the leftover is 59 cents. Therefore, the answer is 59 cents. But maybe the problem is expecting something else?

Alternatively, maybe they had the total amount in dollars and cents, but they wanted to convert the coins (i.e., the cents part) into dollars. But that would be impossible unless they have more coins. Wait, but if they have, for example, $27.47, which is 27 dollars and 47 cents. If they wanted to convert the coins (the 47 cents) into dollars, they would need to exchange them for dollars, but since 47 cents is less than a dollar, they can't. Therefore, the leftover would be 47 cents. But if they had multiple people, maybe they can combine their coins? Wait, but the total amount is 12,659 cents. So if they convert as many as possible into dollars, they can get 126 dollars, and 59 cents left. Therefore, the answer is 59 cents. But why is the problem mentioning converting coins into dollars? Maybe the problem is that they had all their money in coins, and they wanted to convert them into bills, and the leftover is the cents. So the answer is 59 cents. But maybe I need to check if there's a different interpretation.

Alternatively, maybe the problem is that they had the total amount in dollars and cents, but they wanted to convert the coins (i.e., the cents part) into dollars, but since they can't do that, they have to keep the cents. But that would be the same as the cents part. Wait, but the total amount is $100.07? Wait, earlier when I added the dollars, I thought maybe it was $100.07. Let me check that again.

Original amounts:

Linda: $27.47

Sherry: $35.23

June: $37.37

Connie: $26.52

Adding them up:

Let me add them in dollars and cents separately.

Dollars: 27 + 35 + 37 + 26. Let me compute that:

27 + 35 = 62; 37 + 26 = 63. Then 62 + 63 = 125 dollars.

Cents: 47 + 23 + 37 + 52. Let me compute that:

47 + 23 = 70; 37 + 52 = 89. Then 70 + 89 = 159 cents. 159 cents is $1.59. Therefore, total amount is 125 + 1.59 = $126.59. Wait, so total amount is $126.59. Which is 126 dollars and 59 cents. Therefore, if they converted as many coins into bills as possible, they would have 126 dollars and 59 cents left. But the question says "how much money, in cents, was left after they converted as many of the coins into bills as they could?" So if they converted all the coins into bills, then they would have 126 dollars and 59 cents. But the question is about cents left. Wait, but if they converted all the coins into bills, then they would have no coins left. But the total amount is $126.59, which is 126 dollars and 59 cents. So if they converted all the coins into bills, they would have 126 bills of $1 and 59 cents. But the question is asking for how much money in cents was left. If they converted as many coins into bills as possible, then the leftover would be the cents part, which is 59 cents. Therefore, the answer is 59 cents. Therefore, the answer is 59. But let me check again.

Alternatively, maybe the problem is that they had all their money in coins, so total is 12,659 cents. Converting as many as possible into bills would be 126 dollars (which is 12,600 cents) and 59 cents left. Therefore, the answer is 59 cents. So both methods give the same answer. Therefore, the answer is 59 cents. Therefore, the answer is \boxed{59}.

But let me check once more. Maybe there's a trick here. For example, maybe they had different coins, and when converting into bills, they can't just take the cents part. For example, if they have coins of different denominations, maybe they can't make exact change? But the problem says "convert the coins into dollars", which is a bit ambiguous. But if they have all their money in coins, then converting into bills would just be exchanging all coins for bills, leaving the cents part. But if they have, for example, 59 cents, they can't convert that into a dollar, so that's the leftover. Therefore, the answer is 59 cents. Therefore, I think the answer is 59.

But let me check the total cents again. Earlier, I converted each person's earnings into cents and added them up to get 12,659 cents. Let me check that again:

Linda: 27.47 = 27*100 + 47 = 2747

Sherry: 35.23 = 35*100 + 23 = 3523

June: 37.37 = 37*100 + 37 = 3737

Connie: 26.52 = 26*100 + 52 = 2652

Adding them:

2747 + 3523 = 6270

6270 + 3737 = 10007

10007 + 2652 = 12659. Yes, that's correct. 12,659 cents.

Divide by 100: 12,659 / 100 = 126.59. So 126 dollars and 59 cents. Therefore, the leftover cents is 59. So the answer is 59 cents. Therefore, the answer is \boxed{59}.

But let me think again. Maybe the problem is not about converting all the money into bills, but converting coins into bills, meaning that they had some coins and some bills, and they wanted to convert the coins into bills, but the problem is not clear. However, given the problem statement, the most logical interpretation is that they had all their money in coins, and they wanted to convert as much as possible into bills, leaving some cents. Therefore, the answer is 59 cents.

Alternatively, maybe the problem is that they had the total amount in dollars and cents, and they wanted to convert the coins (i.e., the cents part) into dollars, but since they can't do that, they have to keep the cents. But that would be the same as the cents part. However, the total amount is $126.59, so the cents part is 59 cents. Therefore, the answer is 59 cents. So either way, the answer is 59. Therefore, I think the answer is 59.

**Final Answer**
\boxed{59}

To determine how much money, in cents, was left after converting as many coins as possible into bills, we begin by calculating the total amount of money the four girls earned.

---

### Step 1: Convert Each Person's Earnings to Cents

We convert each person's earnings from dollars to cents to avoid decimal confusion:

- **Linda**: $27.47 = 27 \times 100 + 47 = 2747$ cents
- **Sherry**: $35.23 = 35 \times 100 + 23 = 3523$ cents
- **June**: $37.37 = 37 \times 100 + 37 = 3737$ cents
- **Connie**: $26.52 = 26 \times 100 + 52 = 2652$ cents

---

### Step 2: Add All the Cents Together

Now, we sum up all the cents:

$$
2747 + 3523 + 3737 + 2652 = 12659 \text{ cents}
$$

---

### Step 3: Convert as Many Cents as Possible into Dollars

Since 1 dollar = 100 cents, we divide the total number of cents by 100 to find how many full dollars can be made:

$$
12659 \div 100 = 126 \text{ dollars with a remainder of } 59 \text{ cents}
$$

This means that after converting as many coins as possible into bills (i.e., full dollars), **59 cents** remain unconverted.

---

### Final Answer

$$
\boxed{59}
$$

[Assistant]
Explanation:
Okay, so Linda, Sherry, June, and Connie sold cookies and earned some money. Now they want to convert all their coins into dollars, but they need to know how much money in cents is left after they convert as many coins as possible into bills. Hmm, let me think.

First, I need to figure out the total amount of money they have. Let me add up each person's earnings. Let me write down the amounts:

- Linda: $27.47
- Sherry: $35.23
- June: $37.37
- Connie: $26.52

So adding these up. Let me do it step by step. Let me add Linda and Sherry first. 27.47 + 35.23. Let me add the dollars first: 27 + 35 = 62. Then the cents: 47 cents + 23 cents = 70 cents. So that's $62.70.

Now add June's amount: $37.37. So adding $62.70 + $37.37. Again, dollars: 62 + 37 = 99. Cents: 70 + 37 = 107 cents. Wait, 107 cents is equal to $1.07. So that would make the total dollars 99 + 1 = 100, and then 0.07. So total is $100.07?

Wait, let me check that again. Maybe I should add them all together properly.

Alternatively, maybe convert all amounts to cents first to avoid decimal confusion. Let me try that.

Linda: $27.47 is 27 dollars and 47 cents. So 27*100 + 47 = 2700 + 47 = 2747 cents.

Sherry: $35.23 is 35*100 + 23 = 3500 + 23 = 3523 cents.

June: $37.37 is 37*100 + 37 = 3700 + 37 = 3737 cents.

Connie: $26.52 is 26*100 + 52 = 2600 + 52 = 2652 cents.

Now total cents is 2747 + 3523 + 3737 + 2652. Let me add them step by step.

First, 2747 + 3523. Let me compute that. 2747 + 3523. Let me add 2700 + 3500 = 6200. Then 47 + 23 = 70. So total is 6200 + 70 = 6270 cents.

Next, add 3737. So 6270 + 3737. Let me do 6270 + 3000 = 9270, then +737 = 9270 + 700 = 9970, +37 = 10007 cents.

Then add 2652. So 10007 + 2652. Let me compute that. 10007 + 2000 = 12007, then +652 = 12007 + 600 = 12607, +52 = 12659 cents.

So total money is 12,659 cents. Wait, let me check that again because adding step by step can have errors.

Wait, first step: Linda + Sherry: 2747 + 3523. Let me do 2747 + 3523:

2747 + 3523:

Let me add 2747 + 3523:

Start with 2747 + 3500 = 6247, then +23 = 6270. That's correct.

Then adding June's 3737: 6270 + 3737.

Let me do 6270 + 3737:

Breakdown: 6270 + 3000 = 9270, 9270 + 700 = 9970, 9970 + 37 = 10007. Correct.

Then adding Connie's 2652: 10007 + 2652.

Let me do 10007 + 2000 = 12007, 12007 + 600 = 12607, 12607 + 52 = 12659. Yes, that's 12,659 cents total.

So total money is 12,659 cents. Now, they want to convert as many coins into bills as possible. So the question is, how much money in cents is left after converting as many coins as possible into bills?

Wait, but when you convert coins into bills, you can only convert whole bills. So, for example, if you have a certain number of cents, you can convert as many dollars as possible, and the leftover cents would be the remainder.

But wait, but the problem says they pooled their money together and went to the bank to convert the coins into dollars. So maybe they have a mix of coins and bills? Wait, but the problem says they "converted the coins into dollars." Wait, maybe they had all their money in coins, and they want to convert all coins into bills, and find out how much is left in cents? But if they convert all coins into bills, then there should be no cents left. Wait, that doesn't make sense. Maybe there's a misunderstanding here.

Wait, let me re-read the problem:

"After the sale, they pooled their money together and went to the bank to convert the coins into dollars. How much money, in cents, was left after they converted as many of the coins into bills as they could?"

Wait, so maybe they had some coins and some bills, and they wanted to convert the coins into bills. But the problem says they "pooled their money together" which might mean that they had all their money in coins? Or maybe they had some coins and some bills, but they wanted to convert the coins into bills. But the problem says "convert the coins into dollars", so maybe they had coins, and they wanted to exchange them for bills, and the question is how much money in cents was left after converting as many coins as possible into bills. But if they converted all coins into bills, then there would be no coins left. But the question says "how much money, in cents, was left after they converted as many of the coins into bills as they could?"

Wait, maybe they had some coins and some bills, and they wanted to convert the coins into bills, but the total amount of money they had was in dollars and cents. Wait, but the problem says they "pooled their money together" which is the total amount of money they earned, which is in dollars and cents. But maybe they had that money in coins? Or maybe they had some coins and some bills, but the problem is not clear. Wait, but the problem says "convert the coins into dollars", so maybe they had coins, and they wanted to convert them into dollars, and the question is how much money in cents was left after converting as many coins as possible into bills. But if they converted all coins into bills, then there would be no coins left. But the problem says "how much money, in cents, was left after they converted as many of the coins into bills as they could?"

Wait, maybe the problem is that they had all their money in coins, and they wanted to convert them into bills, but they couldn't convert all of them because they couldn't make exact bills? Wait, but if you have a certain amount of cents, you can convert as many dollars as possible, and the leftover cents would be the remainder. For example, if you have 123 cents, you can convert 1 dollar (100 cents) and have 23 cents left. So maybe the question is, if they had all their money in cents (i.e., all coins), then converting as many as possible into dollars, how much is left in cents?

But the problem says they "pooled their money together" which is the total amount they earned, which is in dollars and cents. But maybe they had that amount in coins? For example, if they had $27.47, that would be 27 dollars and 47 cents, but if they had that in coins, then they would have 27*100 + 47 = 2747 cents. But if they had all their money in coins, then the total is 12,659 cents as calculated before. Then converting as many coins into bills as possible would mean converting as many dollars as possible, and the leftover cents would be the remainder when dividing total cents by 100. Wait, but that would be 12,659 divided by 100, which is 126 dollars and 59 cents. So the leftover cents would be 59 cents. But that seems too straightforward. But maybe that's the answer? But why would the problem mention converting coins into dollars? Maybe there's a trick here.

Wait, but maybe they had some coins and some bills, and they wanted to convert the coins into bills, but the problem is not clear. However, given that they earned money, which is in dollars and cents, but the problem says they converted the coins into dollars. Maybe they had all their money in coins, so the total amount is 12,659 cents. Then converting as many coins into bills as possible would mean converting 12,659 cents into dollars. Since 1 dollar is 100 cents, the number of dollars is 12,659 divided by 100, which is 126 dollars with a remainder of 59 cents. Therefore, the leftover cents would be 59 cents. Therefore, the answer is 59 cents. But that seems too easy. However, maybe there's more to it?

Wait, but maybe the problem is not about converting all the cents into dollars, but converting coins into bills, meaning that they have coins of different denominations, and they want to exchange them for bills, but the question is how much is left in cents after exchanging as much as possible. But if they have all their money in coins, then converting into bills would just be exchanging all coins for bills, leaving no coins. But the problem says "how much money, in cents, was left after they converted as many of the coins into bills as they could?" So maybe they had some coins and some bills, and they wanted to convert the coins into bills, but the problem is not clear. However, given the problem statement, maybe the answer is 59 cents. But let me check again.

Alternatively, maybe the problem is that they had the total amount of money, which is in dollars and cents, and they wanted to convert the coins (i.e., the cents part) into dollars. Wait, but if they have, for example, $27.47, that is 27 dollars and 47 cents. If they wanted to convert the coins (the 47 cents) into dollars, but you can't convert 47 cents into a dollar. So maybe the question is about converting the cents part into dollars? But that doesn't make sense. Alternatively, maybe they had all their money in coins, and they wanted to convert as much as possible into bills, so the leftover is the cents part. Which would be 59 cents. But maybe the problem is more complicated?

Wait, let me check the problem again:

"Linda, Sherry, June, and Connie walked around their neighborhoods selling girl scout cookies. Linda earned $27.47, Sherry earned $35.23, June earned $37.37, and Connie earned $26.52. After the sale, they pooled their money together and went to the bank to convert the coins into dollars. How much money, in cents, was left after they converted as many of the coins into bills as they could?"

So they pooled their money together, which is the total amount of money they earned, which is in dollars and cents. Then they went to the bank to convert the coins into dollars. So maybe they had some coins and some bills, and they wanted to convert the coins into bills. But the problem is, how much money in cents was left after converting as many coins as possible into bills. But if they had some coins, converting them into bills would leave them with bills, but the question is about cents left. Wait, maybe they had all their money in coins, so they had 12,659 cents. Then converting as many coins into bills as possible would mean converting 12,659 cents into 126 dollars and 59 cents. Therefore, the leftover is 59 cents. Therefore, the answer is 59 cents. But maybe the problem is expecting something else?

Alternatively, maybe they had the total amount in dollars and cents, but they wanted to convert the coins (i.e., the cents part) into dollars. But that would be impossible unless they have more coins. Wait, but if they have, for example, $27.47, which is 27 dollars and 47 cents. If they wanted to convert the coins (the 47 cents) into dollars, they would need to exchange them for dollars, but since 47 cents is less than a dollar, they can't. Therefore, the leftover would be 47 cents. But if they had multiple people, maybe they can combine their coins? Wait, but the total amount is 12,659 cents. So if they convert as many as possible into dollars, they can get 126 dollars, and 59 cents left. Therefore, the answer is 59 cents. But why is the problem mentioning converting coins into dollars? Maybe the problem is that they had all their money in coins, and they wanted to convert them into bills, and the leftover is the cents. So the answer is 59 cents. But maybe I need to check if there's a different interpretation.

Alternatively, maybe the problem is that they had the total amount in dollars and cents, but they wanted to convert the coins (i.e., the cents part) into dollars, but since they can't do that, they have to keep the cents. But that would be the same as the cents part. Wait, but the total amount is $100.07? Wait, earlier when I added the dollars, I thought maybe it was $100.07. Let me check that again.

Original amounts:

Linda: $27.47

Sherry: $35.23

June: $37.37

Connie: $26.52

Adding them up:

Let me add them in dollars and cents separately.

Dollars: 27 + 35 + 37 + 26. Let me compute that:

27 + 35 = 62; 37 + 26 = 63. Then 62 + 63 = 125 dollars.

Cents: 47 + 23 + 37 + 52. Let me compute that:

47 + 23 = 70; 37 + 52 = 89. Then 70 + 89 = 159 cents. 159 cents is $1.59. Therefore, total amount is 125 + 1.59 = $126.59. Wait, so total amount is $126.59. Which is 126 dollars and 59 cents. Therefore, if they converted as many coins into bills as possible, they would have 126 dollars and 59 cents left. But the question says "how much money, in cents, was left after they converted as many of the coins into bills as they could?" So if they converted all the coins into bills, then they would have 126 dollars and 59 cents. But the question is about cents left. Wait, but if they converted all the coins into bills, then they would have no coins left. But the total amount is $126.59, which is 126 dollars and 59 cents. So if they converted all the coins into bills, they would have 126 bills of $1 and 59 cents. But the question is asking for how much money in cents was left. If they converted as many coins into bills as possible, then the leftover would be the cents part, which is 59 cents. Therefore, the answer is 59 cents. Therefore, the answer is 59. But let me check again.

Alternatively, maybe the problem is that they had all their money in coins, so total is 12,659 cents. Converting as many as possible into bills would be 126 dollars (which is 12,600 cents) and 59 cents left. Therefore, the answer is 59 cents. So both methods give the same answer. Therefore, the answer is 59 cents. Therefore, the answer is \boxed{59}.

But let me check once more. Maybe there's a trick here. For example, maybe they had different coins, and when converting into bills, they can't just take the cents part. For example, if they have coins of different denominations, maybe they can't make exact change? But the problem says "convert the coins into dollars", which is a bit ambiguous. But if they have all their money in coins, then converting into bills would just be exchanging all coins for bills, leaving the cents part. But if they have, for example, 59 cents, they can't convert that into a dollar, so that's the leftover. Therefore, the answer is 59 cents. Therefore, I think the answer is 59.

But let me check the total cents again. Earlier, I converted each person's earnings into cents and added them up to get 12,659 cents. Let me check that again:

Linda: 27.47 = 27*100 + 47 = 2747

Sherry: 35.23 = 35*100 + 23 = 3523

June: 37.37 = 37*100 + 37 = 3737

Connie: 26.52 = 26*100 + 52 = 2652

Adding them:

2747 + 3523 = 6270

6270 + 3737 = 10007

10007 + 2652 = 12659. Yes, that's correct. 12,659 cents.

Divide by 100: 12,659 / 100 = 126.59. So 126 dollars and 59 cents. Therefore, the leftover cents is 59. So the answer is 59 cents. Therefore, the answer is \boxed{59}.

But let me think again. Maybe the problem is not about converting all the money into bills, but converting coins into bills, meaning that they had some coins and some bills, and they wanted to convert the coins into bills, but the problem is not clear. However, given the problem statement, the most logical interpretation is that they had all their money in coins, and they wanted to convert as much as possible into bills, leaving some cents. Therefore, the answer is 59 cents.

Alternatively, maybe the problem is that they had the total amount in dollars and cents, and they wanted to convert the coins (i.e., the cents part) into dollars, but since they can't do that, they have to keep the cents. But that would be the same as the cents part. However, the total amount is $126.59, so the cents part is 59 cents. Therefore, the answer is 59 cents. So either way, the answer is 59. Therefore, I think the answer is 59.

**Final Answer**
\boxed{59}

To determine how much money, in cents, was left after converting as many coins as possible into bills, we begin by calculating the total amount of money the four girls earned.

---

### Step 1: Convert Each Person's Earnings to Cents

We convert each person's earnings from dollars to cents to avoid decimal confusion:

- **Linda**: $27.47 = 27 \times 100 + 47 = 2747$ cents
- **Sherry**: $35.23 = 35 \times 100 + 23 = 3523$ cents
- **June**: $37.37 = 37 \times 100 + 37 = 3737$ cents
- **Connie**: $26.52 = 26 \times 100 + 52 = 2652$ cents

---

### Step 2: Add All the Cents Together

Now, we sum up all the cents:

$$
2747 + 3523 + 3737 + 2652 = 12659 \text{ cents}
$$

---

### Step 3: Convert as Many Cents as Possible into Dollars

Since 1 dollar = 100 cents, we divide the total number of cents by 100 to find how many full dollars can be made:

$$
12659 \div 100 = 126 \text{ dollars with a remainder of } 59 \text{ cents}
$$

This means that after converting as many coins as possible into bills (i.e., full dollars), **59 cents** remain unconverted.

---

### Final Answer

$$
\boxed{59}
$$

Solution:
Instead of adding large numbers together, we can find the residue for each person for easier computation. We convert the amount they earned to cents and find the modulo $100$ for each.  \begin{align*}
2747 &\equiv 47 \pmod{100},\\
3523 &\equiv 23 \pmod{100},\\
3737 &\equiv 37 \pmod{100},\\
2652 &\equiv 52 \pmod{100}
\end{align*}We want to find the modulo $100$ of the total number of cents. We can add the separate residues to get $$47+23+37+52 \equiv 159 \equiv 59 \pmod{100}$$Therefore, they have $\boxed{59}$ cents left after converting as much of the money into bills as possible.