Explanation:
Okay, so I need to figure out the cost of a pencil in cents. The problem says that three pencils and a jumbo eraser cost $1.24, and five pencils and a jumbo eraser cost $1.82. Hmm, let me think about how to approach this.

First, I remember that when you have two equations with two variables, you can solve them using substitution or elimination. Since both equations have a jumbo eraser, maybe I can subtract one equation from the other to eliminate the eraser cost. Let me write down the equations to make it clearer.

Let me denote the cost of a pencil as "p" cents and the cost of a jumbo eraser as "e" cents. Wait, but the problem mentions dollars, so maybe I should convert everything to cents to avoid dealing with decimals? That might make calculations easier. Let me check the original prices:

Three pencils and a jumbo eraser cost $1.24. Converting that to cents, it's 124 cents. Similarly, five pencils and a jumbo eraser cost $1.82, which is 182 cents.

So, the equations would be:

1) 3p + e = 124

2) 5p + e = 182

Okay, so now I have two equations with two variables. If I subtract the first equation from the second, the e terms should cancel out. Let me try that.

Subtracting equation 1 from equation 2:

(5p + e) - (3p + e) = 182 - 124

Let me compute the left side:

5p + e - 3p - e = (5p - 3p) + (e - e) = 2p + 0 = 2p

On the right side, 182 - 124. Let me calculate that. 182 minus 124... 182 - 100 is 82, then minus 24 more is 58. So 58.

Therefore, 2p = 58. Then, solving for p, divide both sides by 2: p = 58 / 2 = 29.

Wait, so the cost of a pencil is 29 cents? Let me check if that makes sense. Let me plug this back into one of the original equations to find the cost of the eraser and verify.

Let me use equation 1: 3p + e = 124. If p is 29, then 3*29 is 87. So 87 + e = 124. Therefore, e = 124 - 87 = 37. So the eraser would cost 37 cents.

Let me check with equation 2: 5p + e = 5*29 + 37. 5*29 is 145, plus 37 is 182. Which matches the second equation. Perfect! So that seems to check out.

But wait, the problem says "In cents, what is the cost of a pencil?" So since we found p = 29 cents, that should be the answer. But let me just make sure I didn't make any calculation errors.

Let me recheck the subtraction step. Original equations:

Equation 1: 3p + e = 124

Equation 2: 5p + e = 182

Subtract equation 1 from equation 2:

(5p + e) - (3p + e) = 182 - 124

Which simplifies to 2p = 58, so p = 29. Yes, that's correct.

Alternatively, maybe I can solve using substitution. Let me try that method to confirm.

From equation 1: e = 124 - 3p

Then substitute into equation 2:

5p + (124 - 3p) = 182

Simplify: 5p + 124 - 3p = 182 => 2p + 124 = 182 => 2p = 182 - 124 = 58 => p = 29. Same result.

So that's consistent. Therefore, the cost of a pencil is 29 cents.

But just to be thorough, let me check if there's any possibility that the problem expects something else. For example, maybe the prices are in dollars and cents, but the question says "No prices include tax," so we don't have to worry about tax. Also, the problem says "In cents," so the answer should be in cents, which we have as 29.

Wait, but let me check if converting dollars to cents was done correctly. $1.24 is 124 cents, yes. $1.82 is 182 cents, yes. So that's correct.

Alternatively, if I had kept the prices in dollars, would that have made a difference? Let me try that approach as well.

Let me denote p as dollars and e as dollars. Then:

Equation 1: 3p + e = 1.24

Equation 2: 5p + e = 1.82

Subtract equation 1 from equation 2:

(5p + e) - (3p + e) = 1.82 - 1.24 => 2p = 0.58 => p = 0.29 dollars. Which is 29 cents. So same answer.

Therefore, regardless of whether I convert to cents first or work in dollars, I get the same result.

So, I think that's solid. The answer is 29 cents.

But just to make sure, let me think if there's any other way this could be interpreted. For example, maybe the problem is trying to trick me with something? Like, maybe the eraser is jumbo, so maybe it's a different price? But no, the problem says "jumbo eraser" but doesn't mention any different pricing for jumbo, so I think it's just a name, not affecting the price.

Alternatively, maybe there's a typo in the problem? But assuming the problem is stated correctly, then my answer should be correct.

Another check: if a pencil is 29 cents, then three pencils are 87 cents, plus eraser 37 cents, total 124 cents, which is $1.24. Five pencils would be 145 cents, plus eraser 37 cents, total 182 cents, which is $1.82. Perfect.

So, I think that's correct. Therefore, the cost of a pencil is 29 cents.

**Final Answer**
The cost of a pencil is \boxed{29} cents.

To determine the cost of a pencil, we are given two equations based on the total cost of pencils and a jumbo eraser:

- Three pencils and a jumbo eraser cost \$1.24.
- Five pencils and a jumbo eraser cost \$1.82.

---

### Step 1: Convert Dollars to Cents

To simplify calculations, we convert the dollar amounts to cents:

- \$1.24 = 124 cents
- \$1.82 = 182 cents

Let:
- $ p $ = cost of one pencil (in cents)
- $ e $ = cost of one jumbo eraser (in cents)

We can now write the system of equations:

$$
\begin{align*}
(1) & \quad 3p + e = 124 \\
(2) & \quad 5p + e = 182
\end{align*}
$$

---

### Step 2: Eliminate the Eraser Cost

Subtract equation (1) from equation (2):

$$
(5p + e) - (3p + e) = 182 - 124
$$

Simplify both sides:

$$
2p = 58
$$

Solve for $ p $:

$$
p = \frac{58}{2} = 29
$$

---

### Step 3: Verify the Solution

Plug $ p = 29 $ back into equation (1) to find $ e $:

$$
3(29) + e = 124 \Rightarrow 87 + e = 124 \Rightarrow e = 37
$$

Now check in equation (2):

$$
5(29) + 37 = 145 + 37 = 182
$$

Both equations are satisfied, confirming the solution is correct.

---

### Final Answer

The cost of a pencil is:

$$
\boxed{29}
$$

[Assistant]
Explanation:
Okay, so I need to figure out the cost of a pencil in cents. The problem says that three pencils and a jumbo eraser cost $1.24, and five pencils and a jumbo eraser cost $1.82. Hmm, let me think about how to approach this.

First, I remember that when you have two equations with two variables, you can solve them using substitution or elimination. Since both equations have a jumbo eraser, maybe I can subtract one equation from the other to eliminate the eraser cost. Let me write down the equations to make it clearer.

Let me denote the cost of a pencil as "p" cents and the cost of a jumbo eraser as "e" cents. Wait, but the problem mentions dollars, so maybe I should convert everything to cents to avoid dealing with decimals? That might make calculations easier. Let me check the original prices:

Three pencils and a jumbo eraser cost $1.24. Converting that to cents, it's 124 cents. Similarly, five pencils and a jumbo eraser cost $1.82, which is 182 cents.

So, the equations would be:

1) 3p + e = 124

2) 5p + e = 182

Okay, so now I have two equations with two variables. If I subtract the first equation from the second, the e terms should cancel out. Let me try that.

Subtracting equation 1 from equation 2:

(5p + e) - (3p + e) = 182 - 124

Let me compute the left side:

5p + e - 3p - e = (5p - 3p) + (e - e) = 2p + 0 = 2p

On the right side, 182 - 124. Let me calculate that. 182 minus 124... 182 - 100 is 82, then minus 24 more is 58. So 58.

Therefore, 2p = 58. Then, solving for p, divide both sides by 2: p = 58 / 2 = 29.

Wait, so the cost of a pencil is 29 cents? Let me check if that makes sense. Let me plug this back into one of the original equations to find the cost of the eraser and verify.

Let me use equation 1: 3p + e = 124. If p is 29, then 3*29 is 87. So 87 + e = 124. Therefore, e = 124 - 87 = 37. So the eraser would cost 37 cents.

Let me check with equation 2: 5p + e = 5*29 + 37. 5*29 is 145, plus 37 is 182. Which matches the second equation. Perfect! So that seems to check out.

But wait, the problem says "In cents, what is the cost of a pencil?" So since we found p = 29 cents, that should be the answer. But let me just make sure I didn't make any calculation errors.

Let me recheck the subtraction step. Original equations:

Equation 1: 3p + e = 124

Equation 2: 5p + e = 182

Subtract equation 1 from equation 2:

(5p + e) - (3p + e) = 182 - 124

Which simplifies to 2p = 58, so p = 29. Yes, that's correct.

Alternatively, maybe I can solve using substitution. Let me try that method to confirm.

From equation 1: e = 124 - 3p

Then substitute into equation 2:

5p + (124 - 3p) = 182

Simplify: 5p + 124 - 3p = 182 => 2p + 124 = 182 => 2p = 182 - 124 = 58 => p = 29. Same result.

So that's consistent. Therefore, the cost of a pencil is 29 cents.

But just to be thorough, let me check if there's any possibility that the problem expects something else. For example, maybe the prices are in dollars and cents, but the question says "No prices include tax," so we don't have to worry about tax. Also, the problem says "In cents," so the answer should be in cents, which we have as 29.

Wait, but let me check if converting dollars to cents was done correctly. $1.24 is 124 cents, yes. $1.82 is 182 cents, yes. So that's correct.

Alternatively, if I had kept the prices in dollars, would that have made a difference? Let me try that approach as well.

Let me denote p as dollars and e as dollars. Then:

Equation 1: 3p + e = 1.24

Equation 2: 5p + e = 1.82

Subtract equation 1 from equation 2:

(5p + e) - (3p + e) = 1.82 - 1.24 => 2p = 0.58 => p = 0.29 dollars. Which is 29 cents. So same answer.

Therefore, regardless of whether I convert to cents first or work in dollars, I get the same result.

So, I think that's solid. The answer is 29 cents.

But just to make sure, let me think if there's any other way this could be interpreted. For example, maybe the problem is trying to trick me with something? Like, maybe the eraser is jumbo, so maybe it's a different price? But no, the problem says "jumbo eraser" but doesn't mention any different pricing for jumbo, so I think it's just a name, not affecting the price.

Alternatively, maybe there's a typo in the problem? But assuming the problem is stated correctly, then my answer should be correct.

Another check: if a pencil is 29 cents, then three pencils are 87 cents, plus eraser 37 cents, total 124 cents, which is $1.24. Five pencils would be 145 cents, plus eraser 37 cents, total 182 cents, which is $1.82. Perfect.

So, I think that's correct. Therefore, the cost of a pencil is 29 cents.

**Final Answer**
The cost of a pencil is \boxed{29} cents.

To determine the cost of a pencil, we are given two equations based on the total cost of pencils and a jumbo eraser:

- Three pencils and a jumbo eraser cost \$1.24.
- Five pencils and a jumbo eraser cost \$1.82.

---

### Step 1: Convert Dollars to Cents

To simplify calculations, we convert the dollar amounts to cents:

- \$1.24 = 124 cents
- \$1.82 = 182 cents

Let:
- $ p $ = cost of one pencil (in cents)
- $ e $ = cost of one jumbo eraser (in cents)

We can now write the system of equations:

$$
\begin{align*}
(1) & \quad 3p + e = 124 \\
(2) & \quad 5p + e = 182
\end{align*}
$$

---

### Step 2: Eliminate the Eraser Cost

Subtract equation (1) from equation (2):

$$
(5p + e) - (3p + e) = 182 - 124
$$

Simplify both sides:

$$
2p = 58
$$

Solve for $ p $:

$$
p = \frac{58}{2} = 29
$$

---

### Step 3: Verify the Solution

Plug $ p = 29 $ back into equation (1) to find $ e $:

$$
3(29) + e = 124 \Rightarrow 87 + e = 124 \Rightarrow e = 37
$$

Now check in equation (2):

$$
5(29) + 37 = 145 + 37 = 182
$$

Both equations are satisfied, confirming the solution is correct.

---

### Final Answer

The cost of a pencil is:

$$
\boxed{29}
$$

Solution:
Call the price of a pencil $p$ and the price of a jumbo eraser $e$, in terms of cents. We can use the following system of equations to represent the information given: \begin{align*}
3p + e &= 124 \\
5p + e &= 182 \\
\end{align*} Subtracting the first equation from the second gives $2p = 58$, or $p = 29$. Thus, the cost of a pencil is $\boxed{29}$ cents.


Explanation:
Okay, so I need to figure out what 9 divided by 2 is as a decimal. Let me think. Hmm, I remember that dividing by 2 is the same as multiplying by 0.5, right? So maybe I can just do 9 times 0.5? Let me check that.

Wait, 9 times 0.5 is the same as half of 9. Half of 9 is 4.5, so is that the answer? But let me make sure I'm not making a mistake here. Maybe I should do the division step by step.

Alright, let's try dividing 9 by 2. So 2 goes into 9 how many times? Well, 2 times 4 is 8, which is less than 9. Then 2 times 5 is 10, which is too much. So it goes in 4 times with a remainder. So 4 times 2 is 8, subtract that from 9, and I get a remainder of 1.

Now, since we're dealing with decimals, I can add a decimal point and a zero to the 9, making it 9.0. Then bring down the zero, so now it's 10. Now, 2 goes into 10 exactly 5 times. So that gives me 4.5.

Wait, so that's the same as before. So 9 divided by 2 is 4.5. Let me verify with another method. Maybe converting the fraction to a decimal by expanding it.

The fraction is 9/2. If I multiply numerator and denominator by 5, that would make the denominator 10, right? Because 2 times 5 is 10. So if I do that, the numerator becomes 9 times 5, which is 45. So 45/10 is 4.5. Yep, that checks out.

Another way to think about it is that 1/2 is 0.5, so 9 times 0.5 is 4.5. All these methods are giving me the same answer. Maybe I can also think in terms of money. If I have $9 and I split it into two equal parts, each part would be $4.50. That makes sense.

Is there any chance I could have messed up the decimal placement? Let me check. If I have 9 divided by 2, the decimal point comes after the 4, so 4.5. If I had put the decimal in the wrong place, like 45, that would be wrong. But since 9 divided by 2 is less than 9, it should be 4.5.

Alternatively, if I use long division properly. Let me write it out:

2 | 9.0

2 goes into 9 four times (4*2=8). Subtract 8 from 9, get 1. Bring down the 0, making it 10. 2 goes into 10 five times. So the answer is 4.5.

Yep, that's consistent. So I think I can be confident that 9/2 as a decimal is 4.5.

Wait, just to make sure, maybe I can use a calculator? But since I don't have one here, I need to rely on my calculations. Let me do another check.

If I take 4.5 and multiply it by 2, I should get back to 9. 4.5 times 2 is 9. Exactly. So that confirms it.

Therefore, after all these different methods and checks, I can be sure that 9 divided by 2 is indeed 4.5.

**Final Answer**
The decimal representation of $\frac{9}{2}$ is \boxed{4.5}.

To express the fraction $\frac{9}{2}$ as a decimal, we can approach the problem through division or by converting the fraction into a decimal form using known equivalences.

---

### Step-by-Step Division Method

We begin by performing the division $9 \div 2$:

1. **Divide 9 by 2**:
   - 2 goes into 9 **4 times** (since $2 \times 4 = 8$).
   - Subtract 8 from 9, leaving a **remainder of 1**.

2. **Add a decimal point and a zero** to continue the division:
   - Bring down a zero, making the new number **10**.
   - 2 goes into 10 **5 times** (since $2 \times 5 = 10$).

3. **Final result**:
   - The quotient is **4.5**.

---

### Alternative Method: Fraction to Decimal Conversion

We can also convert the fraction $\frac{9}{2}$ into a decimal by multiplying both the numerator and denominator by 5 to make the denominator 10:

$$
\frac{9}{2} = \frac{9 \times 5}{2 \times 5} = \frac{45}{10} = 4.5
$$

---

### Verification

To ensure the result is correct, we can reverse the process:

- Multiply $4.5$ by 2:
  $$
  4.5 \times 2 = 9
  $$
  This confirms that the decimal representation is accurate.

---

### Final Answer

$$
\boxed{4.5}
$$

[Assistant]
Explanation:
Okay, so I need to figure out what 9 divided by 2 is as a decimal. Let me think. Hmm, I remember that dividing by 2 is the same as multiplying by 0.5, right? So maybe I can just do 9 times 0.5? Let me check that.

Wait, 9 times 0.5 is the same as half of 9. Half of 9 is 4.5, so is that the answer? But let me make sure I'm not making a mistake here. Maybe I should do the division step by step.

Alright, let's try dividing 9 by 2. So 2 goes into 9 how many times? Well, 2 times 4 is 8, which is less than 9. Then 2 times 5 is 10, which is too much. So it goes in 4 times with a remainder. So 4 times 2 is 8, subtract that from 9, and I get a remainder of 1.

Now, since we're dealing with decimals, I can add a decimal point and a zero to the 9, making it 9.0. Then bring down the zero, so now it's 10. Now, 2 goes into 10 exactly 5 times. So that gives me 4.5.

Wait, so that's the same as before. So 9 divided by 2 is 4.5. Let me verify with another method. Maybe converting the fraction to a decimal by expanding it.

The fraction is 9/2. If I multiply numerator and denominator by 5, that would make the denominator 10, right? Because 2 times 5 is 10. So if I do that, the numerator becomes 9 times 5, which is 45. So 45/10 is 4.5. Yep, that checks out.

Another way to think about it is that 1/2 is 0.5, so 9 times 0.5 is 4.5. All these methods are giving me the same answer. Maybe I can also think in terms of money. If I have $9 and I split it into two equal parts, each part would be $4.50. That makes sense.

Is there any chance I could have messed up the decimal placement? Let me check. If I have 9 divided by 2, the decimal point comes after the 4, so 4.5. If I had put the decimal in the wrong place, like 45, that would be wrong. But since 9 divided by 2 is less than 9, it should be 4.5.

Alternatively, if I use long division properly. Let me write it out:

2 | 9.0

2 goes into 9 four times (4*2=8). Subtract 8 from 9, get 1. Bring down the 0, making it 10. 2 goes into 10 five times. So the answer is 4.5.

Yep, that's consistent. So I think I can be confident that 9/2 as a decimal is 4.5.

Wait, just to make sure, maybe I can use a calculator? But since I don't have one here, I need to rely on my calculations. Let me do another check.

If I take 4.5 and multiply it by 2, I should get back to 9. 4.5 times 2 is 9. Exactly. So that confirms it.

Therefore, after all these different methods and checks, I can be sure that 9 divided by 2 is indeed 4.5.

**Final Answer**
The decimal representation of $\frac{9}{2}$ is \boxed{4.5}.

To express the fraction $\frac{9}{2}$ as a decimal, we can approach the problem through division or by converting the fraction into a decimal form using known equivalences.

---

### Step-by-Step Division Method

We begin by performing the division $9 \div 2$:

1. **Divide 9 by 2**:
   - 2 goes into 9 **4 times** (since $2 \times 4 = 8$).
   - Subtract 8 from 9, leaving a **remainder of 1**.

2. **Add a decimal point and a zero** to continue the division:
   - Bring down a zero, making the new number **10**.
   - 2 goes into 10 **5 times** (since $2 \times 5 = 10$).

3. **Final result**:
   - The quotient is **4.5**.

---

### Alternative Method: Fraction to Decimal Conversion

We can also convert the fraction $\frac{9}{2}$ into a decimal by multiplying both the numerator and denominator by 5 to make the denominator 10:

$$
\frac{9}{2} = \frac{9 \times 5}{2 \times 5} = \frac{45}{10} = 4.5
$$

---

### Verification

To ensure the result is correct, we can reverse the process:

- Multiply $4.5$ by 2:
  $$
  4.5 \times 2 = 9
  $$
  This confirms that the decimal representation is accurate.

---

### Final Answer

$$
\boxed{4.5}
$$

Solution: We can solve this problem by division.  Alternatively, we can multiply the numerator and denominator by 5, yielding $\frac{45}{10}$.  Since dividing a number by 10 shifts the decimal point to the left by one place, this yields $\boxed{4.5}$.


Explanation:
Okay, so I need to convert the fraction 57/160 into a terminating decimal. Hmm, let me think about how to do that. First, I remember that a fraction will terminate if the denominator, after simplifying the fraction, has only 2 and/or 5 as prime factors. Let me check if 160 meets that condition.

Breaking down 160 into prime factors: 160 divided by 2 is 80, divided by 2 again is 40, again is 20, again is 10, again is 5. So that's 2^5 * 5^1. Yep, only 2s and 5s. So that means 57/160 should be a terminating decimal. Good.

Now, how do I convert it? Well, one way is to divide 57 by 160 directly. But maybe there's a smarter way? Let me think. Since the denominator is 160, which is 16 * 10, or 2^5 * 5^1. To make it a power of 10, I can multiply numerator and denominator by something to get the denominator to be 10^n. Let me try that.

The denominator is 160. Let me see, 160 is 16 * 10. If I want to make it 10^something, maybe I need to multiply by 5? Wait, 160 * 5 = 800, which is 8 * 100. Hmm, not sure. Alternatively, since 160 is 2^5 * 5^1, to make it 2^something * 5^something, so that it's 10^something. The highest power between 2 and 5 is 5. So if I multiply numerator and denominator by 5^4, which is 625, then denominator becomes 2^5 * 5^1 * 5^4 = 2^5 * 5^5 = 10^5. That would make the denominator 10^5, which is 100000. Let me check that.

So 160 * 5^4 = 160 * 625. Let me compute that. 160 * 625. Well, 160 * 625. Since 625 is 5^4, and 160 is 16*10. Maybe I can think of 625 * 160. Let me compute 625 * 160. 625 * 100 = 62500, 625 * 60 = 37500. So total is 62500 + 37500 = 100000. Oh, that's nice! So denominator becomes 100000. Therefore, numerator is 57 * 625. Let me compute that.

57 * 625. Hmm. Let me break it down. 57 * 600 = 34,200. 57 * 25 = 1,425. So total is 34,200 + 1,425 = 35,625. Therefore, 57/160 = 35,625 / 100,000. Which is 0.35625. Let me check that. Because 35,625 divided by 100,000 moves the decimal five places to the left: 0.35625. So that's the decimal. Let me verify this by another method to be sure.

Alternatively, I can perform the division 57 divided by 160. Let me do that. Since 57 is less than 160, it's 0. something. Let me write it as 57.00000 divided by 160.

First, 160 goes into 570 three times (since 160*3=480). Subtract 480 from 570, get 90. Bring down a zero: 900. 160 goes into 900 five times (160*5=800). Subtract 800 from 900, get 100. Bring down a zero: 1000. 160 goes into 1000 six times (160*6=960). Subtract 960 from 1000, get 40. Bring down a zero: 400. 160 goes into 400 two times (160*2=320). Subtract 320 from 400, get 80. Bring down a zero: 800. 160 goes into 800 five times (160*5=800). Subtract 800, get 0. So the decimal is 0.35625. Yep, same result. So that's correct.

Alternatively, maybe I can use another approach. Since 1/160 is equal to... Well, 1/160 is 0.00625. Wait, let me check that. 1/160. Since 1/10 is 0.1, 1/16 is 0.0625. So 1/160 is 1/(16*10) = (1/16)/10 = 0.0625 / 10 = 0.00625. So 1/160 is 0.00625. Then 57/160 is 57 * 0.00625. Let me compute that. 57 * 0.00625. Let me break it down. 50 * 0.00625 = 0.3125, and 7 * 0.00625 = 0.04375. Adding them together: 0.3125 + 0.04375 = 0.35625. Same result again. So that's consistent.

So all methods lead to 0.35625. Therefore, the decimal is terminating, as expected, and the value is 0.35625. Let me just check once more with another method. Maybe convert the fraction step by step.

Alternatively, since 160 is 2^5 * 5, and to convert to decimal, we can multiply numerator and denominator by 5^4 to get denominator as 10^5. Which we did earlier. So that gives 57*625 / 100000 = 35625 / 100000 = 0.35625. Yep. So that's correct.

Alternatively, maybe I can use decimal expansion by multiplying numerator and denominator by something else. For example, 160 is 16 * 10. So 57/160 = 57/(16*10) = (57/16)/10. Let me compute 57 divided by 16 first. 16*3=48, so 57-48=9. So 57/16 = 3 + 9/16. Then divide by 10: 3/10 + 9/(16*10) = 0.3 + 0.05625 = 0.35625. Wait, how did I get 9/(16*10) as 0.05625? Let me check. 9/16 is 0.5625, so divided by 10 is 0.05625. Then adding to 0.3 gives 0.35625. Yep, same answer. So that works too.

So all methods confirm that 57/160 is 0.35625. Therefore, the terminating decimal is 0.35625. Let me just check if there's any possibility of error here. For example, did I miscalculate 57*625? Let me verify that again. 57*625. Let me think of 625 as 500 + 125. So 57*500 = 28,500. 57*125. Let me compute 57*100=5,700; 57*25=1,425. So 5,700 + 1,425 = 7,125. Then total is 28,500 + 7,125 = 35,625. Yes, that's correct. So 35,625 divided by 100,000 is indeed 0.35625.

Another check: 0.35625 multiplied by 160 should give 57. Let me compute that. 0.35625 * 160. Let me break it down. 0.35625 * 100 = 35.625. 0.35625 * 60 = ?

Well, 0.35625 * 60. Let me compute 0.35625 * 6 * 10. 0.35625 * 6: 0.3*6=1.8, 0.05*6=0.3, 0.00625*6=0.0375. Adding them: 1.8 + 0.3 = 2.1 + 0.0375 = 2.1375. Then times 10 is 21.375. So total of 35.625 + 21.375 = 57. Exactly. Perfect. So that confirms that 0.35625 * 160 = 57. Therefore, the decimal is correct.

So, after all these checks, I can be confident that 57/160 as a decimal is 0.35625, which is a terminating decimal. Therefore, the answer is 0.35625.

**Final Answer**
The decimal representation of $\frac{57}{160}$ is \boxed{0.35625}.

To convert the fraction $\frac{57}{160}$ into a **terminating decimal**, we begin by analyzing the denominator.

---

### Step 1: Check for Terminating Decimal

A fraction will have a **terminating decimal** if, after simplifying, its **denominator** has **only the prime factors 2 and/or 5**.

Let’s factor the denominator:

$$
160 = 2^5 \times 5
$$

Since the prime factors are only 2 and 5, the fraction **$\frac{57}{160}$** will indeed result in a **terminating decimal**.

---

### Step 2: Convert to a Decimal

We can convert the fraction by multiplying both the numerator and denominator by a power of 10 to make the denominator a power of 10. Since the denominator is $2^5 \times 5$, we can multiply by $5^4 = 625$ to make the denominator $10^5 = 100000$:

$$
\frac{57}{160} = \frac{57 \times 625}{160 \times 625} = \frac{35625}{100000}
$$

Now, divide the numerator by the denominator:

$$
\frac{35625}{100000} = 0.35625
$$

---

### Step 3: Verification

To confirm, we can perform the division $57 \div 160$ directly:

- $160$ goes into $570$ three times → $3$
- $160$ goes into $900$ five times → $5$
- $160$ goes into $1000$ six times → $6$
- $160$ goes into $400$ two times → $2$
- $160$ goes into $800$ five times → $5$

So the decimal is:

$$
0.35625
$$

This matches our earlier result.

---

### Final Answer

$$
\boxed{0.35625}
$$

[Assistant]
Explanation:
Okay, so I need to convert the fraction 57/160 into a terminating decimal. Hmm, let me think about how to do that. First, I remember that a fraction will terminate if the denominator, after simplifying the fraction, has only 2 and/or 5 as prime factors. Let me check if 160 meets that condition.

Breaking down 160 into prime factors: 160 divided by 2 is 80, divided by 2 again is 40, again is 20, again is 10, again is 5. So that's 2^5 * 5^1. Yep, only 2s and 5s. So that means 57/160 should be a terminating decimal. Good.

Now, how do I convert it? Well, one way is to divide 57 by 160 directly. But maybe there's a smarter way? Let me think. Since the denominator is 160, which is 16 * 10, or 2^5 * 5^1. To make it a power of 10, I can multiply numerator and denominator by something to get the denominator to be 10^n. Let me try that.

The denominator is 160. Let me see, 160 is 16 * 10. If I want to make it 10^something, maybe I need to multiply by 5? Wait, 160 * 5 = 800, which is 8 * 100. Hmm, not sure. Alternatively, since 160 is 2^5 * 5^1, to make it 2^something * 5^something, so that it's 10^something. The highest power between 2 and 5 is 5. So if I multiply numerator and denominator by 5^4, which is 625, then denominator becomes 2^5 * 5^1 * 5^4 = 2^5 * 5^5 = 10^5. That would make the denominator 10^5, which is 100000. Let me check that.

So 160 * 5^4 = 160 * 625. Let me compute that. 160 * 625. Well, 160 * 625. Since 625 is 5^4, and 160 is 16*10. Maybe I can think of 625 * 160. Let me compute 625 * 160. 625 * 100 = 62500, 625 * 60 = 37500. So total is 62500 + 37500 = 100000. Oh, that's nice! So denominator becomes 100000. Therefore, numerator is 57 * 625. Let me compute that.

57 * 625. Hmm. Let me break it down. 57 * 600 = 34,200. 57 * 25 = 1,425. So total is 34,200 + 1,425 = 35,625. Therefore, 57/160 = 35,625 / 100,000. Which is 0.35625. Let me check that. Because 35,625 divided by 100,000 moves the decimal five places to the left: 0.35625. So that's the decimal. Let me verify this by another method to be sure.

Alternatively, I can perform the division 57 divided by 160. Let me do that. Since 57 is less than 160, it's 0. something. Let me write it as 57.00000 divided by 160.

First, 160 goes into 570 three times (since 160*3=480). Subtract 480 from 570, get 90. Bring down a zero: 900. 160 goes into 900 five times (160*5=800). Subtract 800 from 900, get 100. Bring down a zero: 1000. 160 goes into 1000 six times (160*6=960). Subtract 960 from 1000, get 40. Bring down a zero: 400. 160 goes into 400 two times (160*2=320). Subtract 320 from 400, get 80. Bring down a zero: 800. 160 goes into 800 five times (160*5=800). Subtract 800, get 0. So the decimal is 0.35625. Yep, same result. So that's correct.

Alternatively, maybe I can use another approach. Since 1/160 is equal to... Well, 1/160 is 0.00625. Wait, let me check that. 1/160. Since 1/10 is 0.1, 1/16 is 0.0625. So 1/160 is 1/(16*10) = (1/16)/10 = 0.0625 / 10 = 0.00625. So 1/160 is 0.00625. Then 57/160 is 57 * 0.00625. Let me compute that. 57 * 0.00625. Let me break it down. 50 * 0.00625 = 0.3125, and 7 * 0.00625 = 0.04375. Adding them together: 0.3125 + 0.04375 = 0.35625. Same result again. So that's consistent.

So all methods lead to 0.35625. Therefore, the decimal is terminating, as expected, and the value is 0.35625. Let me just check once more with another method. Maybe convert the fraction step by step.

Alternatively, since 160 is 2^5 * 5, and to convert to decimal, we can multiply numerator and denominator by 5^4 to get denominator as 10^5. Which we did earlier. So that gives 57*625 / 100000 = 35625 / 100000 = 0.35625. Yep. So that's correct.

Alternatively, maybe I can use decimal expansion by multiplying numerator and denominator by something else. For example, 160 is 16 * 10. So 57/160 = 57/(16*10) = (57/16)/10. Let me compute 57 divided by 16 first. 16*3=48, so 57-48=9. So 57/16 = 3 + 9/16. Then divide by 10: 3/10 + 9/(16*10) = 0.3 + 0.05625 = 0.35625. Wait, how did I get 9/(16*10) as 0.05625? Let me check. 9/16 is 0.5625, so divided by 10 is 0.05625. Then adding to 0.3 gives 0.35625. Yep, same answer. So that works too.

So all methods confirm that 57/160 is 0.35625. Therefore, the terminating decimal is 0.35625. Let me just check if there's any possibility of error here. For example, did I miscalculate 57*625? Let me verify that again. 57*625. Let me think of 625 as 500 + 125. So 57*500 = 28,500. 57*125. Let me compute 57*100=5,700; 57*25=1,425. So 5,700 + 1,425 = 7,125. Then total is 28,500 + 7,125 = 35,625. Yes, that's correct. So 35,625 divided by 100,000 is indeed 0.35625.

Another check: 0.35625 multiplied by 160 should give 57. Let me compute that. 0.35625 * 160. Let me break it down. 0.35625 * 100 = 35.625. 0.35625 * 60 = ?

Well, 0.35625 * 60. Let me compute 0.35625 * 6 * 10. 0.35625 * 6: 0.3*6=1.8, 0.05*6=0.3, 0.00625*6=0.0375. Adding them: 1.8 + 0.3 = 2.1 + 0.0375 = 2.1375. Then times 10 is 21.375. So total of 35.625 + 21.375 = 57. Exactly. Perfect. So that confirms that 0.35625 * 160 = 57. Therefore, the decimal is correct.

So, after all these checks, I can be confident that 57/160 as a decimal is 0.35625, which is a terminating decimal. Therefore, the answer is 0.35625.

**Final Answer**
The decimal representation of $\frac{57}{160}$ is \boxed{0.35625}.

To convert the fraction $\frac{57}{160}$ into a **terminating decimal**, we begin by analyzing the denominator.

---

### Step 1: Check for Terminating Decimal

A fraction will have a **terminating decimal** if, after simplifying, its **denominator** has **only the prime factors 2 and/or 5**.

Let’s factor the denominator:

$$
160 = 2^5 \times 5
$$

Since the prime factors are only 2 and 5, the fraction **$\frac{57}{160}$** will indeed result in a **terminating decimal**.

---

### Step 2: Convert to a Decimal

We can convert the fraction by multiplying both the numerator and denominator by a power of 10 to make the denominator a power of 10. Since the denominator is $2^5 \times 5$, we can multiply by $5^4 = 625$ to make the denominator $10^5 = 100000$:

$$
\frac{57}{160} = \frac{57 \times 625}{160 \times 625} = \frac{35625}{100000}
$$

Now, divide the numerator by the denominator:

$$
\frac{35625}{100000} = 0.35625
$$

---

### Step 3: Verification

To confirm, we can perform the division $57 \div 160$ directly:

- $160$ goes into $570$ three times → $3$
- $160$ goes into $900$ five times → $5$
- $160$ goes into $1000$ six times → $6$
- $160$ goes into $400$ two times → $2$
- $160$ goes into $800$ five times → $5$

So the decimal is:

$$
0.35625
$$

This matches our earlier result.

---

### Final Answer

$$
\boxed{0.35625}
$$

Solution: A terminating decimal can be written in the form $\frac{a}{10^b}$, where $a$ and $b$ are integers. So we try to get a denominator of the form $10^b$: $$\frac{57}{160}=\frac{57}{2^5\cdot5}\cdot\frac{5^4}{5^4}=\frac{57\cdot5^4}{10^5}=\frac{35625}{10^5}=\boxed{.35625}.$$


Explanation:
Okay, so I need to find the last nonzero digit to the right of the decimal point in the decimal expansion of 137 divided by 500. Hmm, let me think about how to approach this.

First, maybe I should just do the division and see what the decimal expansion looks like. Let me try that.

So, 137 divided by 500. Well, 500 goes into 137 zero times. Then, adding a decimal point and some zeros, it would be 137.000... So, 500 goes into 1370 two times because 2 times 500 is 1000. Subtract 1000 from 1370, and I get 370. Bring down the next zero, making it 3700. Now, how many times does 500 go into 3700? Let me calculate that. 500 times 7 is 3500, right? So 7 times. Subtract 3500 from 3700, and I get 200. Bring down another zero, making it 2000. 500 goes into 2000 exactly 4 times. So that's 4.

So putting that all together, the decimal expansion is 0.274... Wait, let me check again.

Starting with 137 divided by 500. Let me write it as a fraction: 137/500. Maybe I can simplify this fraction or convert it to a decimal by multiplying numerator and denominator by something to make the denominator a power of 10?

Since 500 is 5 times 100, which is 5*10^2. So, if I multiply numerator and denominator by 2, I get (137*2)/(500*2) = 274/1000. Oh! That's easier. So 274 divided by 1000 is 0.274. So the decimal expansion is 0.274.

Wait, so the decimal expansion is 0.274, right? So the digits after the decimal are 2, 7, 4, and then zeros? So the last nonzero digit to the right of the decimal would be 4? But wait, the question says "the last nonzero digit to the right of the decimal point". But in 0.274, after the decimal, there are three digits: 2, 7, 4. All of them are nonzero. So the last nonzero digit is 4?

But maybe I need to check if there are more digits? Wait, 274/1000 is exactly 0.274, right? Because 274 divided by 1000 is moving the decimal three places to the left. So 0.274. So there are no more digits after that. So the decimal expansion terminates here. Therefore, the last nonzero digit is 4.

But let me make sure I didn't make a mistake in my initial division. Let me verify by multiplying 500 by 0.274. 500 * 0.274. Let's compute that.

First, 500 * 0.2 = 100. Then 500 * 0.07 = 35. Then 500 * 0.004 = 2. So adding those together: 100 + 35 = 135, plus 2 is 137. So yes, that gives 137. So 500 * 0.274 = 137. Therefore, 137/500 is indeed 0.274.

Therefore, the decimal expansion is 0.274, so the digits after the decimal are 2, 7, 4. All non-zero. Therefore, the last nonzero digit is 4.

But wait, the question says "the last nonzero digit to the right of the decimal point". So if there are no zeros after the decimal, then the last digit is 4. But maybe I need to check if there are any trailing zeros? For example, if the decimal was 0.274000..., then the last nonzero digit would still be 4. But in this case, since it's terminating, there are no zeros after.

Alternatively, maybe the question is trying to trick me? Let me think again.

Alternatively, maybe I need to consider the decimal expansion in a different way? For example, if the fraction wasn't terminating, but in this case, it is terminating.

Wait, 500 factors into 2^2 * 5^3. So, since the denominator has only 2s and 5s, the decimal terminates. The number of decimal places is the maximum of the exponents of 2 and 5 in the denominator. Here, the denominator is 500 = 2^2 * 5^3, so the number of decimal places is 3. Therefore, the decimal terminates at the third decimal place. Therefore, 0.274 is correct.

Therefore, the last nonzero digit is 4.

But maybe I need to check if there's a different interpretation? For example, maybe the question is referring to something else? Let me check the original question again: "What is the last nonzero digit to the right of the decimal point in the decimal expansion of 137/500?"

So, decimal expansion is 0.274, so digits after decimal are 2, 7, 4. All non-zero. Therefore, the last one is 4.

But maybe I need to think about it in terms of multiplying numerator and denominator by something else? For example, if I wanted to write it as a decimal with more digits? But since it's terminating, there are no more digits.

Alternatively, maybe the question is expecting me to consider something else? For example, if there were trailing zeros, but in this case, there are none.

Alternatively, maybe I made a mistake in simplifying the fraction? Let me check again.

Original fraction: 137/500. Multiply numerator and denominator by 2: 137*2=274, 500*2=1000. So 274/1000. Yes, that's correct. So 274 divided by 1000 is 0.274.

Alternatively, maybe I need to consider the decimal expansion in another base? But no, the question says "decimal expansion", so base 10.

Alternatively, maybe the question is a trick question? For example, if the decimal expansion was something like 0.274000..., then the last nonzero digit would be 4, but if there were more digits, but in this case, there are not.

Alternatively, maybe I need to think about the decimal expansion in terms of continued fractions or something else? But no, that seems overcomplicating.

Alternatively, maybe I need to check if 137/500 can be simplified further? Let me check.

137 is a prime number? Let me check. 137 divided by 2? No. 3? 1+3+7=11, not divisible by 3. 5? Doesn't end with 0 or 5. 7? 7*19=133, 137-133=4, not divisible by 7. 11? 11*12=132, 137-132=5, not divisible by 11. 13? 13*10=130, 137-130=7, not divisible by 13. So 137 is a prime number. Therefore, 137/500 cannot be simplified further.

Therefore, the decimal expansion is indeed 0.274.

Therefore, the last nonzero digit to the right of the decimal is 4.

But wait, let me check once more. Maybe I need to write out the decimal expansion step by step.

Dividing 137 by 500.

Since 500 is larger than 137, we write it as 0. and then proceed.

Multiply 137 by 10: 1370. How many times does 500 go into 1370?

As before, 500*2=1000, 500*3=1500 which is too big. So 2 times.

Subtract 1000 from 1370: 370.

Bring down a zero: 3700.

How many times does 500 go into 3700? 500*7=3500, 500*8=4000 which is too big. So 7 times.

Subtract 3500 from 3700: 200.

Bring down a zero: 2000.

How many times does 500 go into 2000? Exactly 4 times.

Subtract 4*500=2000, remainder 0.

So the decimal expansion is 0.274.

So, digits after decimal: 2, 7, 4. No more digits.

Therefore, the last nonzero digit is 4.

Therefore, the answer is 4.

But wait, maybe the question is from a source where sometimes they consider something else? For example, maybe they want the last nonzero digit before the decimal? But no, the question says "to the right of the decimal point".

Alternatively, maybe there's a misunderstanding in the question? For example, if the fraction was something else? But no, the question is about 137/500.

Alternatively, maybe I need to consider negative numbers? But no, 137 and 500 are both positive.

Alternatively, maybe the question is in another language and "right of the decimal point" is mistranslated? But assuming the question is correctly translated.

Alternatively, maybe I need to think about the decimal expansion in terms of repeating decimals? But since the denominator factors into 2s and 5s, it's a terminating decimal.

Therefore, I think my conclusion is correct. The last nonzero digit to the right of the decimal point is 4.

But just to be thorough, let me think if there's another way to approach this problem.

Another way: Since 137/500 = 137/(5*100) = (137/5)/100. Let me compute 137 divided by 5 first.

137 divided by 5 is 27.4. Then divide by 100, which moves the decimal two places to the left: 0.274. So same result.

Alternatively, maybe using fractions. Let me think.

But I think all methods lead to the same decimal expansion.

Alternatively, maybe using modular arithmetic? For example, to find the last nonzero digit, but since the decimal terminates, maybe not necessary.

But just for practice, let me try.

If I want to find the decimal expansion of 137/500, which is 0.274.

But if I wanted to find the last nonzero digit, maybe I can think of multiplying numerator and denominator by something to make the denominator 1, but that's not helpful.

Alternatively, think of the decimal expansion as follows:

The decimal expansion of a fraction with denominator 10^n is just the numerator divided by 10^n, so 274/1000 is 0.274.

Therefore, the last nonzero digit is 4.

Alternatively, if the decimal was longer, say, if it was 0.274000..., then the last nonzero digit is still 4.

Therefore, regardless of the approach, I keep getting 4.

Therefore, I think the answer is 4.

But just to make sure, let me check with another method.

Let me write 137 divided by 500.

Since 500 is 5*100, so 137 divided by 5 is 27.4, then divided by 100 is 0.274.

Alternatively, 137 divided by 500. Let me divide numerator and denominator by 5: 137/500 = (137/5)/100 = 27.4/100 = 0.274.

Same result.

Alternatively, think of 137/500 as (137/5)/100.

137 divided by 5: 5*27=135, remainder 2. So 27 + 2/5 = 27.4. Then divide by 100: 0.274.

Same thing.

Alternatively, think of 137/500 as 137*(2/1000) = 274/1000 = 0.274.

Same result.

Therefore, all methods confirm that the decimal expansion is 0.274.

Therefore, the last nonzero digit to the right of the decimal is 4.

Therefore, I think the answer is 4.

But just to be 100% sure, let me check with another approach.

Suppose I didn't know how to divide 137 by 500. Let me think about decimal expansions.

The decimal expansion of a fraction can be found by multiplying numerator and denominator by 2^k to make the denominator a power of 10.

Since 500 = 5^3 * 2^2. To make the denominator a power of 10, we need to multiply numerator and denominator by 2^(number of 5s) to get 10^3.

Wait, denominator is 5^3 * 2^2. To make it 10^3, we need to multiply by 2^1, because 5^3 * 2^2 * 2^1 = 5^3 * 2^3 = 10^3.

Therefore, multiply numerator and denominator by 2:

137 * 2 = 274

500 * 2 = 1000

Therefore, 274/1000 = 0.274.

Same result.

Therefore, the decimal expansion is 0.274.

Therefore, the last nonzero digit is 4.

Therefore, I think I can confidently say that the answer is 4.

But just to check once more, maybe I can convert 137/500 into a decimal using another method.

Let me think of 1/500.

1/500 is 0.002. Therefore, 137/500 is 137 * 0.002.

Calculating 137 * 0.002:

First, 100 * 0.002 = 0.2

30 * 0.002 = 0.06

7 * 0.002 = 0.014

Adding them together: 0.2 + 0.06 = 0.26 + 0.014 = 0.274.

Same result.

Therefore, 137/500 = 0.274.

Therefore, the decimal expansion is 0.274, so the last nonzero digit is 4.

Therefore, the answer is 4.

I think I've checked this multiple ways, and all methods lead to the same conclusion. Therefore, I can be confident that the last nonzero digit to the right of the decimal point in the decimal expansion of 137/500 is 4.

**Final Answer**
The last nonzero digit to the right of the decimal point is \boxed{4}.

To determine the **last nonzero digit to the right of the decimal point** in the decimal expansion of the fraction $\frac{137}{500}$, we begin by analyzing the structure of the fraction and its decimal representation.

---

### Step 1: Simplify the Fraction

We start by simplifying the fraction:

$$
\frac{137}{500}
$$

Note that 500 can be factored as:

$$
500 = 2^2 \times 5^3
$$

This means the denominator is composed only of powers of 2 and 5, which guarantees that the decimal expansion of the fraction will **terminate**. The number of decimal places is determined by the **maximum exponent** of 2 or 5 in the denominator, which is 3 in this case.

---

### Step 2: Convert to a Decimal

To convert the fraction to a decimal, we can multiply both numerator and denominator by 2 to make the denominator a power of 10:

$$
\frac{137}{500} = \frac{137 \times 2}{500 \times 2} = \frac{274}{1000}
$$

Now, dividing 274 by 1000 simply moves the decimal point three places to the left:

$$
\frac{274}{1000} = 0.274
$$

So, the decimal expansion of $\frac{137}{500}$ is:

$$
0.274
$$

---

### Step 3: Identify the Last Nonzero Digit

Looking at the decimal expansion $0.274$, the digits to the right of the decimal point are:

- 2
- 7
- 4

All of these digits are **nonzero**, and since the decimal terminates, there are no additional digits after the 4. Therefore, the **last nonzero digit to the right of the decimal point** is:

$$
\boxed{4}
$$

---

### Final Answer

$$
\boxed{4}
$$

[Assistant]
Explanation:
Okay, so I need to find the last nonzero digit to the right of the decimal point in the decimal expansion of 137 divided by 500. Hmm, let me think about how to approach this.

First, maybe I should just do the division and see what the decimal expansion looks like. Let me try that.

So, 137 divided by 500. Well, 500 goes into 137 zero times. Then, adding a decimal point and some zeros, it would be 137.000... So, 500 goes into 1370 two times because 2 times 500 is 1000. Subtract 1000 from 1370, and I get 370. Bring down the next zero, making it 3700. Now, how many times does 500 go into 3700? Let me calculate that. 500 times 7 is 3500, right? So 7 times. Subtract 3500 from 3700, and I get 200. Bring down another zero, making it 2000. 500 goes into 2000 exactly 4 times. So that's 4.

So putting that all together, the decimal expansion is 0.274... Wait, let me check again.

Starting with 137 divided by 500. Let me write it as a fraction: 137/500. Maybe I can simplify this fraction or convert it to a decimal by multiplying numerator and denominator by something to make the denominator a power of 10?

Since 500 is 5 times 100, which is 5*10^2. So, if I multiply numerator and denominator by 2, I get (137*2)/(500*2) = 274/1000. Oh! That's easier. So 274 divided by 1000 is 0.274. So the decimal expansion is 0.274.

Wait, so the decimal expansion is 0.274, right? So the digits after the decimal are 2, 7, 4, and then zeros? So the last nonzero digit to the right of the decimal would be 4? But wait, the question says "the last nonzero digit to the right of the decimal point". But in 0.274, after the decimal, there are three digits: 2, 7, 4. All of them are nonzero. So the last nonzero digit is 4?

But maybe I need to check if there are more digits? Wait, 274/1000 is exactly 0.274, right? Because 274 divided by 1000 is moving the decimal three places to the left. So 0.274. So there are no more digits after that. So the decimal expansion terminates here. Therefore, the last nonzero digit is 4.

But let me make sure I didn't make a mistake in my initial division. Let me verify by multiplying 500 by 0.274. 500 * 0.274. Let's compute that.

First, 500 * 0.2 = 100. Then 500 * 0.07 = 35. Then 500 * 0.004 = 2. So adding those together: 100 + 35 = 135, plus 2 is 137. So yes, that gives 137. So 500 * 0.274 = 137. Therefore, 137/500 is indeed 0.274.

Therefore, the decimal expansion is 0.274, so the digits after the decimal are 2, 7, 4. All non-zero. Therefore, the last nonzero digit is 4.

But wait, the question says "the last nonzero digit to the right of the decimal point". So if there are no zeros after the decimal, then the last digit is 4. But maybe I need to check if there are any trailing zeros? For example, if the decimal was 0.274000..., then the last nonzero digit would still be 4. But in this case, since it's terminating, there are no zeros after.

Alternatively, maybe the question is trying to trick me? Let me think again.

Alternatively, maybe I need to consider the decimal expansion in a different way? For example, if the fraction wasn't terminating, but in this case, it is terminating.

Wait, 500 factors into 2^2 * 5^3. So, since the denominator has only 2s and 5s, the decimal terminates. The number of decimal places is the maximum of the exponents of 2 and 5 in the denominator. Here, the denominator is 500 = 2^2 * 5^3, so the number of decimal places is 3. Therefore, the decimal terminates at the third decimal place. Therefore, 0.274 is correct.

Therefore, the last nonzero digit is 4.

But maybe I need to check if there's a different interpretation? For example, maybe the question is referring to something else? Let me check the original question again: "What is the last nonzero digit to the right of the decimal point in the decimal expansion of 137/500?"

So, decimal expansion is 0.274, so digits after decimal are 2, 7, 4. All non-zero. Therefore, the last one is 4.

But maybe I need to think about it in terms of multiplying numerator and denominator by something else? For example, if I wanted to write it as a decimal with more digits? But since it's terminating, there are no more digits.

Alternatively, maybe the question is expecting me to consider something else? For example, if there were trailing zeros, but in this case, there are none.

Alternatively, maybe I made a mistake in simplifying the fraction? Let me check again.

Original fraction: 137/500. Multiply numerator and denominator by 2: 137*2=274, 500*2=1000. So 274/1000. Yes, that's correct. So 274 divided by 1000 is 0.274.

Alternatively, maybe I need to consider the decimal expansion in another base? But no, the question says "decimal expansion", so base 10.

Alternatively, maybe the question is a trick question? For example, if the decimal expansion was something like 0.274000..., then the last nonzero digit would be 4, but if there were more digits, but in this case, there are not.

Alternatively, maybe I need to think about the decimal expansion in terms of continued fractions or something else? But no, that seems overcomplicating.

Alternatively, maybe I need to check if 137/500 can be simplified further? Let me check.

137 is a prime number? Let me check. 137 divided by 2? No. 3? 1+3+7=11, not divisible by 3. 5? Doesn't end with 0 or 5. 7? 7*19=133, 137-133=4, not divisible by 7. 11? 11*12=132, 137-132=5, not divisible by 11. 13? 13*10=130, 137-130=7, not divisible by 13. So 137 is a prime number. Therefore, 137/500 cannot be simplified further.

Therefore, the decimal expansion is indeed 0.274.

Therefore, the last nonzero digit to the right of the decimal is 4.

But wait, let me check once more. Maybe I need to write out the decimal expansion step by step.

Dividing 137 by 500.

Since 500 is larger than 137, we write it as 0. and then proceed.

Multiply 137 by 10: 1370. How many times does 500 go into 1370?

As before, 500*2=1000, 500*3=1500 which is too big. So 2 times.

Subtract 1000 from 1370: 370.

Bring down a zero: 3700.

How many times does 500 go into 3700? 500*7=3500, 500*8=4000 which is too big. So 7 times.

Subtract 3500 from 3700: 200.

Bring down a zero: 2000.

How many times does 500 go into 2000? Exactly 4 times.

Subtract 4*500=2000, remainder 0.

So the decimal expansion is 0.274.

So, digits after decimal: 2, 7, 4. No more digits.

Therefore, the last nonzero digit is 4.

Therefore, the answer is 4.

But wait, maybe the question is from a source where sometimes they consider something else? For example, maybe they want the last nonzero digit before the decimal? But no, the question says "to the right of the decimal point".

Alternatively, maybe there's a misunderstanding in the question? For example, if the fraction was something else? But no, the question is about 137/500.

Alternatively, maybe I need to consider negative numbers? But no, 137 and 500 are both positive.

Alternatively, maybe the question is in another language and "right of the decimal point" is mistranslated? But assuming the question is correctly translated.

Alternatively, maybe I need to think about the decimal expansion in terms of repeating decimals? But since the denominator factors into 2s and 5s, it's a terminating decimal.

Therefore, I think my conclusion is correct. The last nonzero digit to the right of the decimal point is 4.

But just to be thorough, let me think if there's another way to approach this problem.

Another way: Since 137/500 = 137/(5*100) = (137/5)/100. Let me compute 137 divided by 5 first.

137 divided by 5 is 27.4. Then divide by 100, which moves the decimal two places to the left: 0.274. So same result.

Alternatively, maybe using fractions. Let me think.

But I think all methods lead to the same decimal expansion.

Alternatively, maybe using modular arithmetic? For example, to find the last nonzero digit, but since the decimal terminates, maybe not necessary.

But just for practice, let me try.

If I want to find the decimal expansion of 137/500, which is 0.274.

But if I wanted to find the last nonzero digit, maybe I can think of multiplying numerator and denominator by something to make the denominator 1, but that's not helpful.

Alternatively, think of the decimal expansion as follows:

The decimal expansion of a fraction with denominator 10^n is just the numerator divided by 10^n, so 274/1000 is 0.274.

Therefore, the last nonzero digit is 4.

Alternatively, if the decimal was longer, say, if it was 0.274000..., then the last nonzero digit is still 4.

Therefore, regardless of the approach, I keep getting 4.

Therefore, I think the answer is 4.

But just to make sure, let me check with another method.

Let me write 137 divided by 500.

Since 500 is 5*100, so 137 divided by 5 is 27.4, then divided by 100 is 0.274.

Alternatively, 137 divided by 500. Let me divide numerator and denominator by 5: 137/500 = (137/5)/100 = 27.4/100 = 0.274.

Same result.

Alternatively, think of 137/500 as (137/5)/100.

137 divided by 5: 5*27=135, remainder 2. So 27 + 2/5 = 27.4. Then divide by 100: 0.274.

Same thing.

Alternatively, think of 137/500 as 137*(2/1000) = 274/1000 = 0.274.

Same result.

Therefore, all methods confirm that the decimal expansion is 0.274.

Therefore, the last nonzero digit to the right of the decimal is 4.

Therefore, I think the answer is 4.

But just to be 100% sure, let me check with another approach.

Suppose I didn't know how to divide 137 by 500. Let me think about decimal expansions.

The decimal expansion of a fraction can be found by multiplying numerator and denominator by 2^k to make the denominator a power of 10.

Since 500 = 5^3 * 2^2. To make the denominator a power of 10, we need to multiply numerator and denominator by 2^(number of 5s) to get 10^3.

Wait, denominator is 5^3 * 2^2. To make it 10^3, we need to multiply by 2^1, because 5^3 * 2^2 * 2^1 = 5^3 * 2^3 = 10^3.

Therefore, multiply numerator and denominator by 2:

137 * 2 = 274

500 * 2 = 1000

Therefore, 274/1000 = 0.274.

Same result.

Therefore, the decimal expansion is 0.274.

Therefore, the last nonzero digit is 4.

Therefore, I think I can confidently say that the answer is 4.

But just to check once more, maybe I can convert 137/500 into a decimal using another method.

Let me think of 1/500.

1/500 is 0.002. Therefore, 137/500 is 137 * 0.002.

Calculating 137 * 0.002:

First, 100 * 0.002 = 0.2

30 * 0.002 = 0.06

7 * 0.002 = 0.014

Adding them together: 0.2 + 0.06 = 0.26 + 0.014 = 0.274.

Same result.

Therefore, 137/500 = 0.274.

Therefore, the decimal expansion is 0.274, so the last nonzero digit is 4.

Therefore, the answer is 4.

I think I've checked this multiple ways, and all methods lead to the same conclusion. Therefore, I can be confident that the last nonzero digit to the right of the decimal point in the decimal expansion of 137/500 is 4.

**Final Answer**
The last nonzero digit to the right of the decimal point is \boxed{4}.

To determine the **last nonzero digit to the right of the decimal point** in the decimal expansion of the fraction $\frac{137}{500}$, we begin by analyzing the structure of the fraction and its decimal representation.

---

### Step 1: Simplify the Fraction

We start by simplifying the fraction:

$$
\frac{137}{500}
$$

Note that 500 can be factored as:

$$
500 = 2^2 \times 5^3
$$

This means the denominator is composed only of powers of 2 and 5, which guarantees that the decimal expansion of the fraction will **terminate**. The number of decimal places is determined by the **maximum exponent** of 2 or 5 in the denominator, which is 3 in this case.

---

### Step 2: Convert to a Decimal

To convert the fraction to a decimal, we can multiply both numerator and denominator by 2 to make the denominator a power of 10:

$$
\frac{137}{500} = \frac{137 \times 2}{500 \times 2} = \frac{274}{1000}
$$

Now, dividing 274 by 1000 simply moves the decimal point three places to the left:

$$
\frac{274}{1000} = 0.274
$$

So, the decimal expansion of $\frac{137}{500}$ is:

$$
0.274
$$

---

### Step 3: Identify the Last Nonzero Digit

Looking at the decimal expansion $0.274$, the digits to the right of the decimal point are:

- 2
- 7
- 4

All of these digits are **nonzero**, and since the decimal terminates, there are no additional digits after the 4. Therefore, the **last nonzero digit to the right of the decimal point** is:

$$
\boxed{4}
$$

---

### Final Answer

$$
\boxed{4}
$$

Solution: We can rewrite $\frac{137}{500}$ in the form $\frac{274}{1000}$, so $\frac{137}{500} = \frac{274}{1000} = 0.274$ and the last nonzero digit is $\boxed{4}$.


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: 80 + 50 = 130, 5 + 8 = 13, so total is 130 + 13 = 143. Wait, let me check that again. 85 + 58. 85 + 50 is 135, then +8 is 143. Yes, that's correct.

But let me verify if all the conditions are satisfied. 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, 35 - 8 = 27. Yes, that's correct. So, the original number is 85, reversed is 58, their sum is 143. Therefore, the answer should be 143.

But let me think again if there's another way to approach this problem, just to make sure I didn't make any mistakes. Maybe using algebraic methods or checking if there are other possible numbers.

Wait, the equations were a + b = 13 and a - b = 3. Solving them gives a unique solution, right? Because two equations with two variables. So, unless there's some constraints on a and b, like digits must be between 0 and 9. Let me check if a and b are valid digits. a is 8, which is okay for a two-digit number (since the tens digit can't be zero). b is 5, which is also okay. So, that's valid.

Alternatively, suppose I didn't think of variables. Let me think. Let me call the original number N, and reversed number M. Then N - M = 27, and digits of N add to 13. But maybe that's more abstract. But since we already solved it with variables, and checked the answer, maybe that's sufficient.

Alternatively, maybe I can think of possible two-digit numbers where digits add up to 13. Let me list them:

Possible pairs (a, b) where a + b = 13. Since a and b are digits (so 0-9), possible values:

a can be from 4 to 9, because if a is 4, then b is 9; a=5, b=8; a=6, b=7; a=7, b=6; a=8, b=5; a=9, b=4. So the numbers would be 49, 58, 67, 76, 85, 94.

Now, check which of these numbers when subtracted by their reversed counterparts gives 27.

Let me check each:

First, 49: reversed is 94. 49 - 94 = -45. Not 27. 94 - 49 = 45. Not 27.

Next, 58: reversed is 85. 58 - 85 = -27. 85 - 58 = 27. Oh, that's exactly the difference given. So, original number is 85, reversed is 58. So, that's the same as before. So, the sum is 85 + 58 = 143. So, same answer.

Alternatively, check others for thoroughness:

67 reversed is 76. 67 - 76 = -9. 76 - 67 = 9. Not 27.

76 reversed is 67. 76 - 67 = 9. Same as above.

85 reversed is 58. 85 - 58 = 27. Which matches.

94 reversed is 49. 94 - 49 = 45. Not 27.

So, only 85 and 58 satisfy the difference condition. Therefore, the answer is 85 + 58 = 143. So, that's consistent with my previous result.

Therefore, I think the answer is 143. Let me just check once again the sum: 85 + 58. 80 + 50 is 130, 5 + 8 is 13, so 130 + 13 = 143. Yep. So, that's correct.

Alternatively, maybe there's a way to find the sum without finding the original numbers? Let me think. Let me denote original number as N = 10a + b, reversed number as M = 10b + a. Then, the problem says N - M = 27, and a + b = 13. We need to find N + M.

So, N + M = (10a + b) + (10b + a) = 11a + 11b = 11(a + b). Since a + b = 13, then N + M = 11 * 13 = 143. Oh! That's a smarter way. Instead of solving for a and b, we can directly compute N + M as 11*(a + b). Since a + b is given as 13, then the answer is 11*13 = 143. That's a much quicker way. So, that's another way to think about it. So, that confirms the answer again.

Therefore, regardless of the method, the answer is 143. So, I think that's solid.

**Final Answer**
The sum of the original number and the number with its digits reversed is \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:

$$
N = 10a + b
$$

When the digits are reversed, the number becomes:

$$
M = 10b + a
$$

---

### Step 1: Use the given conditions

**Condition 1:** The sum of the digits is 13:

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

**Condition 2:** 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 2: Solve the system of equations

We now solve the system:

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

Add the two equations:

$$
(a + b) + (a - b) = 13 + 3 \Rightarrow 2a = 16 \Rightarrow a = 8
$$

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

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

So, the original number is:

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

The reversed number is:

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

---

### Step 3: Find the sum of the original and reversed numbers

$$
N + M = 85 + 58 = 143
$$

---

### Step 4: Alternative approach (more efficient)

We can also find the sum directly using the expressions for $ N $ and $ M $:

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

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

$$
N + M = 11 \times 13 = 143
$$

---

### 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: 80 + 50 = 130, 5 + 8 = 13, so total is 130 + 13 = 143. Wait, let me check that again. 85 + 58. 85 + 50 is 135, then +8 is 143. Yes, that's correct.

But let me verify if all the conditions are satisfied. 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, 35 - 8 = 27. Yes, that's correct. So, the original number is 85, reversed is 58, their sum is 143. Therefore, the answer should be 143.

But let me think again if there's another way to approach this problem, just to make sure I didn't make any mistakes. Maybe using algebraic methods or checking if there are other possible numbers.

Wait, the equations were a + b = 13 and a - b = 3. Solving them gives a unique solution, right? Because two equations with two variables. So, unless there's some constraints on a and b, like digits must be between 0 and 9. Let me check if a and b are valid digits. a is 8, which is okay for a two-digit number (since the tens digit can't be zero). b is 5, which is also okay. So, that's valid.

Alternatively, suppose I didn't think of variables. Let me think. Let me call the original number N, and reversed number M. Then N - M = 27, and digits of N add to 13. But maybe that's more abstract. But since we already solved it with variables, and checked the answer, maybe that's sufficient.

Alternatively, maybe I can think of possible two-digit numbers where digits add up to 13. Let me list them:

Possible pairs (a, b) where a + b = 13. Since a and b are digits (so 0-9), possible values:

a can be from 4 to 9, because if a is 4, then b is 9; a=5, b=8; a=6, b=7; a=7, b=6; a=8, b=5; a=9, b=4. So the numbers would be 49, 58, 67, 76, 85, 94.

Now, check which of these numbers when subtracted by their reversed counterparts gives 27.

Let me check each:

First, 49: reversed is 94. 49 - 94 = -45. Not 27. 94 - 49 = 45. Not 27.

Next, 58: reversed is 85. 58 - 85 = -27. 85 - 58 = 27. Oh, that's exactly the difference given. So, original number is 85, reversed is 58. So, that's the same as before. So, the sum is 85 + 58 = 143. So, same answer.

Alternatively, check others for thoroughness:

67 reversed is 76. 67 - 76 = -9. 76 - 67 = 9. Not 27.

76 reversed is 67. 76 - 67 = 9. Same as above.

85 reversed is 58. 85 - 58 = 27. Which matches.

94 reversed is 49. 94 - 49 = 45. Not 27.

So, only 85 and 58 satisfy the difference condition. Therefore, the answer is 85 + 58 = 143. So, that's consistent with my previous result.

Therefore, I think the answer is 143. Let me just check once again the sum: 85 + 58. 80 + 50 is 130, 5 + 8 is 13, so 130 + 13 = 143. Yep. So, that's correct.

Alternatively, maybe there's a way to find the sum without finding the original numbers? Let me think. Let me denote original number as N = 10a + b, reversed number as M = 10b + a. Then, the problem says N - M = 27, and a + b = 13. We need to find N + M.

So, N + M = (10a + b) + (10b + a) = 11a + 11b = 11(a + b). Since a + b = 13, then N + M = 11 * 13 = 143. Oh! That's a smarter way. Instead of solving for a and b, we can directly compute N + M as 11*(a + b). Since a + b is given as 13, then the answer is 11*13 = 143. That's a much quicker way. So, that's another way to think about it. So, that confirms the answer again.

Therefore, regardless of the method, the answer is 143. So, I think that's solid.

**Final Answer**
The sum of the original number and the number with its digits reversed is \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:

$$
N = 10a + b
$$

When the digits are reversed, the number becomes:

$$
M = 10b + a
$$

---

### Step 1: Use the given conditions

**Condition 1:** The sum of the digits is 13:

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

**Condition 2:** 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 2: Solve the system of equations

We now solve the system:

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

Add the two equations:

$$
(a + b) + (a - b) = 13 + 3 \Rightarrow 2a = 16 \Rightarrow a = 8
$$

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

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

So, the original number is:

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

The reversed number is:

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

---

### Step 3: Find the sum of the original and reversed numbers

$$
N + M = 85 + 58 = 143
$$

---

### Step 4: Alternative approach (more efficient)

We can also find the sum directly using the expressions for $ N $ and $ M $:

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

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

$$
N + M = 11 \times 13 = 143
$$

---

### 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 representatives should match 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. So, the total parts would be 4 + 5 = 9 parts.

But how does this relate to the number of representatives? If the total number of representatives is 18, then each part would correspond to 18 divided by 9, which is 2. So, each part is 2 representatives. Therefore, the 7th grade, which has 4 parts, would have 4*2=8 representatives, and the 8th grade with 5 parts would have 5*2=10 representatives.

Wait, let me verify that. If 7th grade has 8 and 8th has 10, then total is 18, which matches the given total. So that seems to check out. But let me think again if this is the correct approach.

Alternatively, maybe I should calculate the proportion of each grade's enrollment relative to the total enrollment. Let me try that method as well to confirm.

Total enrollment is 520 + 650 = 1170 students.

So, the proportion of 7th grade is 520 / 1170, and 8th grade is 650 / 1170.

Then, the number of representatives for each grade would be that proportion multiplied by the total number of representatives, which is 18.

Calculating for 7th grade: (520 / 1170) * 18. Let me compute that.

First, simplify 520 / 1170. Let me divide numerator and denominator by 130 again. 520 ÷ 130 = 4, 1170 ÷ 130 = 9. So, 4/9. Therefore, 4/9 * 18 = (4 * 18)/9 = 4 * 2 = 8.

Similarly for 8th grade: 650 / 1170 = 5/9 (since 650 ÷ 130 = 5, 1170 ÷ 130 = 9). Then 5/9 * 18 = 10.

So, same result. Therefore, the 8th grade should have 10 representatives.

But wait, the question says "how many representatives should the 8th grade have if there is to be a fair representation of the two grades?" So, the answer is 10? Let me check once more.

Alternatively, maybe there's another way to think about it. For example, using the concept of weighted averages or something else? Let me see.

Suppose we let x be the number of representatives for 8th grade, then the 7th grade would have 18 - x representatives.

We need the ratio of representatives to be proportional to the enrollments. So:

(18 - x)/x = 520/650

Wait, is that correct? Wait, the ratio of representatives should be equal to the ratio of enrollments. So, (number of representatives for 7th grade) / (number for 8th grade) = (enrollment of 7th grade) / (enrollment of 8th grade).

So, (18 - x)/x = 520/650. Let me solve this equation.

First, simplify 520/650. As before, divide numerator and denominator by 130: 4/5. So:

(18 - x)/x = 4/5

Cross-multiplying:

5*(18 - x) = 4*x

90 - 5x = 4x

90 = 9x

x = 10

So, same answer. Therefore, the 8th grade should have 10 representatives.

Therefore, all methods lead to 10. So, I think that's the answer.

But just to make sure, let me think if there's any other way this could be interpreted. For example, sometimes in some contexts, you might have to round numbers if they don't divide evenly, but in this case, since 520:650 simplifies to 4:5, and 4+5=9, which divides evenly into 18 (since 18/9=2), so there's no rounding needed. Therefore, exact numbers.

Alternatively, if the numbers didn't divide evenly, we might have to use some kind of rounding method, but here it's straightforward.

Another check: total enrollment is 1170. If we have 18 representatives, then the number of representatives per student would be 18 / 1170 ≈ 0.01538 representatives per student. Then, the 7th grade has 520 students, so 520 * 0.01538 ≈ 8, and 650 * 0.01538 ≈ 10. So that also gives the same numbers.

Therefore, all methods confirm that the 8th grade should have 10 representatives.

I think that's solid. So, the answer is 10.

**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 enrollment** of each grade.

---

### Step 1: Understand the Given Data

- **7th grade enrollment**: 520 students
- **8th grade enrollment**: 650 students
- **Total representatives**: 18

We are to find how many of these 18 representatives should be from the **8th grade**.

---

### Step 2: Determine the Ratio of Enrollments

The ratio of 7th to 8th grade enrollments is:

$$
\frac{520}{650}
$$

Simplify this ratio by dividing both numbers by their greatest common divisor (GCD), which is 130:

$$
\frac{520 \div 130}{650 \div 130} = \frac{4}{5}
$$

So, the ratio of 7th to 8th grade enrollments is **4:5**.

---

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

The total number of parts in the ratio is:

$$
4 + 5 = 9
$$

Since there are 18 total representatives, each part of the ratio corresponds to:

$$
\frac{18}{9} = 2 \text{ representatives per part}
$$

Now, the 8th grade corresponds to **5 parts**, so the number of representatives it should have is:

$$
5 \times 2 = 10
$$

---

### Step 4: Verify the Result

- 7th grade: 4 parts → $4 \times 2 = 8$ representatives
- 8th grade: 5 parts → $5 \times 2 = 10$ representatives
- Total: $8 + 10 = 18$ representatives ✔

This matches the total number of representatives, confirming the allocation is correct.

---

### 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 representatives should match 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. So, the total parts would be 4 + 5 = 9 parts.

But how does this relate to the number of representatives? If the total number of representatives is 18, then each part would correspond to 18 divided by 9, which is 2. So, each part is 2 representatives. Therefore, the 7th grade, which has 4 parts, would have 4*2=8 representatives, and the 8th grade with 5 parts would have 5*2=10 representatives.

Wait, let me verify that. If 7th grade has 8 and 8th has 10, then total is 18, which matches the given total. So that seems to check out. But let me think again if this is the correct approach.

Alternatively, maybe I should calculate the proportion of each grade's enrollment relative to the total enrollment. Let me try that method as well to confirm.

Total enrollment is 520 + 650 = 1170 students.

So, the proportion of 7th grade is 520 / 1170, and 8th grade is 650 / 1170.

Then, the number of representatives for each grade would be that proportion multiplied by the total number of representatives, which is 18.

Calculating for 7th grade: (520 / 1170) * 18. Let me compute that.

First, simplify 520 / 1170. Let me divide numerator and denominator by 130 again. 520 ÷ 130 = 4, 1170 ÷ 130 = 9. So, 4/9. Therefore, 4/9 * 18 = (4 * 18)/9 = 4 * 2 = 8.

Similarly for 8th grade: 650 / 1170 = 5/9 (since 650 ÷ 130 = 5, 1170 ÷ 130 = 9). Then 5/9 * 18 = 10.

So, same result. Therefore, the 8th grade should have 10 representatives.

But wait, the question says "how many representatives should the 8th grade have if there is to be a fair representation of the two grades?" So, the answer is 10? Let me check once more.

Alternatively, maybe there's another way to think about it. For example, using the concept of weighted averages or something else? Let me see.

Suppose we let x be the number of representatives for 8th grade, then the 7th grade would have 18 - x representatives.

We need the ratio of representatives to be proportional to the enrollments. So:

(18 - x)/x = 520/650

Wait, is that correct? Wait, the ratio of representatives should be equal to the ratio of enrollments. So, (number of representatives for 7th grade) / (number for 8th grade) = (enrollment of 7th grade) / (enrollment of 8th grade).

So, (18 - x)/x = 520/650. Let me solve this equation.

First, simplify 520/650. As before, divide numerator and denominator by 130: 4/5. So:

(18 - x)/x = 4/5

Cross-multiplying:

5*(18 - x) = 4*x

90 - 5x = 4x

90 = 9x

x = 10

So, same answer. Therefore, the 8th grade should have 10 representatives.

Therefore, all methods lead to 10. So, I think that's the answer.

But just to make sure, let me think if there's any other way this could be interpreted. For example, sometimes in some contexts, you might have to round numbers if they don't divide evenly, but in this case, since 520:650 simplifies to 4:5, and 4+5=9, which divides evenly into 18 (since 18/9=2), so there's no rounding needed. Therefore, exact numbers.

Alternatively, if the numbers didn't divide evenly, we might have to use some kind of rounding method, but here it's straightforward.

Another check: total enrollment is 1170. If we have 18 representatives, then the number of representatives per student would be 18 / 1170 ≈ 0.01538 representatives per student. Then, the 7th grade has 520 students, so 520 * 0.01538 ≈ 8, and 650 * 0.01538 ≈ 10. So that also gives the same numbers.

Therefore, all methods confirm that the 8th grade should have 10 representatives.

I think that's solid. So, the answer is 10.

**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 enrollment** of each grade.

---

### Step 1: Understand the Given Data

- **7th grade enrollment**: 520 students
- **8th grade enrollment**: 650 students
- **Total representatives**: 18

We are to find how many of these 18 representatives should be from the **8th grade**.

---

### Step 2: Determine the Ratio of Enrollments

The ratio of 7th to 8th grade enrollments is:

$$
\frac{520}{650}
$$

Simplify this ratio by dividing both numbers by their greatest common divisor (GCD), which is 130:

$$
\frac{520 \div 130}{650 \div 130} = \frac{4}{5}
$$

So, the ratio of 7th to 8th grade enrollments is **4:5**.

---

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

The total number of parts in the ratio is:

$$
4 + 5 = 9
$$

Since there are 18 total representatives, each part of the ratio corresponds to:

$$
\frac{18}{9} = 2 \text{ representatives per part}
$$

Now, the 8th grade corresponds to **5 parts**, so the number of representatives it should have is:

$$
5 \times 2 = 10
$$

---

### Step 4: Verify the Result

- 7th grade: 4 parts → $4 \times 2 = 8$ representatives
- 8th grade: 5 parts → $5 \times 2 = 10$ representatives
- Total: $8 + 10 = 18$ representatives ✔

This matches the total number of representatives, confirming the allocation is correct.

---

### 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 we want 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:

C(6, 5) = (6 × 5!) / (5! × 1!)

Yes, so the 5! cancels out from numerator and denominator, leaving me with 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), so C(6, 5) should be equal to C(6, 1). And C(6, 1) is 6, which matches. 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 combination formula, I could think of it as follows: For the first student, I have 6 choices, then 5, then 4, then 3, then 2. But wait, that would be permutations, right? Because order matters here. So if I do 6 × 5 × 4 × 3 × 2, that's 720. But since order doesn't matter, I need to divide by the number of ways the 5 students can be arranged, which is 5!. So, 720 divided by 120 (which is 5!) equals 6. Yep, that also gives me 6. So that's another way to get the same answer.

So, all these different methods lead me to the conclusion that the number of ways is 6. Therefore, the answer should be 6. Let me just make sure there isn't a trick or something I'm missing here. The problem says "selected from a group of 6 students," and there's no mention of any restrictions or special conditions. So, it's a straightforward combination problem.

Another way to think about it: if you have 6 students, and you want to choose 5, then each selection is missing exactly one student. Since there are 6 students, each one can be the one left out, hence 6 different selections. That's very intuitive. So, yeah, that's another confirmation.

I think I've verified it multiple ways: using the combination formula, using the property that C(n, k) = C(n, n - k), using permutations and then adjusting for overcounting, and thinking about it in terms of exclusion. All of them point to the answer being 6. So I feel confident that 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: Understanding the Problem

We are selecting **5 students** from a group of **6**, and the order in which we select them is **not important**. For example, selecting students A, B, C, D, E is the same as selecting B, A, C, D, E. 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, the product of all positive integers up to that number.

---

### Step 2: Apply the Formula

Here, $ n = 6 $ and $ k = 5 $. Plugging into the formula:

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

We can simplify this by recognizing that:

$$
6! = 6 \times 5!
$$

So:

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

---

### Step 3: Alternative Interpretation

Another way to think about this is: selecting 5 students out of 6 is the same as **excluding 1 student**. Since there are 6 students, there are **6 different ways** to leave out one student, which corresponds to 6 different groups of 5 students.

This confirms the result.

---

### 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 we want 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:

C(6, 5) = (6 × 5!) / (5! × 1!)

Yes, so the 5! cancels out from numerator and denominator, leaving me with 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), so C(6, 5) should be equal to C(6, 1). And C(6, 1) is 6, which matches. 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 combination formula, I could think of it as follows: For the first student, I have 6 choices, then 5, then 4, then 3, then 2. But wait, that would be permutations, right? Because order matters here. So if I do 6 × 5 × 4 × 3 × 2, that's 720. But since order doesn't matter, I need to divide by the number of ways the 5 students can be arranged, which is 5!. So, 720 divided by 120 (which is 5!) equals 6. Yep, that also gives me 6. So that's another way to get the same answer.

So, all these different methods lead me to the conclusion that the number of ways is 6. Therefore, the answer should be 6. Let me just make sure there isn't a trick or something I'm missing here. The problem says "selected from a group of 6 students," and there's no mention of any restrictions or special conditions. So, it's a straightforward combination problem.

Another way to think about it: if you have 6 students, and you want to choose 5, then each selection is missing exactly one student. Since there are 6 students, each one can be the one left out, hence 6 different selections. That's very intuitive. So, yeah, that's another confirmation.

I think I've verified it multiple ways: using the combination formula, using the property that C(n, k) = C(n, n - k), using permutations and then adjusting for overcounting, and thinking about it in terms of exclusion. All of them point to the answer being 6. So I feel confident that 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: Understanding the Problem

We are selecting **5 students** from a group of **6**, and the order in which we select them is **not important**. For example, selecting students A, B, C, D, E is the same as selecting B, A, C, D, E. 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, the product of all positive integers up to that number.

---

### Step 2: Apply the Formula

Here, $ n = 6 $ and $ k = 5 $. Plugging into the formula:

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

We can simplify this by recognizing that:

$$
6! = 6 \times 5!
$$

So:

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

---

### Step 3: Alternative Interpretation

Another way to think about this is: selecting 5 students out of 6 is the same as **excluding 1 student**. Since there are 6 students, there are **6 different ways** to leave out one student, which corresponds to 6 different groups of 5 students.

This confirms the result.

---

### 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.