Explanation:
Okay, so I need to convert the fraction 57/160 into a terminating decimal. Hmm, let me recall what a terminating decimal is. From what I remember, a terminating decimal is a decimal that ends after a finite number of digits, right? Unlike repeating decimals, which go on forever with a repeating pattern. So, the question is, how do I convert this fraction into such a decimal?

First, I think the key here is understanding when a fraction will convert to a terminating decimal. There's something about the denominator, isn't there? If I remember correctly, a fraction will have a terminating decimal expansion if the denominator, after simplifying the fraction to its lowest terms, has no prime factors other than 2 and 5. Let me check if that's right.

So, let's see. The denominator here is 160. Let me factor that into primes. Let me divide 160 by 2 first. 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 times 5^1. Therefore, the prime factors are 2 and 5. Since there are no other prime factors, this should result in a terminating decimal. Okay, that seems good. So, 57/160 should convert to a terminating decimal.

Now, how do I actually perform the conversion? Well, there are a couple of ways. One is to do the long division of 57 divided by 160. Another way is to manipulate the fraction so that the denominator becomes a power of 10, which would make it straightforward to write as a decimal. Let me try both approaches and see which one works better.

Starting with the second method, since the denominator is 160, which is 2^5 * 5^1. To convert this into a power of 10, we need to multiply numerator and denominator by 5^4, which would make the denominator 2^5 * 5^5 = 10^5. Let me verify that.

Original denominator is 160 = 2^5 * 5^1. If I multiply numerator and denominator by 5^4, then the denominator becomes 2^5 * 5^(1+4) = 2^5 * 5^5 = (2*5)^5 = 10^5. Yes, that works. So, multiplying numerator and denominator by 5^4, which is 625, should give me a denominator of 10^5, which is 100000.

So let me compute that. The numerator is 57 * 625. Let me calculate that.

First, 57 * 600 is 34,200. Then 57 * 25 is 1,425. Adding those together: 34,200 + 1,425 = 35,625. Therefore, the fraction becomes 35,625 / 100,000.

So, 35,625 divided by 100,000 is equal to moving the decimal point five places to the left, which would be 0.35625. Let me check that.

Yes, because 35,625 divided by 100,000. Let's write that out. 35,625 ÷ 100,000. Since there are five zeros in the denominator, I move the decimal five places to the left. Starting from 35,625, which is the same as 35625.0, moving the decimal five places left: 0.35625. Perfect.

Alternatively, I could perform the long division of 57 divided by 160. Let me try that method as a check.

So, dividing 57 by 160. Since 160 is larger than 57, the decimal will start with 0. and then we proceed.

Let me write this out.

160 | 57.000000...

First, 160 goes into 570 how many times? Let me calculate that. 160 * 3 is 480, 160 * 4 is 640, which is too big. So 3 times.

3 * 160 is 480. Subtract 480 from 570: 570 - 480 = 90. Bring down the next 0, making it 900.

Now, how many times does 160 go into 900? Let's see: 160 * 5 = 800, 160 * 6 = 960. 960 is too big. So 5 times.

5 * 160 = 800. Subtract from 900: 900 - 800 = 100. Bring down the next 0, making it 1000.

Now, how many times does 160 go into 1000? Let me divide 1000 by 160. 160 * 6 = 960, 160 * 6.25 = 1000. Wait, but we need integer division here. So 160 * 6 = 960, subtract that from 1000 gives 40. Bring down the next 0, making it 400.

How many times does 160 go into 400? That's exactly 2 times, because 160 * 2 = 320. Wait, 160 * 2 is 320. Wait, 400 - 320 = 80. Wait, but 160 * 2.5 = 400. But again, integer division. So 160 goes into 400 two times.

Wait, 160 * 2 = 320. Subtract 320 from 400: 80. Bring down a 0, making it 800.

Now, how many times does 160 go into 800? 160 * 5 = 800. So that's 5 times.

Subtract 800 from 800, remainder is 0. So we're done.

Let me compile the decimal digits we have so far.

Starting with 0., then the first division gave us 3 (from 570 ÷ 160), then 5 (from 900 ÷ 160), then 6 (from 1000 ÷ 160), then 2 (from 400 ÷ 160), then 5 (from 800 ÷ 160). Wait, let me check again:

Wait, step by step:

First step: 0.3 (because 3*160=480, remainder 90)

Second step: bring down 0, making 900. 5*160=800, so write 5, remainder 100.

Third step: bring down 0, making 1000. 6*160=960, so write 6, remainder 40.

Fourth step: bring down 0, making 400. 2*160=320, so write 2, remainder 80.

Fifth step: bring down 0, making 800. 5*160=800, so write 5, remainder 0.

So the decimal is 0.35625.

So that matches the previous result. Therefore, both methods confirm that 57/160 is 0.35625.

Wait, let me confirm once more. If I do 57 divided by 160 on a calculator, what do I get? Well, since I don't have a calculator here, but let me verify with multiplication. Let me check 0.35625 multiplied by 160 equals 57.

Calculating 0.35625 * 160. Let me break this down.

First, 0.35625 * 100 = 35.625

0.35625 * 60 = ?

Well, 0.35625 * 60. Let me compute 0.35625 * 6 * 10.

0.35625 * 6. Let's compute 0.35625 * 6.

Multiply 0.35625 by 2: 0.7125. Multiply by 3: 2.1375. So 0.35625 * 6 = 2.1375. Then times 10 is 21.375.

So total is 35.625 + 21.375 = 57. So yes, that checks out.

Therefore, the decimal is indeed 0.35625.

So both methods agree. Therefore, 57/160 as a decimal is 0.35625, which is a terminating decimal.

But just to recap, the key steps were:

1. Recognize that since the denominator factors into only 2s and 5s, the decimal terminates.

2. Either convert the fraction to have a denominator of a power of 10 by multiplying numerator and denominator by the necessary factors (in this case, 5^4) to get 10^5.

3. Alternatively, perform the long division of 57 divided by 160, keeping track of the remainders and the decimal places.

Either way, we arrive at the same result.

Another way to think about it: 160 is 16 * 10, so maybe we can convert the fraction into something over 10, then over 100, etc. Let me try that approach as a third method.

So, 57/160. Since 160 is 16*10, maybe divide numerator and denominator by 16 first? Wait, 57 divided by 16? Let's see. 16*3 = 48, 16*3.5 = 56, so 57/16 is 3.5625. Then, since we divided denominator by 16, which is 2^4, so original denominator was 160 = 16*10. Therefore, if we have 57/160 = (57/16)/10. So that would be 3.5625 / 10 = 0.35625.

Oh, that's another way. So that's clever. Breaking down the denominator into 16 and 10, then dividing the numerator by 16 first and then dividing by 10.

So, 57 divided by 16 is 3.5625, then divided by 10 is 0.35625.

That's a neat method as well.

So, multiple approaches confirm that the decimal is 0.35625.

Therefore, I can be confident that the answer is 0.35625.

But let me just check once more for errors in calculation. Let me verify the multiplication step again.

If 0.35625 is the decimal, then multiplying by 160 should give 57.

Let me do 0.35625 * 160.

First, note that 0.35625 * 160 = 0.35625 * 16 * 10.

Compute 0.35625 * 16:

Well, 0.35625 * 10 = 3.5625

0.35625 * 6 = 2.1375

Adding those together: 3.5625 + 2.1375 = 5.7.

Then multiply by 10: 5.7 * 10 = 57.

Yes, perfect.

Therefore, that's correct.

Alternatively, breaking down the decimal:

0.35625 can be thought of as 0.3 + 0.05 + 0.006 + 0.00025.

Calculating each part:

0.3 is 3/10,

0.05 is 5/100,

0.006 is 6/1000,

0.00025 is 25/100000.

Adding them all up:

3/10 + 5/100 + 6/1000 + 25/100000.

Converting to decimal places:

3/10 = 0.3

5/100 = 0.05

6/1000 = 0.006

25/100000 = 0.00025

Adding them gives 0.3 + 0.05 = 0.35; 0.35 + 0.006 = 0.356; 0.356 + 0.00025 = 0.35625.

So, 0.35625.

Therefore, regardless of the method, I keep arriving at the same decimal.

Another verification: convert the decimal back to a fraction. Let me check if 0.35625 equals 57/160.

0.35625 can be written as 35625/100000. Simplify that fraction.

Divide numerator and denominator by 25:

35625 ÷ 25 = 1425

100000 ÷ 25 = 4000

So 1425/4000.

Divide numerator and denominator by 5:

1425 ÷ 5 = 285

4000 ÷ 5 = 800

So 285/800.

Divide numerator and denominator by 5 again:

285 ÷ 5 = 57

800 ÷ 5 = 160

So we get back to 57/160.

Perfect! So that confirms that 0.35625 is indeed equal to 57/160.

Therefore, all methods converge to the same result, and multiple checks confirm the accuracy.

So, I think that's thorough enough. There's no doubt now that 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.

---

### Understanding Terminating Decimals

A fraction will have a **terminating decimal** expansion if, after simplifying the fraction, its **denominator** only contains the prime factors **2 and/or 5**. This is because powers of 10 are products of 2 and 5 (i.e., $10 = 2 \cdot 5$), and any denominator that is a product of these primes can be made into a power of 10 through multiplication.

---

### Step 1: Factor the Denominator

We factor $160$:

$$
160 = 2^5 \cdot 5
$$

There are no other prime factors. So, **$\frac{57}{160}$ is a terminating decimal**.

---

### Step 2: Convert to a Decimal

To convert this into a decimal, we can either:

#### Option A: Long Division
Perform the division of $57$ by $160$:

- $57.0000 ÷ 160 ≈ 0.35625$

#### Option B: Multiply to get a Denominator of a Power of 10

Since $160 = 2^5 \cdot 5$, to make the denominator a power of 10, we need to multiply both numerator and denominator by $5^4$ (which is $625$):

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

Now, simplify:

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

---

### Final Verification

Recall that:

$$
0.35625 \cdot 160 = 57 \quad \checkmark
$$

This confirms that the decimal conversion is correct.

---

### Final Answer

$$
\boxed{0.35625}
$$

[Assistant]
Explanation:
Okay, so I need to convert the fraction 57/160 into a terminating decimal. Hmm, let me recall what a terminating decimal is. From what I remember, a terminating decimal is a decimal that ends after a finite number of digits, right? Unlike repeating decimals, which go on forever with a repeating pattern. So, the question is, how do I convert this fraction into such a decimal?

First, I think the key here is understanding when a fraction will convert to a terminating decimal. There's something about the denominator, isn't there? If I remember correctly, a fraction will have a terminating decimal expansion if the denominator, after simplifying the fraction to its lowest terms, has no prime factors other than 2 and 5. Let me check if that's right.

So, let's see. The denominator here is 160. Let me factor that into primes. Let me divide 160 by 2 first. 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 times 5^1. Therefore, the prime factors are 2 and 5. Since there are no other prime factors, this should result in a terminating decimal. Okay, that seems good. So, 57/160 should convert to a terminating decimal.

Now, how do I actually perform the conversion? Well, there are a couple of ways. One is to do the long division of 57 divided by 160. Another way is to manipulate the fraction so that the denominator becomes a power of 10, which would make it straightforward to write as a decimal. Let me try both approaches and see which one works better.

Starting with the second method, since the denominator is 160, which is 2^5 * 5^1. To convert this into a power of 10, we need to multiply numerator and denominator by 5^4, which would make the denominator 2^5 * 5^5 = 10^5. Let me verify that.

Original denominator is 160 = 2^5 * 5^1. If I multiply numerator and denominator by 5^4, then the denominator becomes 2^5 * 5^(1+4) = 2^5 * 5^5 = (2*5)^5 = 10^5. Yes, that works. So, multiplying numerator and denominator by 5^4, which is 625, should give me a denominator of 10^5, which is 100000.

So let me compute that. The numerator is 57 * 625. Let me calculate that.

First, 57 * 600 is 34,200. Then 57 * 25 is 1,425. Adding those together: 34,200 + 1,425 = 35,625. Therefore, the fraction becomes 35,625 / 100,000.

So, 35,625 divided by 100,000 is equal to moving the decimal point five places to the left, which would be 0.35625. Let me check that.

Yes, because 35,625 divided by 100,000. Let's write that out. 35,625 ÷ 100,000. Since there are five zeros in the denominator, I move the decimal five places to the left. Starting from 35,625, which is the same as 35625.0, moving the decimal five places left: 0.35625. Perfect.

Alternatively, I could perform the long division of 57 divided by 160. Let me try that method as a check.

So, dividing 57 by 160. Since 160 is larger than 57, the decimal will start with 0. and then we proceed.

Let me write this out.

160 | 57.000000...

First, 160 goes into 570 how many times? Let me calculate that. 160 * 3 is 480, 160 * 4 is 640, which is too big. So 3 times.

3 * 160 is 480. Subtract 480 from 570: 570 - 480 = 90. Bring down the next 0, making it 900.

Now, how many times does 160 go into 900? Let's see: 160 * 5 = 800, 160 * 6 = 960. 960 is too big. So 5 times.

5 * 160 = 800. Subtract from 900: 900 - 800 = 100. Bring down the next 0, making it 1000.

Now, how many times does 160 go into 1000? Let me divide 1000 by 160. 160 * 6 = 960, 160 * 6.25 = 1000. Wait, but we need integer division here. So 160 * 6 = 960, subtract that from 1000 gives 40. Bring down the next 0, making it 400.

How many times does 160 go into 400? That's exactly 2 times, because 160 * 2 = 320. Wait, 160 * 2 is 320. Wait, 400 - 320 = 80. Wait, but 160 * 2.5 = 400. But again, integer division. So 160 goes into 400 two times.

Wait, 160 * 2 = 320. Subtract 320 from 400: 80. Bring down a 0, making it 800.

Now, how many times does 160 go into 800? 160 * 5 = 800. So that's 5 times.

Subtract 800 from 800, remainder is 0. So we're done.

Let me compile the decimal digits we have so far.

Starting with 0., then the first division gave us 3 (from 570 ÷ 160), then 5 (from 900 ÷ 160), then 6 (from 1000 ÷ 160), then 2 (from 400 ÷ 160), then 5 (from 800 ÷ 160). Wait, let me check again:

Wait, step by step:

First step: 0.3 (because 3*160=480, remainder 90)

Second step: bring down 0, making 900. 5*160=800, so write 5, remainder 100.

Third step: bring down 0, making 1000. 6*160=960, so write 6, remainder 40.

Fourth step: bring down 0, making 400. 2*160=320, so write 2, remainder 80.

Fifth step: bring down 0, making 800. 5*160=800, so write 5, remainder 0.

So the decimal is 0.35625.

So that matches the previous result. Therefore, both methods confirm that 57/160 is 0.35625.

Wait, let me confirm once more. If I do 57 divided by 160 on a calculator, what do I get? Well, since I don't have a calculator here, but let me verify with multiplication. Let me check 0.35625 multiplied by 160 equals 57.

Calculating 0.35625 * 160. Let me break this down.

First, 0.35625 * 100 = 35.625

0.35625 * 60 = ?

Well, 0.35625 * 60. Let me compute 0.35625 * 6 * 10.

0.35625 * 6. Let's compute 0.35625 * 6.

Multiply 0.35625 by 2: 0.7125. Multiply by 3: 2.1375. So 0.35625 * 6 = 2.1375. Then times 10 is 21.375.

So total is 35.625 + 21.375 = 57. So yes, that checks out.

Therefore, the decimal is indeed 0.35625.

So both methods agree. Therefore, 57/160 as a decimal is 0.35625, which is a terminating decimal.

But just to recap, the key steps were:

1. Recognize that since the denominator factors into only 2s and 5s, the decimal terminates.

2. Either convert the fraction to have a denominator of a power of 10 by multiplying numerator and denominator by the necessary factors (in this case, 5^4) to get 10^5.

3. Alternatively, perform the long division of 57 divided by 160, keeping track of the remainders and the decimal places.

Either way, we arrive at the same result.

Another way to think about it: 160 is 16 * 10, so maybe we can convert the fraction into something over 10, then over 100, etc. Let me try that approach as a third method.

So, 57/160. Since 160 is 16*10, maybe divide numerator and denominator by 16 first? Wait, 57 divided by 16? Let's see. 16*3 = 48, 16*3.5 = 56, so 57/16 is 3.5625. Then, since we divided denominator by 16, which is 2^4, so original denominator was 160 = 16*10. Therefore, if we have 57/160 = (57/16)/10. So that would be 3.5625 / 10 = 0.35625.

Oh, that's another way. So that's clever. Breaking down the denominator into 16 and 10, then dividing the numerator by 16 first and then dividing by 10.

So, 57 divided by 16 is 3.5625, then divided by 10 is 0.35625.

That's a neat method as well.

So, multiple approaches confirm that the decimal is 0.35625.

Therefore, I can be confident that the answer is 0.35625.

But let me just check once more for errors in calculation. Let me verify the multiplication step again.

If 0.35625 is the decimal, then multiplying by 160 should give 57.

Let me do 0.35625 * 160.

First, note that 0.35625 * 160 = 0.35625 * 16 * 10.

Compute 0.35625 * 16:

Well, 0.35625 * 10 = 3.5625

0.35625 * 6 = 2.1375

Adding those together: 3.5625 + 2.1375 = 5.7.

Then multiply by 10: 5.7 * 10 = 57.

Yes, perfect.

Therefore, that's correct.

Alternatively, breaking down the decimal:

0.35625 can be thought of as 0.3 + 0.05 + 0.006 + 0.00025.

Calculating each part:

0.3 is 3/10,

0.05 is 5/100,

0.006 is 6/1000,

0.00025 is 25/100000.

Adding them all up:

3/10 + 5/100 + 6/1000 + 25/100000.

Converting to decimal places:

3/10 = 0.3

5/100 = 0.05

6/1000 = 0.006

25/100000 = 0.00025

Adding them gives 0.3 + 0.05 = 0.35; 0.35 + 0.006 = 0.356; 0.356 + 0.00025 = 0.35625.

So, 0.35625.

Therefore, regardless of the method, I keep arriving at the same decimal.

Another verification: convert the decimal back to a fraction. Let me check if 0.35625 equals 57/160.

0.35625 can be written as 35625/100000. Simplify that fraction.

Divide numerator and denominator by 25:

35625 ÷ 25 = 1425

100000 ÷ 25 = 4000

So 1425/4000.

Divide numerator and denominator by 5:

1425 ÷ 5 = 285

4000 ÷ 5 = 800

So 285/800.

Divide numerator and denominator by 5 again:

285 ÷ 5 = 57

800 ÷ 5 = 160

So we get back to 57/160.

Perfect! So that confirms that 0.35625 is indeed equal to 57/160.

Therefore, all methods converge to the same result, and multiple checks confirm the accuracy.

So, I think that's thorough enough. There's no doubt now that 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.

---

### Understanding Terminating Decimals

A fraction will have a **terminating decimal** expansion if, after simplifying the fraction, its **denominator** only contains the prime factors **2 and/or 5**. This is because powers of 10 are products of 2 and 5 (i.e., $10 = 2 \cdot 5$), and any denominator that is a product of these primes can be made into a power of 10 through multiplication.

---

### Step 1: Factor the Denominator

We factor $160$:

$$
160 = 2^5 \cdot 5
$$

There are no other prime factors. So, **$\frac{57}{160}$ is a terminating decimal**.

---

### Step 2: Convert to a Decimal

To convert this into a decimal, we can either:

#### Option A: Long Division
Perform the division of $57$ by $160$:

- $57.0000 ÷ 160 ≈ 0.35625$

#### Option B: Multiply to get a Denominator of a Power of 10

Since $160 = 2^5 \cdot 5$, to make the denominator a power of 10, we need to multiply both numerator and denominator by $5^4$ (which is $625$):

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

Now, simplify:

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

---

### Final Verification

Recall that:

$$
0.35625 \cdot 160 = 57 \quad \checkmark
$$

This confirms that the decimal conversion is correct.

---

### 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. Let me think about how to approach this.

First, I know that dividing by 500 is the same as dividing by 5 and then by 100, right? Because 500 is 5 times 100. So maybe if I can simplify the fraction first? Let me try that.

So 137 divided by 500. Let me write that as 137/500. If I divide both numerator and denominator by something... Well, 500 is 5*100, so maybe divide numerator and denominator by 5 first? Let me check if 137 is divisible by 5. 5 times 27 is 135, so 137 divided by 5 would be 27.4. Wait, but that's not helpful for simplifying the fraction. Maybe I should just convert the fraction to a decimal directly?

Alternatively, since dividing by 500 is the same as multiplying by 2/1000, because 1/500 is 2/1000. Wait, yes! Because 500 times 2 is 1000, so 1/500 is 2/1000. Therefore, 137 divided by 500 would be 137 times 2 divided by 1000. Let me compute that.

So 137 * 2 is 274. Therefore, 274 divided by 1000 is 0.274. Wait, so the decimal expansion is 0.274? Then the digits to the right of the decimal are 2, 7, 4. The last nonzero digit would be 4? But wait, is there any more digits after that? Because if it's exactly 0.274, then there are no more digits. But maybe I need to check if the decimal terminates or if there are more digits?

But 500 factors into 2^2 * 5^3. So the denominator can be written as 2^2*5^3. To convert a fraction to a decimal, we can write it with denominator as a power of 10. Since 2 and 5 are factors of 10, we can multiply numerator and denominator by 2^something to make the denominator 10^n.

Let me recall that the number of decimal places is determined by the maximum of the exponents of 2 and 5 in the denominator. So since the original denominator is 500 = 2^2 * 5^3. The maximum exponent is 3 (from 5^3). Therefore, we need to multiply numerator and denominator by 2^(3-2) = 2^1 to make the denominator 10^3.

Therefore, 137/500 = (137 * 2) / (500 * 2) = 274 / 1000 = 0.274. Therefore, indeed, the decimal expansion is terminating, and it is 0.274. So the digits after the decimal are 2, 7, 4, and then it ends. Therefore, the last nonzero digit to the right of the decimal is 4.

Wait, but the question says "the last nonzero digit to the right of the decimal point". Since the decimal expansion is 0.274, there are three digits after the decimal: 2, 7, 4. All of them are nonzero. So the last one is 4.

But maybe I made a mistake here? Let me check with another method. Maybe by long division?

Let me try dividing 137 by 500.

Since 500 goes into 137 zero times, we write 0. and then add decimal places.

So 500 goes into 1370 two times (since 2*500=1000). Subtract 1000 from 1370, we get 370. Bring down a zero (making it 3700). 500 goes into 3700 seven times (7*500=3500). Subtract 3500 from 3700, get 200. Bring down a zero, making it 2000. 500 goes into 2000 four times (4*500=2000). Subtract, get zero. So the decimal expansion is 0.274. Yep, that matches.

So then, the decimal expansion is exactly 0.274 with no more digits. Therefore, the last nonzero digit is 4.

But wait, the problem says "the last nonzero digit to the right of the decimal point". If all digits are nonzero, then the last one is just the last digit, which is 4.

But maybe there's a trick here? Like, maybe if the decimal had trailing zeros? For example, if it was 0.2740, then the last nonzero digit would still be 4. But since in this case, the decimal terminates at 0.274, there are no trailing zeros.

Alternatively, maybe the problem is more complicated? Let me check if I did the calculation correctly.

Wait, 137 divided by 500. Let me do decimal division step by step.

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

Multiply numerator by 10: 1370 divided by 500. 500*2=1000, 500*3=1500. So 2 times. 1370 - 1000 = 370. Bring down a zero: 3700.

Divide 3700 by 500. 500*7=3500. 3700 - 3500=200. Bring down a zero: 2000.

Divide 2000 by 500. That's 4. 2000 - 2000 = 0. So done. So decimal is 0.274.

Therefore, I think my conclusion is correct. The last nonzero digit is 4.

But let me think again. Maybe the question is a trick question? For example, if the decimal had more digits, but because of some repeating? But 500 is 2^2*5^3. Since the prime factors are only 2 and 5, the decimal expansion terminates. Therefore, there's no repeating part. Hence, 0.274 is exact.

Alternatively, maybe the question is referring to something else? For example, maybe there's a different way to interpret the question?

Wait, the question is: "the last nonzero digit to the right of the decimal point in the decimal expansion of 137/500". So, if the decimal expansion is 0.274, then the digits after decimal are 2, 7, 4. All non-zero. Therefore, the last nonzero digit is 4.

Alternatively, perhaps I need to consider that when written as a decimal, maybe there are more digits? But no. Since 137/500 is exactly 0.274. There's no need for more digits.

But to be thorough, let me check with another approach. Let me compute 137 divided by 500.

Alternatively, since 1/500 is 0.002, so 137*(0.002) would be...

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

Calculating 137 * 0.002: 137 * 2 = 274, then divided by 1000, so 0.274. Same as before.

So that also gives me 0.274.

Alternatively, maybe there's a mistake in my initial thought? Let me check with a calculator.

If I compute 137 divided by 500. Well, 500 goes into 137 zero times. Then 500 into 1370 is 2 times (2*500=1000), remainder 370. Then 500 into 3700 is 7 times (7*500=3500), remainder 200. Then 500 into 2000 is 4 times, remainder 0. So decimal is 0.274.

So yes, that's correct.

Therefore, I think that the answer is 4.

But just to make sure, maybe there's a different interpretation. For example, maybe the question is referring to the last non-zero digit before the decimal? But no, it specifically says "to the right of the decimal point".

So, in the decimal expansion, the digits to the right of the decimal are 2, 7, 4. All non-zero. Therefore, the last one is 4.

Alternatively, if there were trailing zeros, like if it was 0.2740, then the last non-zero digit is still 4.

Alternatively, if there was a decimal like 0.274000..., then the last non-zero digit is 4.

So, in any case, since the decimal terminates at 4, that's the last non-zero digit.

Therefore, the answer is 4.

But let me think again. Maybe I need to consider something else? For example, maybe the problem is in another base? But no, the question says "decimal expansion", which is base 10.

Alternatively, maybe there's a mistake in my calculation? Let me check once again.

Compute 137 divided by 500. Let me compute 137 divided by 5 first, then divided by 100.

137 divided by 5 is 27.4. Then divide by 100: 27.4 / 100 = 0.274. Correct.

So that gives me 0.274.

Alternatively, maybe I can think in terms of multiplying numerator and denominator by something to make the denominator a power of 10.

As I thought before, 500 = 2^2 * 5^3. To make denominator 10^3, we need to multiply numerator and denominator by 2^(3-2) = 2^1.

So, 137 * 2 = 274. Denominator 500 * 2 = 1000. Hence 274 / 1000 = 0.274.

So again, same result.

Therefore, I think my answer is correct. The last non-zero digit to the right of the decimal is 4.

But hold on, maybe the question is expecting me to consider something else? Let me check if there's another way to interpret the question.

Wait, the problem says "the last nonzero digit to the right of the decimal point in the decimal expansion of 137/500". So maybe there is a misunderstanding here. For example, if the decimal expansion was something like 0.274000..., then the last non-zero digit is 4. If it was 0.2745... with more digits, but since in this case it terminates, it's 0.274.

Alternatively, maybe the problem is a trick question where you have to consider some other decimal expansion?

Alternatively, maybe I need to convert the fraction into a decimal with more precision? But how?

Wait, if you have 137 divided by 500, and you multiply numerator and denominator by 2, you get 274/1000. So 274 divided by 1000 is 0.274. There's no more digits. So the decimal is exact.

Alternatively, suppose I didn't do the math right? Let me compute 137/500.

Since 500*0.2 = 100.

500*0.27 = 500*(0.2 + 0.07) = 100 + 35 = 135.

Then 500*0.274 = 500*(0.27 + 0.004) = 135 + 2 = 137.

Yes, exactly. Therefore, 137/500 is 0.274.

Therefore, I think that the answer is 4.

But just to make sure, maybe I can check an example with similar numbers. For example, if I take 1/500, that is 0.002. The last non-zero digit is 2. Similarly, 2/500 is 0.004, last non-zero digit is 4. 3/500 is 0.006, etc. Then 137/500 is 0.274, so the last non-zero digit is 4.

Alternatively, suppose we have 137 divided by 500. Let me write that as 137 divided by 5 divided by 100.

137 divided by 5 is 27.4. Divided by 100 is 0.274. Yep.

Alternatively, maybe I can think about the decimal places. Since 500 is 5*100, so dividing by 5 and then dividing by 100. Dividing by 100 moves decimal two places to the left.

But dividing by 5 first: 137 divided by 5 is 27.4. Then dividing by 100 is moving decimal two places left: 0.274.

Same result.

Therefore, I think that the answer is definitely 4.

But maybe I need to check with another approach? For example, using fractions. Let me think.

The decimal expansion of a fraction is determined by the numerator over denominator. Since 137 and 500 are co-prime? Let me check.

Find GCD of 137 and 500.

Well, 137 is a prime number? Let me check.

137 divided by 2? No. Divided by 3? 3*45=135, 3*45+2=137. Not divisible by 3. Divided by 5? Ends with 7. Divided by 7? 7*19=133, 137-133=4. Not divisible by 7. 11? 11*12=132, 132+11=143>137. So 137 is prime.

Therefore, 137 and 500 are co-prime? Because 500 factors into 2^2 * 5^3, and 137 is a prime not in that list. Therefore, the fraction is reduced completely.

Therefore, decimal expansion terminates because denominator has no prime factors other than 2 and 5. Which we already know.

So the decimal expansion is exactly 0.274.

Therefore, the last non-zero digit is 4.

So I think all methods confirm that. Therefore, I think the answer is 4.

But just to be absolutely sure, let me think if there's any other possible way that the decimal expansion could have more digits.

Wait, for example, suppose that someone thought that 137/500 is equal to something else? Like, if they thought that 137 divided by 500 is... Well, I can compute 137 divided by 500.

Let me do 500 * 0.2 = 100.

So 0.2 is 100. Then 0.27 would be 0.2 + 0.07 = 100 + 500*0.07 = 100 + 35 = 135.

Then 135 + 500*0.004 = 135 + 2 = 137. So 0.274.

Therefore, 0.274.

Alternatively, if I use a calculator (pretending I have one here). Let me do 137 / 500.

Well, 500 * 0.2 = 100. 137 - 100 = 37.

So 0.2 + (37 / 500). Now, 37 / 500.

37 divided by 500. Well, 37 / 5 is 7.4, then divided by 100 is 0.074.

So 0.2 + 0.074 = 0.274.

Same result.

Therefore, I think my answer is correct. The last non-zero digit is 4.

Therefore, I think the answer is 4.

But just to make sure, let me check if there's a different interpretation of the question. For example, sometimes in some contexts, the "last nonzero digit" might be interpreted as the last digit before the decimal or something. But the question specifically says "to the right of the decimal point", so that is the decimal part.

Alternatively, maybe I need to consider something else?

Wait, maybe the question is a trick question? Let me think. For example, if the decimal expansion had some repeating digits but the last non-zero digit before the repetition? But no, in this case, the decimal terminates.

Alternatively, maybe there is confusion between the last non-zero digit in the entire number versus the last non-zero digit after the decimal. But the question clearly says "to the right of the decimal point", so that's only the decimal part.

Alternatively, maybe the question is in another base? But no, decimal expansion is base 10.

So I think my answer is correct.

Therefore, 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 find the **last nonzero digit to the right of the decimal point** in the decimal expansion of $\frac{137}{500}$, let's proceed step by step, ensuring clarity and correctness in the reasoning.

---

### Step 1: Understand the Nature of the Decimal Expansion

The decimal expansion of a fraction will be **terminating** if the **denominator in the reduced form** has only **factors of 2 and 5**. This is because such denominators can be expressed as powers of 10, which yields a finite number of digits in the decimal expansion.

Given:

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

Let's factor the denominator:

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

Only the primes 2 and 5 are present — therefore, the decimal expansion of this fraction is **terminating**.

---

### Step 2: Convert the Fraction into a Decimal

To make the denominator a power of 10, we need to bring the denominator to $10^3$ (since the largest exponent is 3 from the factor $5^3$):

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

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

Thus, the decimal expansion is:

$$
0.274
$$

---

### Step 3: Identify the Last Nonzero Digit

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

$$
2,\ 7,\ 4
$$

Each of these digits is **nonzero**, and since there are no trailing zeros or repeating digits, the **last nonzero digit** to the right of the decimal 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. Let me think about how to approach this.

First, I know that dividing by 500 is the same as dividing by 5 and then by 100, right? Because 500 is 5 times 100. So maybe if I can simplify the fraction first? Let me try that.

So 137 divided by 500. Let me write that as 137/500. If I divide both numerator and denominator by something... Well, 500 is 5*100, so maybe divide numerator and denominator by 5 first? Let me check if 137 is divisible by 5. 5 times 27 is 135, so 137 divided by 5 would be 27.4. Wait, but that's not helpful for simplifying the fraction. Maybe I should just convert the fraction to a decimal directly?

Alternatively, since dividing by 500 is the same as multiplying by 2/1000, because 1/500 is 2/1000. Wait, yes! Because 500 times 2 is 1000, so 1/500 is 2/1000. Therefore, 137 divided by 500 would be 137 times 2 divided by 1000. Let me compute that.

So 137 * 2 is 274. Therefore, 274 divided by 1000 is 0.274. Wait, so the decimal expansion is 0.274? Then the digits to the right of the decimal are 2, 7, 4. The last nonzero digit would be 4? But wait, is there any more digits after that? Because if it's exactly 0.274, then there are no more digits. But maybe I need to check if the decimal terminates or if there are more digits?

But 500 factors into 2^2 * 5^3. So the denominator can be written as 2^2*5^3. To convert a fraction to a decimal, we can write it with denominator as a power of 10. Since 2 and 5 are factors of 10, we can multiply numerator and denominator by 2^something to make the denominator 10^n.

Let me recall that the number of decimal places is determined by the maximum of the exponents of 2 and 5 in the denominator. So since the original denominator is 500 = 2^2 * 5^3. The maximum exponent is 3 (from 5^3). Therefore, we need to multiply numerator and denominator by 2^(3-2) = 2^1 to make the denominator 10^3.

Therefore, 137/500 = (137 * 2) / (500 * 2) = 274 / 1000 = 0.274. Therefore, indeed, the decimal expansion is terminating, and it is 0.274. So the digits after the decimal are 2, 7, 4, and then it ends. Therefore, the last nonzero digit to the right of the decimal is 4.

Wait, but the question says "the last nonzero digit to the right of the decimal point". Since the decimal expansion is 0.274, there are three digits after the decimal: 2, 7, 4. All of them are nonzero. So the last one is 4.

But maybe I made a mistake here? Let me check with another method. Maybe by long division?

Let me try dividing 137 by 500.

Since 500 goes into 137 zero times, we write 0. and then add decimal places.

So 500 goes into 1370 two times (since 2*500=1000). Subtract 1000 from 1370, we get 370. Bring down a zero (making it 3700). 500 goes into 3700 seven times (7*500=3500). Subtract 3500 from 3700, get 200. Bring down a zero, making it 2000. 500 goes into 2000 four times (4*500=2000). Subtract, get zero. So the decimal expansion is 0.274. Yep, that matches.

So then, the decimal expansion is exactly 0.274 with no more digits. Therefore, the last nonzero digit is 4.

But wait, the problem says "the last nonzero digit to the right of the decimal point". If all digits are nonzero, then the last one is just the last digit, which is 4.

But maybe there's a trick here? Like, maybe if the decimal had trailing zeros? For example, if it was 0.2740, then the last nonzero digit would still be 4. But since in this case, the decimal terminates at 0.274, there are no trailing zeros.

Alternatively, maybe the problem is more complicated? Let me check if I did the calculation correctly.

Wait, 137 divided by 500. Let me do decimal division step by step.

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

Multiply numerator by 10: 1370 divided by 500. 500*2=1000, 500*3=1500. So 2 times. 1370 - 1000 = 370. Bring down a zero: 3700.

Divide 3700 by 500. 500*7=3500. 3700 - 3500=200. Bring down a zero: 2000.

Divide 2000 by 500. That's 4. 2000 - 2000 = 0. So done. So decimal is 0.274.

Therefore, I think my conclusion is correct. The last nonzero digit is 4.

But let me think again. Maybe the question is a trick question? For example, if the decimal had more digits, but because of some repeating? But 500 is 2^2*5^3. Since the prime factors are only 2 and 5, the decimal expansion terminates. Therefore, there's no repeating part. Hence, 0.274 is exact.

Alternatively, maybe the question is referring to something else? For example, maybe there's a different way to interpret the question?

Wait, the question is: "the last nonzero digit to the right of the decimal point in the decimal expansion of 137/500". So, if the decimal expansion is 0.274, then the digits after decimal are 2, 7, 4. All non-zero. Therefore, the last nonzero digit is 4.

Alternatively, perhaps I need to consider that when written as a decimal, maybe there are more digits? But no. Since 137/500 is exactly 0.274. There's no need for more digits.

But to be thorough, let me check with another approach. Let me compute 137 divided by 500.

Alternatively, since 1/500 is 0.002, so 137*(0.002) would be...

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

Calculating 137 * 0.002: 137 * 2 = 274, then divided by 1000, so 0.274. Same as before.

So that also gives me 0.274.

Alternatively, maybe there's a mistake in my initial thought? Let me check with a calculator.

If I compute 137 divided by 500. Well, 500 goes into 137 zero times. Then 500 into 1370 is 2 times (2*500=1000), remainder 370. Then 500 into 3700 is 7 times (7*500=3500), remainder 200. Then 500 into 2000 is 4 times, remainder 0. So decimal is 0.274.

So yes, that's correct.

Therefore, I think that the answer is 4.

But just to make sure, maybe there's a different interpretation. For example, maybe the question is referring to the last non-zero digit before the decimal? But no, it specifically says "to the right of the decimal point".

So, in the decimal expansion, the digits to the right of the decimal are 2, 7, 4. All non-zero. Therefore, the last one is 4.

Alternatively, if there were trailing zeros, like if it was 0.2740, then the last non-zero digit is still 4.

Alternatively, if there was a decimal like 0.274000..., then the last non-zero digit is 4.

So, in any case, since the decimal terminates at 4, that's the last non-zero digit.

Therefore, the answer is 4.

But let me think again. Maybe I need to consider something else? For example, maybe the problem is in another base? But no, the question says "decimal expansion", which is base 10.

Alternatively, maybe there's a mistake in my calculation? Let me check once again.

Compute 137 divided by 500. Let me compute 137 divided by 5 first, then divided by 100.

137 divided by 5 is 27.4. Then divide by 100: 27.4 / 100 = 0.274. Correct.

So that gives me 0.274.

Alternatively, maybe I can think in terms of multiplying numerator and denominator by something to make the denominator a power of 10.

As I thought before, 500 = 2^2 * 5^3. To make denominator 10^3, we need to multiply numerator and denominator by 2^(3-2) = 2^1.

So, 137 * 2 = 274. Denominator 500 * 2 = 1000. Hence 274 / 1000 = 0.274.

So again, same result.

Therefore, I think my answer is correct. The last non-zero digit to the right of the decimal is 4.

But hold on, maybe the question is expecting me to consider something else? Let me check if there's another way to interpret the question.

Wait, the problem says "the last nonzero digit to the right of the decimal point in the decimal expansion of 137/500". So maybe there is a misunderstanding here. For example, if the decimal expansion was something like 0.274000..., then the last non-zero digit is 4. If it was 0.2745... with more digits, but since in this case it terminates, it's 0.274.

Alternatively, maybe the problem is a trick question where you have to consider some other decimal expansion?

Alternatively, maybe I need to convert the fraction into a decimal with more precision? But how?

Wait, if you have 137 divided by 500, and you multiply numerator and denominator by 2, you get 274/1000. So 274 divided by 1000 is 0.274. There's no more digits. So the decimal is exact.

Alternatively, suppose I didn't do the math right? Let me compute 137/500.

Since 500*0.2 = 100.

500*0.27 = 500*(0.2 + 0.07) = 100 + 35 = 135.

Then 500*0.274 = 500*(0.27 + 0.004) = 135 + 2 = 137.

Yes, exactly. Therefore, 137/500 is 0.274.

Therefore, I think that the answer is 4.

But just to make sure, maybe I can check an example with similar numbers. For example, if I take 1/500, that is 0.002. The last non-zero digit is 2. Similarly, 2/500 is 0.004, last non-zero digit is 4. 3/500 is 0.006, etc. Then 137/500 is 0.274, so the last non-zero digit is 4.

Alternatively, suppose we have 137 divided by 500. Let me write that as 137 divided by 5 divided by 100.

137 divided by 5 is 27.4. Divided by 100 is 0.274. Yep.

Alternatively, maybe I can think about the decimal places. Since 500 is 5*100, so dividing by 5 and then dividing by 100. Dividing by 100 moves decimal two places to the left.

But dividing by 5 first: 137 divided by 5 is 27.4. Then dividing by 100 is moving decimal two places left: 0.274.

Same result.

Therefore, I think that the answer is definitely 4.

But maybe I need to check with another approach? For example, using fractions. Let me think.

The decimal expansion of a fraction is determined by the numerator over denominator. Since 137 and 500 are co-prime? Let me check.

Find GCD of 137 and 500.

Well, 137 is a prime number? Let me check.

137 divided by 2? No. Divided by 3? 3*45=135, 3*45+2=137. Not divisible by 3. Divided by 5? Ends with 7. Divided by 7? 7*19=133, 137-133=4. Not divisible by 7. 11? 11*12=132, 132+11=143>137. So 137 is prime.

Therefore, 137 and 500 are co-prime? Because 500 factors into 2^2 * 5^3, and 137 is a prime not in that list. Therefore, the fraction is reduced completely.

Therefore, decimal expansion terminates because denominator has no prime factors other than 2 and 5. Which we already know.

So the decimal expansion is exactly 0.274.

Therefore, the last non-zero digit is 4.

So I think all methods confirm that. Therefore, I think the answer is 4.

But just to be absolutely sure, let me think if there's any other possible way that the decimal expansion could have more digits.

Wait, for example, suppose that someone thought that 137/500 is equal to something else? Like, if they thought that 137 divided by 500 is... Well, I can compute 137 divided by 500.

Let me do 500 * 0.2 = 100.

So 0.2 is 100. Then 0.27 would be 0.2 + 0.07 = 100 + 500*0.07 = 100 + 35 = 135.

Then 135 + 500*0.004 = 135 + 2 = 137. So 0.274.

Therefore, 0.274.

Alternatively, if I use a calculator (pretending I have one here). Let me do 137 / 500.

Well, 500 * 0.2 = 100. 137 - 100 = 37.

So 0.2 + (37 / 500). Now, 37 / 500.

37 divided by 500. Well, 37 / 5 is 7.4, then divided by 100 is 0.074.

So 0.2 + 0.074 = 0.274.

Same result.

Therefore, I think my answer is correct. The last non-zero digit is 4.

Therefore, I think the answer is 4.

But just to make sure, let me check if there's a different interpretation of the question. For example, sometimes in some contexts, the "last nonzero digit" might be interpreted as the last digit before the decimal or something. But the question specifically says "to the right of the decimal point", so that is the decimal part.

Alternatively, maybe I need to consider something else?

Wait, maybe the question is a trick question? Let me think. For example, if the decimal expansion had some repeating digits but the last non-zero digit before the repetition? But no, in this case, the decimal terminates.

Alternatively, maybe there is confusion between the last non-zero digit in the entire number versus the last non-zero digit after the decimal. But the question clearly says "to the right of the decimal point", so that's only the decimal part.

Alternatively, maybe the question is in another base? But no, decimal expansion is base 10.

So I think my answer is correct.

Therefore, 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 find the **last nonzero digit to the right of the decimal point** in the decimal expansion of $\frac{137}{500}$, let's proceed step by step, ensuring clarity and correctness in the reasoning.

---

### Step 1: Understand the Nature of the Decimal Expansion

The decimal expansion of a fraction will be **terminating** if the **denominator in the reduced form** has only **factors of 2 and 5**. This is because such denominators can be expressed as powers of 10, which yields a finite number of digits in the decimal expansion.

Given:

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

Let's factor the denominator:

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

Only the primes 2 and 5 are present — therefore, the decimal expansion of this fraction is **terminating**.

---

### Step 2: Convert the Fraction into a Decimal

To make the denominator a power of 10, we need to bring the denominator to $10^3$ (since the largest exponent is 3 from the factor $5^3$):

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

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

Thus, the decimal expansion is:

$$
0.274
$$

---

### Step 3: Identify the Last Nonzero Digit

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

$$
2,\ 7,\ 4
$$

Each of these digits is **nonzero**, and since there are no trailing zeros or repeating digits, the **last nonzero digit** to the right of the decimal 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 try to understand what it's asking. The problem says that the sum of the digits of a two-digit number is 13. Also, the difference between the number and the number with its digits reversed is 27. Then, I need to find the sum of the original number and the reversed number. Alright, let me break this down step by step.

First, let's denote the original two-digit number. Let me call the tens digit 'a' and the units digit 'b'. So, the original number would be 10a + b because in a two-digit number, the tens place is multiplied by 10. For example, if the number is 23, then a is 2 and b is 3, so the number is 10*2 + 3 = 23. That seems right.

Now, the problem says the sum of the digits is 13. So, that would be a + b = 13. Got that. Equation one: a + b = 13.

Next, the difference between the original number and the reversed number is 27. Let me think. The reversed number would have the digits swapped, so it would be 10b + a. So, the original number is 10a + b, the reversed number is 10b + a. The problem says the difference between the original number and the reversed number is 27. Wait, does it specify which one is subtracted from which? Let me check the problem again.

"The difference between the number and the number with its digits reversed is 27." Hmm. It just says "the difference," but usually, difference can be interpreted as absolute value, but maybe in this case, they mean the original number minus the reversed number is 27? But maybe not. Wait, but maybe we need to consider both possibilities? Let me think.

Wait, but let's suppose that the original number is larger than the reversed number. If that's the case, then (original) - (reversed) = 27. If not, then (reversed) - (original) = 27. But since the problem says "the difference," maybe it's just the absolute value. However, perhaps we can figure it out from the equations. Let me proceed with that assumption first that original number minus reversed number is 27. If that leads to a contradiction, I can check the other way.

So, let me write that as:

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

Simplifying that:

10a + b - 10b - a = 27

Combine like terms:

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

Which is:

9a - 9b = 27

Divide both sides by 9:

a - b = 3

So, that gives another equation: a - b = 3.

So now, we have two equations:

1) a + b = 13

2) a - b = 3

So, we can solve these two equations simultaneously. Let me do that.

Adding equations 1 and 2:

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

Which is:

2a = 16

Therefore, a = 8.

Then, substituting 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, which would make the original number 85.

Let me check if this satisfies the second condition. The reversed number would be 58. The difference between 85 and 58 is 85 - 58 = 27. Yes, that's exactly the given difference. So that works.

Therefore, the original number is 85, reversed number is 58. The problem asks for the sum of the original number and the reversed number. So that's 85 + 58. Let me calculate that.

85 + 58: 80 + 50 = 130, 5 + 8 = 13, so total is 130 + 13 = 143. Therefore, the answer is 143. Wait, let me check that again. 85 + 58. 85 + 50 is 135, plus 8 is 143. Yes, that's right.

But hold on, let me just make sure I didn't make any mistakes in assuming that original number minus reversed number is 27. What if the difference was reversed? Let me check that scenario as well, just to be thorough.

Suppose instead that the reversed number minus original number is 27. Then, (10b + a) - (10a + b) = 27.

Simplifying:

10b + a - 10a - b = 27

Which is:

9b - 9a = 27

Divide by 9:

b - a = 3

So then, we have:

From the first equation, a + b = 13

And from this, b - a = 3

So solving these equations:

Adding them: (a + b) + (b - a) = 13 + 3 => 2b = 16 => b = 8. Then, a = 13 - 8 = 5. Therefore, original number is 10a + b = 58, reversed number is 85. Then, the difference between reversed and original is 85 - 58 = 27, which matches. So in this case, the original number is 58, reversed is 85, and their sum is 58 + 85 = 143 as well. So regardless of which way the difference is, the sum remains the same. Therefore, the answer is 143.

Wait, so even if the original number was 58, which is smaller than 85, the difference is still 27 when you subtract the original from the reversed. So, in both cases, whether the original number is 85 or 58, the sum of the original number and reversed number is 143. Therefore, the answer is 143. So that seems to check out.

Alternatively, maybe there's another way to approach this problem. Let me think. For example, if we let the original number be N, and reversed number be M. Then, N - M = 27 or M - N = 27. Also, digits of N add up to 13. Also, if N is a two-digit number, then N = 10a + b, M = 10b + a. So, their sum is N + M = 10a + b + 10b + a = 11a + 11b = 11(a + b). Since a + b is given as 13, then N + M = 11 * 13 = 143. Wait, that's interesting. So, in fact, regardless of the individual values of a and b, if we know that a + b = 13, then N + M is 11*(a + b) = 11*13 = 143. Therefore, the answer is 143. That's a much more straightforward way. So maybe I didn't need to go through all the equations?

Wait, but why does this work? Because N + M is 11(a + b). That's because N = 10a + b and M = 10b + a. So adding them gives 10a + b + 10b + a = 11a + 11b = 11(a + b). Therefore, if we know that a + b is 13, then N + M is 11*13 = 143. Therefore, regardless of what the individual digits are, as long as their sum is 13, the sum of the number and its reverse is 143. Therefore, even without finding the individual numbers, we can compute the answer directly. But the problem also gave the condition about the difference between the number and its reverse being 27. However, that condition is actually not necessary for computing N + M? Wait, is that possible?

Wait, but if the problem didn't give the difference condition, but only the sum of digits, then the sum N + M is 11*(a + b), which is fixed once a + b is known. However, in this problem, they give the difference condition, which may be redundant? Wait, no. Wait, maybe not. For example, suppose that the sum of digits is 13, but there are multiple numbers that satisfy that. However, the difference between the number and its reverse is 27. So, actually, the difference condition is necessary to confirm that such a number exists, or maybe to find the specific numbers? But in this case, even if we don't use the difference condition, if we know the sum of digits, we can compute N + M. However, perhaps the problem gives both conditions to make sure that there is a unique solution? Because if the sum of digits is 13, then N + M is fixed, but maybe there are multiple numbers with digit sum 13. However, the difference condition is additional information to narrow down the possibilities? Wait, but in reality, with the sum of digits and the difference between the number and its reverse, we can find the numbers. But regardless, the problem is asking for N + M, which is always 11*(sum of digits). Therefore, even if there are multiple numbers with digit sum 13, their sum with their reverse is always 11*13. Wait, but let me check with an example. Let's say the digits add up to 13. Let me take a different example. Suppose a two-digit number with digits adding to 13. Let's pick a number, say 49. Sum of digits 4 + 9 = 13. The reversed number is 94. The sum of original and reversed is 49 + 94 = 143. Same result. Another example: 58. 5 + 8 = 13. Reverse is 85. Sum is 58 + 85 = 143. Another example: 67. 6 + 7 = 13. Reverse is 76. Sum is 67 + 76 = 143. Same answer. Similarly, 76, 85, 94. So all of these pairs add up to 143. Therefore, regardless of the specific number, if the digits add up to 13, the sum of the number and its reverse is 143. Therefore, even if the problem had given different conditions, as long as the sum of digits is 13, the answer would still be 143. However, the problem also gives the difference condition. So why is that given? Maybe to check if the number actually exists? Because there are two-digit numbers with digits adding to 13, but the difference between the number and its reverse is 27. For example, take 49. The difference is 94 - 49 = 45. Not 27. For 58: 85 - 58 = 27. So that works. Similarly, 67: reverse is 76. Difference is 76 - 67 = 9. So that's not 27. Similarly, 85: 85 - 58 = 27. So, the difference is 27 only for some numbers with digits summing to 13. Therefore, the problem gives the difference condition to narrow down the possible numbers. However, since the problem is asking for N + M, which is fixed once the digits sum to 13, regardless of the difference. Therefore, even if the difference condition wasn't given, the answer would still be 143. However, since the problem gives both conditions, we can use them to verify that such numbers exist. But maybe the problem is designed to see if you can find the individual numbers first and then add them, or if you can find the sum directly. But either way, the answer is 143. Therefore, I think the answer is 143. However, let me make sure that I didn't make a mistake in my reasoning.

Wait, but the problem says "the difference between the number and the number with its digits reversed is 27." So if we take the number 58, the difference between 58 and its reversal 85 is 85 - 58 = 27. So that's 27. If we take 85, difference is 85 - 58 = 27. So in both cases, it's 27. So, in this case, both numbers satisfy the difference condition. Therefore, the problem is consistent. Therefore, since the sum is always 143, the answer is 143.

But let me think again. Suppose that the problem had different numbers. For example, if the sum of the digits was something else, say 10, and the difference was something else, then the answer would depend on both conditions. But in this case, because the sum of digits is given, the sum of the number and the reverse is fixed. Therefore, even without the difference, the answer is fixed. Therefore, maybe the difference condition is redundant? However, the problem gives both conditions, so maybe it's to ensure that such a number exists? Because if the sum of digits was 13, but the difference between the number and its reverse is something impossible, then there would be no solution. However, in this case, there are solutions. So maybe the problem is just giving both conditions, but the answer can be found via the sum of digits. However, maybe in the problem's original context, the difference is necessary to find the digits? But since I can find the answer without it, but maybe in some other problems, the difference is necessary.

But in any case, since both approaches lead me to 143, I can be confident that the answer is 143. Therefore, I think the answer is \boxed{143}.

**Final Answer**
\boxed{143}

To solve this problem, we are given two conditions about a two-digit number:

1. The **sum of the digits** of the number is 13.
2. The **difference** between the number and the number with its digits reversed is 27.

We are asked to find the **sum** of the original number and the number with its digits reversed.

---

### Step 1: Represent the Number Algebraically

Let the tens digit be $ a $, and the units digit be $ b $. Then the original number is:

$$
N = 10a + b
$$

When the digits are reversed, the number becomes:

$$
N' = 10b + a
$$

---

### Step 2: Use the Given Conditions

**Condition 1: Sum of the digits is 13**
$$
a + b = 13 \tag{1}
$$

**Condition 2: Difference between the number and its reverse is 27**
$$
N - N' = 27 \Rightarrow (10a + b) - (10b + a) = 27
$$

Simplifying:

$$
10a + b - 10b - a = 27 \Rightarrow 9a - 9b = 27 \Rightarrow a - b = 3 \tag{2}
$$

---

### Step 3: Solve the System of Equations

We now solve equations (1) and (2):

From (1): $ a + b = 13 $
From (2): $ a - b = 3 $

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:

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

Its reverse is:

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

---

### Step 4: Find the Sum of the Original and the Reversed Number

$$
N + N' = 85 + 58 = 143
$$

---

### Step 5: Optional General Insight

We can also solve this directly using the algebraic form of the numbers:

The sum of the original and reversed number is:

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

We are told $ a + b = 13 $, so:

$$
N + N' = 11 \times 13 = \boxed{143}
$$

---

### Final Answer

$$
\boxed{143}
$$

[Assistant]
Explanation:
Okay, so I need to solve this problem about a two-digit number. Let me try to understand what it's asking. The problem says that the sum of the digits of a two-digit number is 13. Also, the difference between the number and the number with its digits reversed is 27. Then, I need to find the sum of the original number and the reversed number. Alright, let me break this down step by step.

First, let's denote the original two-digit number. Let me call the tens digit 'a' and the units digit 'b'. So, the original number would be 10a + b because in a two-digit number, the tens place is multiplied by 10. For example, if the number is 23, then a is 2 and b is 3, so the number is 10*2 + 3 = 23. That seems right.

Now, the problem says the sum of the digits is 13. So, that would be a + b = 13. Got that. Equation one: a + b = 13.

Next, the difference between the original number and the reversed number is 27. Let me think. The reversed number would have the digits swapped, so it would be 10b + a. So, the original number is 10a + b, the reversed number is 10b + a. The problem says the difference between the original number and the reversed number is 27. Wait, does it specify which one is subtracted from which? Let me check the problem again.

"The difference between the number and the number with its digits reversed is 27." Hmm. It just says "the difference," but usually, difference can be interpreted as absolute value, but maybe in this case, they mean the original number minus the reversed number is 27? But maybe not. Wait, but maybe we need to consider both possibilities? Let me think.

Wait, but let's suppose that the original number is larger than the reversed number. If that's the case, then (original) - (reversed) = 27. If not, then (reversed) - (original) = 27. But since the problem says "the difference," maybe it's just the absolute value. However, perhaps we can figure it out from the equations. Let me proceed with that assumption first that original number minus reversed number is 27. If that leads to a contradiction, I can check the other way.

So, let me write that as:

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

Simplifying that:

10a + b - 10b - a = 27

Combine like terms:

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

Which is:

9a - 9b = 27

Divide both sides by 9:

a - b = 3

So, that gives another equation: a - b = 3.

So now, we have two equations:

1) a + b = 13

2) a - b = 3

So, we can solve these two equations simultaneously. Let me do that.

Adding equations 1 and 2:

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

Which is:

2a = 16

Therefore, a = 8.

Then, substituting 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, which would make the original number 85.

Let me check if this satisfies the second condition. The reversed number would be 58. The difference between 85 and 58 is 85 - 58 = 27. Yes, that's exactly the given difference. So that works.

Therefore, the original number is 85, reversed number is 58. The problem asks for the sum of the original number and the reversed number. So that's 85 + 58. Let me calculate that.

85 + 58: 80 + 50 = 130, 5 + 8 = 13, so total is 130 + 13 = 143. Therefore, the answer is 143. Wait, let me check that again. 85 + 58. 85 + 50 is 135, plus 8 is 143. Yes, that's right.

But hold on, let me just make sure I didn't make any mistakes in assuming that original number minus reversed number is 27. What if the difference was reversed? Let me check that scenario as well, just to be thorough.

Suppose instead that the reversed number minus original number is 27. Then, (10b + a) - (10a + b) = 27.

Simplifying:

10b + a - 10a - b = 27

Which is:

9b - 9a = 27

Divide by 9:

b - a = 3

So then, we have:

From the first equation, a + b = 13

And from this, b - a = 3

So solving these equations:

Adding them: (a + b) + (b - a) = 13 + 3 => 2b = 16 => b = 8. Then, a = 13 - 8 = 5. Therefore, original number is 10a + b = 58, reversed number is 85. Then, the difference between reversed and original is 85 - 58 = 27, which matches. So in this case, the original number is 58, reversed is 85, and their sum is 58 + 85 = 143 as well. So regardless of which way the difference is, the sum remains the same. Therefore, the answer is 143.

Wait, so even if the original number was 58, which is smaller than 85, the difference is still 27 when you subtract the original from the reversed. So, in both cases, whether the original number is 85 or 58, the sum of the original number and reversed number is 143. Therefore, the answer is 143. So that seems to check out.

Alternatively, maybe there's another way to approach this problem. Let me think. For example, if we let the original number be N, and reversed number be M. Then, N - M = 27 or M - N = 27. Also, digits of N add up to 13. Also, if N is a two-digit number, then N = 10a + b, M = 10b + a. So, their sum is N + M = 10a + b + 10b + a = 11a + 11b = 11(a + b). Since a + b is given as 13, then N + M = 11 * 13 = 143. Wait, that's interesting. So, in fact, regardless of the individual values of a and b, if we know that a + b = 13, then N + M is 11*(a + b) = 11*13 = 143. Therefore, the answer is 143. That's a much more straightforward way. So maybe I didn't need to go through all the equations?

Wait, but why does this work? Because N + M is 11(a + b). That's because N = 10a + b and M = 10b + a. So adding them gives 10a + b + 10b + a = 11a + 11b = 11(a + b). Therefore, if we know that a + b is 13, then N + M is 11*13 = 143. Therefore, regardless of what the individual digits are, as long as their sum is 13, the sum of the number and its reverse is 143. Therefore, even without finding the individual numbers, we can compute the answer directly. But the problem also gave the condition about the difference between the number and its reverse being 27. However, that condition is actually not necessary for computing N + M? Wait, is that possible?

Wait, but if the problem didn't give the difference condition, but only the sum of digits, then the sum N + M is 11*(a + b), which is fixed once a + b is known. However, in this problem, they give the difference condition, which may be redundant? Wait, no. Wait, maybe not. For example, suppose that the sum of digits is 13, but there are multiple numbers that satisfy that. However, the difference between the number and its reverse is 27. So, actually, the difference condition is necessary to confirm that such a number exists, or maybe to find the specific numbers? But in this case, even if we don't use the difference condition, if we know the sum of digits, we can compute N + M. However, perhaps the problem gives both conditions to make sure that there is a unique solution? Because if the sum of digits is 13, then N + M is fixed, but maybe there are multiple numbers with digit sum 13. However, the difference condition is additional information to narrow down the possibilities? Wait, but in reality, with the sum of digits and the difference between the number and its reverse, we can find the numbers. But regardless, the problem is asking for N + M, which is always 11*(sum of digits). Therefore, even if there are multiple numbers with digit sum 13, their sum with their reverse is always 11*13. Wait, but let me check with an example. Let's say the digits add up to 13. Let me take a different example. Suppose a two-digit number with digits adding to 13. Let's pick a number, say 49. Sum of digits 4 + 9 = 13. The reversed number is 94. The sum of original and reversed is 49 + 94 = 143. Same result. Another example: 58. 5 + 8 = 13. Reverse is 85. Sum is 58 + 85 = 143. Another example: 67. 6 + 7 = 13. Reverse is 76. Sum is 67 + 76 = 143. Same answer. Similarly, 76, 85, 94. So all of these pairs add up to 143. Therefore, regardless of the specific number, if the digits add up to 13, the sum of the number and its reverse is 143. Therefore, even if the problem had given different conditions, as long as the sum of digits is 13, the answer would still be 143. However, the problem also gives the difference condition. So why is that given? Maybe to check if the number actually exists? Because there are two-digit numbers with digits adding to 13, but the difference between the number and its reverse is 27. For example, take 49. The difference is 94 - 49 = 45. Not 27. For 58: 85 - 58 = 27. So that works. Similarly, 67: reverse is 76. Difference is 76 - 67 = 9. So that's not 27. Similarly, 85: 85 - 58 = 27. So, the difference is 27 only for some numbers with digits summing to 13. Therefore, the problem gives the difference condition to narrow down the possible numbers. However, since the problem is asking for N + M, which is fixed once the digits sum to 13, regardless of the difference. Therefore, even if the difference condition wasn't given, the answer would still be 143. However, since the problem gives both conditions, we can use them to verify that such numbers exist. But maybe the problem is designed to see if you can find the individual numbers first and then add them, or if you can find the sum directly. But either way, the answer is 143. Therefore, I think the answer is 143. However, let me make sure that I didn't make a mistake in my reasoning.

Wait, but the problem says "the difference between the number and the number with its digits reversed is 27." So if we take the number 58, the difference between 58 and its reversal 85 is 85 - 58 = 27. So that's 27. If we take 85, difference is 85 - 58 = 27. So in both cases, it's 27. So, in this case, both numbers satisfy the difference condition. Therefore, the problem is consistent. Therefore, since the sum is always 143, the answer is 143.

But let me think again. Suppose that the problem had different numbers. For example, if the sum of the digits was something else, say 10, and the difference was something else, then the answer would depend on both conditions. But in this case, because the sum of digits is given, the sum of the number and the reverse is fixed. Therefore, even without the difference, the answer is fixed. Therefore, maybe the difference condition is redundant? However, the problem gives both conditions, so maybe it's to ensure that such a number exists? Because if the sum of digits was 13, but the difference between the number and its reverse is something impossible, then there would be no solution. However, in this case, there are solutions. So maybe the problem is just giving both conditions, but the answer can be found via the sum of digits. However, maybe in the problem's original context, the difference is necessary to find the digits? But since I can find the answer without it, but maybe in some other problems, the difference is necessary.

But in any case, since both approaches lead me to 143, I can be confident that the answer is 143. Therefore, I think the answer is \boxed{143}.

**Final Answer**
\boxed{143}

To solve this problem, we are given two conditions about a two-digit number:

1. The **sum of the digits** of the number is 13.
2. The **difference** between the number and the number with its digits reversed is 27.

We are asked to find the **sum** of the original number and the number with its digits reversed.

---

### Step 1: Represent the Number Algebraically

Let the tens digit be $ a $, and the units digit be $ b $. Then the original number is:

$$
N = 10a + b
$$

When the digits are reversed, the number becomes:

$$
N' = 10b + a
$$

---

### Step 2: Use the Given Conditions

**Condition 1: Sum of the digits is 13**
$$
a + b = 13 \tag{1}
$$

**Condition 2: Difference between the number and its reverse is 27**
$$
N - N' = 27 \Rightarrow (10a + b) - (10b + a) = 27
$$

Simplifying:

$$
10a + b - 10b - a = 27 \Rightarrow 9a - 9b = 27 \Rightarrow a - b = 3 \tag{2}
$$

---

### Step 3: Solve the System of Equations

We now solve equations (1) and (2):

From (1): $ a + b = 13 $
From (2): $ a - b = 3 $

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:

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

Its reverse is:

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

---

### Step 4: Find the Sum of the Original and the Reversed Number

$$
N + N' = 85 + 58 = 143
$$

---

### Step 5: Optional General Insight

We can also solve this directly using the algebraic form of the numbers:

The sum of the original and reversed number is:

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

We are told $ a + b = 13 $, so:

$$
N + N' = 11 \times 13 = \boxed{143}
$$

---

### 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, let's see. The problem is about figuring 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. The total number of representatives is 18, and we need a fair representation. Hmm, fair representation usually means that the number of representatives each grade gets is proportional to their enrollment, right? So maybe I need to figure out the proportion of each grade's enrollment compared to the total enrollment and then apply that proportion to the total number of representatives.

Let me start by calculating the total enrollment. The 7th grade has 520 students and the 8th grade has 650. So total enrollment is 520 + 650. Let me add those numbers. 520 plus 650... Well, 500 + 600 is 1100, and 20 + 50 is 70, so total is 1170. So total enrollment is 1170 students.

Now, if we need to divide the 18 representatives proportionally, then the 7th grade should get a number of representatives equal to (520 / 1170) * 18, and the 8th grade would get (650 / 1170) * 18. But since we can't have a fraction of a representative, we might need to round or maybe there's another way. Wait, but maybe the question expects a proportional division without rounding? Let me check.

Alternatively, maybe we can find the ratio between the two grades and apply that ratio to the total number of representatives. Let me see. The ratio of 7th to 8th grade enrollments is 520:650. Let me simplify that ratio. Both numbers are divisible by 130, right? 520 divided by 130 is 4, and 650 divided by 130 is 5. So the simplified ratio is 4:5. Therefore, the 7th grade is to the 8th grade as 4 is to 5.

So the total parts in the ratio are 4 + 5 = 9 parts. Therefore, the 7th grade would get 4/9 of the representatives, and the 8th grade would get 5/9. Since there are 18 representatives total, the 8th grade should get (5/9)*18. Let me calculate that. 18 divided by 9 is 2, so 2*5 is 10. So the 8th grade should have 10 representatives. Wait, is that right?

Let me verify. If the ratio is 4:5, then total parts 9. So 18 divided by 9 is 2 per part. Therefore, 4 parts for 7th grade is 8, and 5 parts for 8th grade is 10. So total 8 + 10 is 18. That seems to add up. But let me check with the original enrollment numbers to see if this ratio holds.

Original enrollments: 520 and 650. Let me check if 520 / 650 is equal to 4 / 5. 520 divided by 650, let's divide numerator and denominator by 130, as before, gives 4/5. So that ratio is correct. So the proportion of representatives should be the same as the proportion of enrollments. Therefore, the calculation seems correct.

Alternatively, another way is to compute the proportion for each grade. For the 7th grade, (520 / 1170) * 18. Let me compute that. First, 520 divided by 1170. Let's see, 520 / 1170. Both numbers can be divided by 130 again? 520 divided by 130 is 4, 1170 divided by 130 is 9. So that's 4/9. Therefore, 4/9 * 18 is 8. So that's 8 representatives for 7th grade, and 18 - 8 = 10 for the 8th grade. Same result. So that checks out.

But just to make sure there isn't some other method or consideration. Sometimes, in problems like this, you might have to use some kind of apportionment method, like the one used in the US Congress, but since the numbers are small and the total representatives are 18, which is a manageable number, and since the ratio divides evenly, there's no need for rounding. Since 5/9 of 18 is exactly 10, there's no fractional representative. Therefore, the answer should be 10.

But let me think again. Suppose we didn't simplify the ratio. Let's say we take 520 and 650. The total is 1170. So the proportion for 8th grade is 650/1170. Let me compute that fraction. Dividing numerator and denominator by 650? Wait, 650 divides into 1170 how many times? 650*1 = 650, 1170 - 650 = 520. So 650/1170 = 650/(650 + 520) = 650/1170. If I divide numerator and denominator by 650, that's 1/(1 + 520/650). Wait, that seems more complicated. Alternatively, divide numerator and denominator by 130, which we already did. 650 divided by 130 is 5, 1170 divided by 130 is 9. So again, 5/9. So same as before.

Alternatively, maybe they want us to use some other method, like weighted average? But no, since the total representatives are fixed, and the question is about proportional representation. So, the key is that the number of representatives should be proportional to the size of the enrollment. Therefore, the answer is 10.

Wait, but let me check if there's some possible trick here. For example, sometimes problems say "fair representation" which might mean something else, like equal number of representatives per student or something. But that's not the usual meaning. Fair representation in a legislative body usually refers to proportionality. So, for example, if the student council is a legislative body, then the number of representatives should be proportional to the number of students. So I think the method we used is correct.

Alternatively, if it was a different kind of fair representation, like each grade having the same number of representatives, but that would not be fair because the grades have different sizes. Since the 8th grade is larger, they should have more representatives. So the answer is 10.

Wait, but just to make sure, let me think of another way. Suppose the ratio is 520:650, which reduces to 4:5. So for every 4 students in 7th grade, there are 5 students in 8th grade. So if we have 18 representatives, how many should be for each? The total parts are 4 + 5 = 9. So 18 divided by 9 is 2. Therefore, 4 parts * 2 = 8, 5 parts * 2 = 10. Yep. So that's the same answer.

Therefore, I think the answer is 10. So the 8th grade should have 10 representatives.

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

To determine how many representatives the 8th grade should have on the Student Council, we begin by understanding the principle of **fair representation**, which in this context means that the number of representatives should be **proportional** to the size of each grade's enrollment.

---

### Step 1: Find Total Enrollment

The 7th grade has 520 students, and the 8th grade has 650 students. Therefore, the **total enrollment** is:

$$
520 + 650 = 1170 \text{ students}
$$

---

### Step 2: Determine the Proportion of Each Grade

We now calculate the **proportion of the total enrollment** that each grade represents:

- **7th grade's proportion**:
  $$
  \frac{520}{1170}
  $$

- **8th grade's proportion**:
  $$
  \frac{650}{1170}
  $$

Both fractions can be simplified. Since both numerator and denominator are divisible by 130:

- $$
  \frac{520}{1170} = \frac{4}{9}, \quad \frac{650}{1170} = \frac{5}{9}
  $$

This tells us that the **ratio of enrollments** is $4:5$ between the 7th and 8th grades.

---

### Step 3: Apply the Ratio to the Total Representatives

The Student Council has 18 total representatives, and we want to divide them in the same 4:5 ratio as the enrollments.

First, determine the number of **parts** in the ratio:
$$
4 + 5 = 9 \text{ parts}
$$

Each part corresponds to:
$$
\frac{18}{9} = 2 \text{ representatives}
$$

Thus, the **8th grade**, which corresponds to **5 parts**, should have:
$$
5 \times 2 = 10 \text{ representatives}
$$

---

### Final Answer

$$
\boxed{10}
$$

[Assistant]
Explanation:
Okay, let's see. The problem is about figuring 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. The total number of representatives is 18, and we need a fair representation. Hmm, fair representation usually means that the number of representatives each grade gets is proportional to their enrollment, right? So maybe I need to figure out the proportion of each grade's enrollment compared to the total enrollment and then apply that proportion to the total number of representatives.

Let me start by calculating the total enrollment. The 7th grade has 520 students and the 8th grade has 650. So total enrollment is 520 + 650. Let me add those numbers. 520 plus 650... Well, 500 + 600 is 1100, and 20 + 50 is 70, so total is 1170. So total enrollment is 1170 students.

Now, if we need to divide the 18 representatives proportionally, then the 7th grade should get a number of representatives equal to (520 / 1170) * 18, and the 8th grade would get (650 / 1170) * 18. But since we can't have a fraction of a representative, we might need to round or maybe there's another way. Wait, but maybe the question expects a proportional division without rounding? Let me check.

Alternatively, maybe we can find the ratio between the two grades and apply that ratio to the total number of representatives. Let me see. The ratio of 7th to 8th grade enrollments is 520:650. Let me simplify that ratio. Both numbers are divisible by 130, right? 520 divided by 130 is 4, and 650 divided by 130 is 5. So the simplified ratio is 4:5. Therefore, the 7th grade is to the 8th grade as 4 is to 5.

So the total parts in the ratio are 4 + 5 = 9 parts. Therefore, the 7th grade would get 4/9 of the representatives, and the 8th grade would get 5/9. Since there are 18 representatives total, the 8th grade should get (5/9)*18. Let me calculate that. 18 divided by 9 is 2, so 2*5 is 10. So the 8th grade should have 10 representatives. Wait, is that right?

Let me verify. If the ratio is 4:5, then total parts 9. So 18 divided by 9 is 2 per part. Therefore, 4 parts for 7th grade is 8, and 5 parts for 8th grade is 10. So total 8 + 10 is 18. That seems to add up. But let me check with the original enrollment numbers to see if this ratio holds.

Original enrollments: 520 and 650. Let me check if 520 / 650 is equal to 4 / 5. 520 divided by 650, let's divide numerator and denominator by 130, as before, gives 4/5. So that ratio is correct. So the proportion of representatives should be the same as the proportion of enrollments. Therefore, the calculation seems correct.

Alternatively, another way is to compute the proportion for each grade. For the 7th grade, (520 / 1170) * 18. Let me compute that. First, 520 divided by 1170. Let's see, 520 / 1170. Both numbers can be divided by 130 again? 520 divided by 130 is 4, 1170 divided by 130 is 9. So that's 4/9. Therefore, 4/9 * 18 is 8. So that's 8 representatives for 7th grade, and 18 - 8 = 10 for the 8th grade. Same result. So that checks out.

But just to make sure there isn't some other method or consideration. Sometimes, in problems like this, you might have to use some kind of apportionment method, like the one used in the US Congress, but since the numbers are small and the total representatives are 18, which is a manageable number, and since the ratio divides evenly, there's no need for rounding. Since 5/9 of 18 is exactly 10, there's no fractional representative. Therefore, the answer should be 10.

But let me think again. Suppose we didn't simplify the ratio. Let's say we take 520 and 650. The total is 1170. So the proportion for 8th grade is 650/1170. Let me compute that fraction. Dividing numerator and denominator by 650? Wait, 650 divides into 1170 how many times? 650*1 = 650, 1170 - 650 = 520. So 650/1170 = 650/(650 + 520) = 650/1170. If I divide numerator and denominator by 650, that's 1/(1 + 520/650). Wait, that seems more complicated. Alternatively, divide numerator and denominator by 130, which we already did. 650 divided by 130 is 5, 1170 divided by 130 is 9. So again, 5/9. So same as before.

Alternatively, maybe they want us to use some other method, like weighted average? But no, since the total representatives are fixed, and the question is about proportional representation. So, the key is that the number of representatives should be proportional to the size of the enrollment. Therefore, the answer is 10.

Wait, but let me check if there's some possible trick here. For example, sometimes problems say "fair representation" which might mean something else, like equal number of representatives per student or something. But that's not the usual meaning. Fair representation in a legislative body usually refers to proportionality. So, for example, if the student council is a legislative body, then the number of representatives should be proportional to the number of students. So I think the method we used is correct.

Alternatively, if it was a different kind of fair representation, like each grade having the same number of representatives, but that would not be fair because the grades have different sizes. Since the 8th grade is larger, they should have more representatives. So the answer is 10.

Wait, but just to make sure, let me think of another way. Suppose the ratio is 520:650, which reduces to 4:5. So for every 4 students in 7th grade, there are 5 students in 8th grade. So if we have 18 representatives, how many should be for each? The total parts are 4 + 5 = 9. So 18 divided by 9 is 2. Therefore, 4 parts * 2 = 8, 5 parts * 2 = 10. Yep. So that's the same answer.

Therefore, I think the answer is 10. So the 8th grade should have 10 representatives.

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

To determine how many representatives the 8th grade should have on the Student Council, we begin by understanding the principle of **fair representation**, which in this context means that the number of representatives should be **proportional** to the size of each grade's enrollment.

---

### Step 1: Find Total Enrollment

The 7th grade has 520 students, and the 8th grade has 650 students. Therefore, the **total enrollment** is:

$$
520 + 650 = 1170 \text{ students}
$$

---

### Step 2: Determine the Proportion of Each Grade

We now calculate the **proportion of the total enrollment** that each grade represents:

- **7th grade's proportion**:
  $$
  \frac{520}{1170}
  $$

- **8th grade's proportion**:
  $$
  \frac{650}{1170}
  $$

Both fractions can be simplified. Since both numerator and denominator are divisible by 130:

- $$
  \frac{520}{1170} = \frac{4}{9}, \quad \frac{650}{1170} = \frac{5}{9}
  $$

This tells us that the **ratio of enrollments** is $4:5$ between the 7th and 8th grades.

---

### Step 3: Apply the Ratio to the Total Representatives

The Student Council has 18 total representatives, and we want to divide them in the same 4:5 ratio as the enrollments.

First, determine the number of **parts** in the ratio:
$$
4 + 5 = 9 \text{ parts}
$$

Each part corresponds to:
$$
\frac{18}{9} = 2 \text{ representatives}
$$

Thus, the **8th grade**, which corresponds to **5 parts**, should have:
$$
5 \times 2 = 10 \text{ representatives}
$$

---

### 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.