Explanation:
Okay, let me try to solve this equation. The problem is 3x plus 2 times (1 plus x) equals 17, and then I need to find the value of 6x plus 5. Alright, let me start by writing down the equation again to make sure I have it right: 3x + 2(1 + x) = 17.

First, I remember that when you have parentheses like this, you should probably distribute the 2 into the (1 + x). Let me do that. So, 2 times 1 is 2, and 2 times x is 2x. So the equation becomes 3x + 2 + 2x = 17. Wait, let me check that again. Yeah, distributing the 2 gives me 2*1 and 2*x, so that's correct.

Now, combining like terms. The terms with x are 3x and 2x. So, 3x plus 2x is 5x. Then, I still have that +2. So the equation simplifies to 5x + 2 = 17. Okay, that seems right. Let me double-check my steps so far. Original equation: 3x + 2(1 + x) = 17. Distribute the 2: 3x + 2*1 + 2*x = 17. Which is 3x + 2 + 2x = 17. Combine like terms: 3x + 2x is 5x, so 5x + 2 = 17. Yeah, that seems correct.

Now, I need to solve for x. So, I have 5x + 2 = 17. To isolate x, I should subtract 2 from both sides. Let me do that: 5x + 2 - 2 = 17 - 2. Which simplifies to 5x = 15. Then, divide both sides by 5: 5x / 5 = 15 / 5. So x = 3. Let me verify that. If x is 3, then plugging back into the original equation: 3*(3) + 2*(1 + 3) = 9 + 2*(4) = 9 + 8 = 17. Yep, that checks out. So x is indeed 3.

But the question isn't asking for x; it's asking for the value of 6x + 5. Alright, so once I have x = 3, I can substitute that into 6x + 5. Let me calculate that. 6 times 3 is 18, and 18 plus 5 is 23. So, 6x + 5 equals 23.

Wait, let me make sure I didn't skip any steps or make a mistake. So x is 3, so 6*3 is 18. 18 + 5 is 23. That seems straightforward. Let me just check if there's another way to approach this problem, maybe by manipulating the original equation to get 6x + 5 directly instead of solving for x first.

Looking at the original equation: 3x + 2(1 + x) = 17. Let me expand that again. 3x + 2 + 2x = 17, which is 5x + 2 = 17. So 5x = 15, so x = 3. If I wanted to get 6x + 5, maybe I could relate 6x to the existing terms. Since 5x is 15, then 6x would be 5x + x, which is 15 + 3 = 18. Then 6x + 5 is 18 + 5 = 23. So that's the same result.

Alternatively, maybe there's another algebraic manipulation. Let me see. Suppose I didn't solve for x. Let's say I have the original equation: 3x + 2(1 + x) = 17. If I want to find 6x + 5, perhaps I can express 6x + 5 in terms of the original equation.

Let me try that. Let me note that 6x + 5 is equal to 2*(3x) + 5. Hmm. From the original equation, 3x + 2(1 + x) = 17. Let me rearrange that. Let me solve for 3x. If I subtract 2(1 + x) from both sides, I get 3x = 17 - 2(1 + x). But maybe that complicates things. Alternatively, since I have 3x + 2x + 2 = 17, which is 5x + 2 = 17. If I multiply both sides of the equation by something? Let me think.

Alternatively, if I want 6x + 5, perhaps I can take the equation 5x + 2 = 17 and add x + 3 to both sides. Let me try that. Adding x + 3 to both sides: 5x + 2 + x + 3 = 17 + x + 3. Simplifies to 6x + 5 = 20 + x. Hmm, but that introduces x on the right side. Not helpful.

Alternatively, maybe if I take 5x + 2 = 17 and add x + 3 to both sides? Wait, that's similar to before. Maybe not helpful.

Alternatively, since we found x = 3, maybe that's the simplest way. So, perhaps there's no shortcut, and just solving for x and substituting is the way to go.

But let me try again. Suppose I didn't solve for x. Let's see. Starting with 3x + 2(1 + x) = 17. Let me expand that again: 3x + 2 + 2x = 17. Combine like terms: 5x + 2 = 17. So 5x = 15. Therefore, x = 3. So then, 6x + 5 is 6*3 +5 = 18 + 5 = 23.

Alternatively, maybe I can find 6x + 5 in terms of 5x + 2. Let's see. 6x + 5 = (5x + 2) + x + 3. Since 5x + 2 is 17, then 17 + x + 3 = 20 + x. But we know that x = 3, so 20 + 3 = 23. So same result.

Alternatively, if I want to find 6x + 5, and I know that 5x = 15, then 6x = 5x + x = 15 + 3 = 18, so 6x + 5 = 18 + 5 = 23.

So regardless of the method, it seems that 6x +5 equals 23. Let me just check once again with substitution. If x = 3, then 6x is 18, plus 5 is 23. Yep.

But maybe I should check if there's an alternative approach, just to make sure I didn't miss anything. For example, could I have manipulated the original equation to get 6x + 5 directly?

Let me think. Starting again: 3x + 2(1 + x) = 17. Let me expand the 2*(1 +x): 2 + 2x. So the equation is 3x + 2 + 2x = 17. Combine 3x and 2x to get 5x + 2 = 17. So 5x = 15, x=3.

Alternatively, maybe I can write 6x + 5 as something. Let me think. If I have 5x + 2 = 17, then multiplying both sides by some number... Let me think. Suppose I multiply both sides by something to get 6x + 5.

Alternatively, if I take 5x + 2 = 17 and add x + 3 to both sides: 5x +2 +x +3 = 17 +x +3. Which is 6x +5 = 20 +x. Then, subtract x from both sides: 5x +5 = 20. Then divide by 5: x +1 = 4. So x = 3. So that's again circular.

Alternatively, if I want 6x +5, maybe express in terms of original equation. Let me think. Since original equation is 3x + 2(1 +x) = 17. Let me try to write 6x +5 as something. Let me note that 6x +5 is twice 3x plus 5. So 2*(3x) + 5. From original equation, 3x is equal to 17 - 2(1 +x). Wait, maybe that's not helpful. Let me write:

From original equation: 3x = 17 - 2(1 +x). Then 2*(3x) = 2*17 - 4(1 +x). Which is 34 - 4 -4x = 30 -4x. So 2*(3x) +5 = 30 -4x +5 = 35 -4x. But then 6x +5 is equal to 35 -4x. But then how does that help me?

Hmm, maybe I need another equation. If I have 5x +2 =17, then 5x =15, so 4x = 15 -x. So substituting back into 35 -4x, we have 35 - (15 -x) = 35 -15 +x = 20 +x. Which again brings us back to 20 +x. And since x=3, that is 23.

But that's again going in circles. So maybe this isn't helpful.

Therefore, the simplest way is just solving for x first, then substituting. Which gives x=3, so 6x +5=23.

Alternatively, maybe there's a trick where you can avoid solving for x directly. For example, if we have the original equation, maybe we can manipulate it to get 6x +5. Let me think.

Original equation: 3x + 2(1 +x) =17. Let me expand the 2(1 +x): 2 +2x. So equation is 3x +2 +2x =17. Combine terms: 5x +2=17. So 5x =15. Let me note that 5x is 15, so 6x is 15 +x. Therefore, 6x +5 is 15 +x +5 =20 +x. Then, since 5x =15, x=3. Therefore, 20 +3=23.

So that's another way to think about it. So if you recognize that 6x +5 can be written as 5x + x +5 = (5x +2) +x +3. Which is 17 +x +3 = 20 +x. Then substitute x=3.

But again, you need to know x=3. So regardless of the method, you still need to find x first. So maybe there's no way around solving for x first.

Alternatively, maybe if you consider the question asks for 6x +5, and the original equation is 3x + 2(1 +x) =17. Maybe express 6x +5 in terms of the original equation.

Let me try. Let me note that 6x +5 can be thought of as 2*(3x) +5. From original equation, 3x =17 - 2(1 +x). So substituting, 2*(17 - 2(1 +x)) +5. Let me compute that.

First, compute 2*(17 -2(1 +x)): 34 -4(1 +x). Then add 5: 34 -4(1 +x) +5 = 39 -4(1 +x). Expanding that: 39 -4 -4x = 35 -4x. So 6x +5 equals 35 -4x.

But now, how do I find this? If 6x +5 =35 -4x, then adding 4x to both sides gives 10x +5 =35. Then subtract 5: 10x=30, so x=3. Then 6x +5=23.

But again, this is going through the same steps. So whether I solve for x first or manipulate the equation, I end up needing to solve for x.

Therefore, it seems that no matter how I approach it, solving for x first is necessary. Therefore, the answer is 23.

Wait, but let me check once again for any calculation errors. Let me recompute 6x +5 when x is 3. 6 times 3 is 18. 18 plus 5 is 23. Correct.

Original equation: 3x +2(1 +x) =17. Plug in x=3: 3*3=9, 2*(1+3)=2*4=8. 9 +8=17. Correct.

So yes, all steps check out. Therefore, the value of 6x +5 is 23.

I think that's solid. No mistakes in my reasoning, and multiple ways of looking at it lead to the same answer. So I'm confident that 23 is the correct answer.

**Final Answer**
The value of $6x + 5$ is \boxed{23}.

To solve the given problem, we start with the equation:

$$
3x + 2(1 + x) = 17
$$

---

### Step 1: Distribute the 2

Apply the distributive property to the term inside the parentheses:

$$
3x + 2 \cdot 1 + 2 \cdot x = 17
$$
$$
3x + 2 + 2x = 17
$$

---

### Step 2: Combine Like Terms

Combine the $x$-terms on the left-hand side:

$$
(3x + 2x) + 2 = 17
$$
$$
5x + 2 = 17
$$

---

### Step 3: Solve for $x$

Isolate $x$ by first subtracting 2 from both sides:

$$
5x = 15
$$

Then divide both sides by 5:

$$
x = 3
$$

---

### Step 4: Find the Value of $6x + 5$

Now that we know $x = 3$, we substitute this value into the expression $6x + 5$:

$$
6x + 5 = 6(3) + 5 = 18 + 5 = 23
$$

---

### Final Answer

$$
\boxed{23}
$$

[Assistant]
Explanation:
Okay, let me try to solve this equation. The problem is 3x plus 2 times (1 plus x) equals 17, and then I need to find the value of 6x plus 5. Alright, let me start by writing down the equation again to make sure I have it right: 3x + 2(1 + x) = 17.

First, I remember that when you have parentheses like this, you should probably distribute the 2 into the (1 + x). Let me do that. So, 2 times 1 is 2, and 2 times x is 2x. So the equation becomes 3x + 2 + 2x = 17. Wait, let me check that again. Yeah, distributing the 2 gives me 2*1 and 2*x, so that's correct.

Now, combining like terms. The terms with x are 3x and 2x. So, 3x plus 2x is 5x. Then, I still have that +2. So the equation simplifies to 5x + 2 = 17. Okay, that seems right. Let me double-check my steps so far. Original equation: 3x + 2(1 + x) = 17. Distribute the 2: 3x + 2*1 + 2*x = 17. Which is 3x + 2 + 2x = 17. Combine like terms: 3x + 2x is 5x, so 5x + 2 = 17. Yeah, that seems correct.

Now, I need to solve for x. So, I have 5x + 2 = 17. To isolate x, I should subtract 2 from both sides. Let me do that: 5x + 2 - 2 = 17 - 2. Which simplifies to 5x = 15. Then, divide both sides by 5: 5x / 5 = 15 / 5. So x = 3. Let me verify that. If x is 3, then plugging back into the original equation: 3*(3) + 2*(1 + 3) = 9 + 2*(4) = 9 + 8 = 17. Yep, that checks out. So x is indeed 3.

But the question isn't asking for x; it's asking for the value of 6x + 5. Alright, so once I have x = 3, I can substitute that into 6x + 5. Let me calculate that. 6 times 3 is 18, and 18 plus 5 is 23. So, 6x + 5 equals 23.

Wait, let me make sure I didn't skip any steps or make a mistake. So x is 3, so 6*3 is 18. 18 + 5 is 23. That seems straightforward. Let me just check if there's another way to approach this problem, maybe by manipulating the original equation to get 6x + 5 directly instead of solving for x first.

Looking at the original equation: 3x + 2(1 + x) = 17. Let me expand that again. 3x + 2 + 2x = 17, which is 5x + 2 = 17. So 5x = 15, so x = 3. If I wanted to get 6x + 5, maybe I could relate 6x to the existing terms. Since 5x is 15, then 6x would be 5x + x, which is 15 + 3 = 18. Then 6x + 5 is 18 + 5 = 23. So that's the same result.

Alternatively, maybe there's another algebraic manipulation. Let me see. Suppose I didn't solve for x. Let's say I have the original equation: 3x + 2(1 + x) = 17. If I want to find 6x + 5, perhaps I can express 6x + 5 in terms of the original equation.

Let me try that. Let me note that 6x + 5 is equal to 2*(3x) + 5. Hmm. From the original equation, 3x + 2(1 + x) = 17. Let me rearrange that. Let me solve for 3x. If I subtract 2(1 + x) from both sides, I get 3x = 17 - 2(1 + x). But maybe that complicates things. Alternatively, since I have 3x + 2x + 2 = 17, which is 5x + 2 = 17. If I multiply both sides of the equation by something? Let me think.

Alternatively, if I want 6x + 5, perhaps I can take the equation 5x + 2 = 17 and add x + 3 to both sides. Let me try that. Adding x + 3 to both sides: 5x + 2 + x + 3 = 17 + x + 3. Simplifies to 6x + 5 = 20 + x. Hmm, but that introduces x on the right side. Not helpful.

Alternatively, maybe if I take 5x + 2 = 17 and add x + 3 to both sides? Wait, that's similar to before. Maybe not helpful.

Alternatively, since we found x = 3, maybe that's the simplest way. So, perhaps there's no shortcut, and just solving for x and substituting is the way to go.

But let me try again. Suppose I didn't solve for x. Let's see. Starting with 3x + 2(1 + x) = 17. Let me expand that again: 3x + 2 + 2x = 17. Combine like terms: 5x + 2 = 17. So 5x = 15. Therefore, x = 3. So then, 6x + 5 is 6*3 +5 = 18 + 5 = 23.

Alternatively, maybe I can find 6x + 5 in terms of 5x + 2. Let's see. 6x + 5 = (5x + 2) + x + 3. Since 5x + 2 is 17, then 17 + x + 3 = 20 + x. But we know that x = 3, so 20 + 3 = 23. So same result.

Alternatively, if I want to find 6x + 5, and I know that 5x = 15, then 6x = 5x + x = 15 + 3 = 18, so 6x + 5 = 18 + 5 = 23.

So regardless of the method, it seems that 6x +5 equals 23. Let me just check once again with substitution. If x = 3, then 6x is 18, plus 5 is 23. Yep.

But maybe I should check if there's an alternative approach, just to make sure I didn't miss anything. For example, could I have manipulated the original equation to get 6x + 5 directly?

Let me think. Starting again: 3x + 2(1 + x) = 17. Let me expand the 2*(1 +x): 2 + 2x. So the equation is 3x + 2 + 2x = 17. Combine 3x and 2x to get 5x + 2 = 17. So 5x = 15, x=3.

Alternatively, maybe I can write 6x + 5 as something. Let me think. If I have 5x + 2 = 17, then multiplying both sides by some number... Let me think. Suppose I multiply both sides by something to get 6x + 5.

Alternatively, if I take 5x + 2 = 17 and add x + 3 to both sides: 5x +2 +x +3 = 17 +x +3. Which is 6x +5 = 20 +x. Then, subtract x from both sides: 5x +5 = 20. Then divide by 5: x +1 = 4. So x = 3. So that's again circular.

Alternatively, if I want 6x +5, maybe express in terms of original equation. Let me think. Since original equation is 3x + 2(1 +x) = 17. Let me try to write 6x +5 as something. Let me note that 6x +5 is twice 3x plus 5. So 2*(3x) + 5. From original equation, 3x is equal to 17 - 2(1 +x). Wait, maybe that's not helpful. Let me write:

From original equation: 3x = 17 - 2(1 +x). Then 2*(3x) = 2*17 - 4(1 +x). Which is 34 - 4 -4x = 30 -4x. So 2*(3x) +5 = 30 -4x +5 = 35 -4x. But then 6x +5 is equal to 35 -4x. But then how does that help me?

Hmm, maybe I need another equation. If I have 5x +2 =17, then 5x =15, so 4x = 15 -x. So substituting back into 35 -4x, we have 35 - (15 -x) = 35 -15 +x = 20 +x. Which again brings us back to 20 +x. And since x=3, that is 23.

But that's again going in circles. So maybe this isn't helpful.

Therefore, the simplest way is just solving for x first, then substituting. Which gives x=3, so 6x +5=23.

Alternatively, maybe there's a trick where you can avoid solving for x directly. For example, if we have the original equation, maybe we can manipulate it to get 6x +5. Let me think.

Original equation: 3x + 2(1 +x) =17. Let me expand the 2(1 +x): 2 +2x. So equation is 3x +2 +2x =17. Combine terms: 5x +2=17. So 5x =15. Let me note that 5x is 15, so 6x is 15 +x. Therefore, 6x +5 is 15 +x +5 =20 +x. Then, since 5x =15, x=3. Therefore, 20 +3=23.

So that's another way to think about it. So if you recognize that 6x +5 can be written as 5x + x +5 = (5x +2) +x +3. Which is 17 +x +3 = 20 +x. Then substitute x=3.

But again, you need to know x=3. So regardless of the method, you still need to find x first. So maybe there's no way around solving for x first.

Alternatively, maybe if you consider the question asks for 6x +5, and the original equation is 3x + 2(1 +x) =17. Maybe express 6x +5 in terms of the original equation.

Let me try. Let me note that 6x +5 can be thought of as 2*(3x) +5. From original equation, 3x =17 - 2(1 +x). So substituting, 2*(17 - 2(1 +x)) +5. Let me compute that.

First, compute 2*(17 -2(1 +x)): 34 -4(1 +x). Then add 5: 34 -4(1 +x) +5 = 39 -4(1 +x). Expanding that: 39 -4 -4x = 35 -4x. So 6x +5 equals 35 -4x.

But now, how do I find this? If 6x +5 =35 -4x, then adding 4x to both sides gives 10x +5 =35. Then subtract 5: 10x=30, so x=3. Then 6x +5=23.

But again, this is going through the same steps. So whether I solve for x first or manipulate the equation, I end up needing to solve for x.

Therefore, it seems that no matter how I approach it, solving for x first is necessary. Therefore, the answer is 23.

Wait, but let me check once again for any calculation errors. Let me recompute 6x +5 when x is 3. 6 times 3 is 18. 18 plus 5 is 23. Correct.

Original equation: 3x +2(1 +x) =17. Plug in x=3: 3*3=9, 2*(1+3)=2*4=8. 9 +8=17. Correct.

So yes, all steps check out. Therefore, the value of 6x +5 is 23.

I think that's solid. No mistakes in my reasoning, and multiple ways of looking at it lead to the same answer. So I'm confident that 23 is the correct answer.

**Final Answer**
The value of $6x + 5$ is \boxed{23}.

To solve the given problem, we start with the equation:

$$
3x + 2(1 + x) = 17
$$

---

### Step 1: Distribute the 2

Apply the distributive property to the term inside the parentheses:

$$
3x + 2 \cdot 1 + 2 \cdot x = 17
$$
$$
3x + 2 + 2x = 17
$$

---

### Step 2: Combine Like Terms

Combine the $x$-terms on the left-hand side:

$$
(3x + 2x) + 2 = 17
$$
$$
5x + 2 = 17
$$

---

### Step 3: Solve for $x$

Isolate $x$ by first subtracting 2 from both sides:

$$
5x = 15
$$

Then divide both sides by 5:

$$
x = 3
$$

---

### Step 4: Find the Value of $6x + 5$

Now that we know $x = 3$, we substitute this value into the expression $6x + 5$:

$$
6x + 5 = 6(3) + 5 = 18 + 5 = 23
$$

---

### Final Answer

$$
\boxed{23}
$$

Solution: Expanding and collecting terms on the left hand side of the first equation gives $5x+2=17$. Subtracting 2 from each side gives $5x=15$, then dividing each side by 5 gives $x=3$. Now that we know what $x$ is, we can substitute it into $6x+5$ and get $6(3)+5=18+5=\boxed{23}$.


Explanation:
Okay, let's see. I need to add 313.9 and 12.6. Hmm, decimals. Alright, so adding decimals. I remember that when you add decimals, you need to line up the decimal points. Let me think. So maybe I should write them down vertically? Let me try that.

First, let me write 313.9. Wait, how many decimal places does it have? It has one decimal place. And 12.6 also has one decimal place. So if I line them up, the decimal points should be aligned. Let me write that out:

 313.9
+ 12.6
--------
?

Okay, so adding the numbers. Let me start from the rightmost digit. The tenths place: 9 plus 6. That's 15. So I write down 5 and carry over 1. Then moving to the units place: 3 plus 2 is 5, plus the carried over 1 makes 6. Then the tens place: 1 plus 2 is 3. Then the hundreds place: 3. Wait, but let me check again.

Wait, maybe I should break it down step by step. Let me write each number with the same number of decimal places. Since both have one decimal place, maybe I can write 313.9 as 313.9 and 12.6 as 12.6. Adding them together.

Alternatively, I can convert them to fractions? Wait, maybe that's more complicated. Let me stick with decimals.

Adding the tenths: 0.9 + 0.6 = 1.5. Then adding the whole numbers: 313 + 12 = 325. Then adding the 1.5 from the decimals. So 325 + 1.5 = 326.5. So the total would be 326.5?

Wait, let me verify. Maybe I should do it column-wise again.

Let me write:

 313.9
+ 12.6
--------
Start from the tenths place: 9 + 6 = 15. So write 5, carry over 1.

Then units place: 3 (from 313.9) + 2 (from 12.6) = 5, plus the carried over 1 is 6.

Then tens place: 1 (from 313.9) + 1 (from 12.6) = 2? Wait, hold on. Wait, the numbers are:

Let me write them properly:

 313.9
+  12.6
--------
So breaking down the digits:

Starting from the rightmost digit:

Tenths place: 9 + 6 = 15. So 5 in the tenths place, carry over 1.

Units place: 3 (from 313.9) + 2 (from 12.6) = 5, plus the carried over 1 makes 6. So 6 in the units place.

Tens place: 1 (from 313.9) + 1 (from 12.6) = 2? Wait, 313.9 is 3 hundreds, 1 ten, 3 units, and 9 tenths. And 12.6 is 0 hundreds, 1 ten, 2 units, and 6 tenths. So when adding:

Hundreds place: 3 + 0 = 3.

Tens place: 1 + 1 = 2.

Units place: 3 + 2 = 5, plus the carryover 1 is 6.

Tenths place: 9 + 6 = 15, so 5 tenths and carryover 1.

Wait, but then the carryover 1 would go to the units place? Wait, no. Wait, when you add tenths, you carry over to the units place? Wait, tenths are one decimal place, so adding them gives a carryover to the units place? Wait, 9 + 6 = 15 tenths, which is 1.5. So when adding decimals, if you have 15 tenths, that is 1 unit and 5 tenths. So that would mean you add 1 to the units place. So the units place would be 3 + 2 + 1 = 6, as before.

So then, putting it all together, the result should be 326.5.

Wait, let me check with another method. Let's convert both numbers to fractions.

313.9 is equal to 313 and 9/10, which is 313 + 9/10.

12.6 is 12 and 6/10, which is 12 + 6/10.

Adding them together: (313 + 12) + (9/10 + 6/10) = 325 + 15/10.

Simplify 15/10 to 1.5, so 325 + 1.5 = 326.5. Yep, same answer.

Alternatively, maybe adding them as decimals directly. Let me think. If I ignore the decimals for a moment, 313.9 is 3139 tenths, and 12.6 is 126 tenths. Adding them together: 3139 + 126 = 3265 tenths, which is 326.5. Exactly. So that works.

Another way: think of 313.9 + 12.6. Let me add 12.6 to 313.9. Maybe break down 12.6 into 10 + 2.6. So first add 10 to 313.9: 313.9 + 10 = 323.9. Then add 2.6: 323.9 + 2.6. Again, breaking it down: 323.9 + 2 = 325.9, then +0.6 = 326.5. Same result.

Alternatively, adding 313.9 + 12.6. Think of 313.9 + 12.6 as 313 + 0.9 + 12 + 0.6 = (313 + 12) + (0.9 + 0.6) = 325 + 1.5 = 326.5.

All methods lead to 326.5. So I think that is correct.

Wait, but let me make sure there are no mistakes in my steps. Sometimes when you carry over numbers, you can make a mistake. Let me do the vertical addition again, carefully:

 313.9
+  12.6
--------
Starting from the tenths place: 9 + 6 = 15. Write down 5, carry over 1.

Units place: 3 + 2 = 5. Plus the carried over 1 makes 6. Write down 6.

Tens place: 1 + 1 = 2. Write down 2.

Hundreds place: 3 + 0 = 3. Write down 3.

So altogether, 326.5. Yes, that's correct. No decimal points in the hundreds place, so the decimal is placed after the 326. So 326.5.

Another check: Use estimation. 313.9 is approximately 314, and 12.6 is approximately 13. Adding 314 + 13 = 327. The exact answer is 326.5, which is close to 327, so that makes sense. The estimation is slightly higher because we rounded up both numbers. So 326.5 is reasonable.

Alternatively, if I subtract 0.5 from 327, I get 326.5, which matches. So all checks out.

Therefore, the result is 326.5.

**Final Answer**
The result of adding 313.9 and 12.6 is \boxed{326.5}.

To add the two decimals $313.9$ and $12.6$, we follow a standard procedure for decimal addition, aligning the decimal points and performing the addition column by column from right to left, including any necessary carryovers.

---

### Step 1: Align the Numbers
We align the numbers based on their decimal points:

```
  313.9
+  12.6
--------
```

---

### Step 2: Add Tenths Place
Add the digits in the tenths (decimal) place:

$$
9 + 6 = 15
$$

Write down **5** in the tenths place and **carry over 1** to the units place.

---

### Step 3: Add Units Place
Now, add the digits in the units place along with the carry-over:

$$
3 + 2 + 1 = 6
$$

Write down **6** in the units place.

---

### Step 4: Add Tens Place
Add the tens digits:

$$
1 + 1 = 2
$$

Write down **2** in the tens place.

---

### Step 5: Add Hundreds Place
Add the hundreds digits:

$$
3 + 0 = 3
$$

Write down **3** in the hundreds place.

---

### Final Sum
Putting it all together, we get:

$$
326.5
$$

---

### Verification
To verify, we can also add the whole numbers and the decimal parts separately:

- Whole numbers: $313 + 12 = 325$
- Decimal parts: $0.9 + 0.6 = 1.5$
- Total: $325 + 1.5 = 326.5$

This matches our earlier result, confirming the accuracy of the addition.

---

### Final Answer
$$
\boxed{326.5}
$$

[Assistant]
Explanation:
Okay, let's see. I need to add 313.9 and 12.6. Hmm, decimals. Alright, so adding decimals. I remember that when you add decimals, you need to line up the decimal points. Let me think. So maybe I should write them down vertically? Let me try that.

First, let me write 313.9. Wait, how many decimal places does it have? It has one decimal place. And 12.6 also has one decimal place. So if I line them up, the decimal points should be aligned. Let me write that out:

 313.9
+ 12.6
--------
?

Okay, so adding the numbers. Let me start from the rightmost digit. The tenths place: 9 plus 6. That's 15. So I write down 5 and carry over 1. Then moving to the units place: 3 plus 2 is 5, plus the carried over 1 makes 6. Then the tens place: 1 plus 2 is 3. Then the hundreds place: 3. Wait, but let me check again.

Wait, maybe I should break it down step by step. Let me write each number with the same number of decimal places. Since both have one decimal place, maybe I can write 313.9 as 313.9 and 12.6 as 12.6. Adding them together.

Alternatively, I can convert them to fractions? Wait, maybe that's more complicated. Let me stick with decimals.

Adding the tenths: 0.9 + 0.6 = 1.5. Then adding the whole numbers: 313 + 12 = 325. Then adding the 1.5 from the decimals. So 325 + 1.5 = 326.5. So the total would be 326.5?

Wait, let me verify. Maybe I should do it column-wise again.

Let me write:

 313.9
+ 12.6
--------
Start from the tenths place: 9 + 6 = 15. So write 5, carry over 1.

Then units place: 3 (from 313.9) + 2 (from 12.6) = 5, plus the carried over 1 is 6.

Then tens place: 1 (from 313.9) + 1 (from 12.6) = 2? Wait, hold on. Wait, the numbers are:

Let me write them properly:

 313.9
+  12.6
--------
So breaking down the digits:

Starting from the rightmost digit:

Tenths place: 9 + 6 = 15. So 5 in the tenths place, carry over 1.

Units place: 3 (from 313.9) + 2 (from 12.6) = 5, plus the carried over 1 makes 6. So 6 in the units place.

Tens place: 1 (from 313.9) + 1 (from 12.6) = 2? Wait, 313.9 is 3 hundreds, 1 ten, 3 units, and 9 tenths. And 12.6 is 0 hundreds, 1 ten, 2 units, and 6 tenths. So when adding:

Hundreds place: 3 + 0 = 3.

Tens place: 1 + 1 = 2.

Units place: 3 + 2 = 5, plus the carryover 1 is 6.

Tenths place: 9 + 6 = 15, so 5 tenths and carryover 1.

Wait, but then the carryover 1 would go to the units place? Wait, no. Wait, when you add tenths, you carry over to the units place? Wait, tenths are one decimal place, so adding them gives a carryover to the units place? Wait, 9 + 6 = 15 tenths, which is 1.5. So when adding decimals, if you have 15 tenths, that is 1 unit and 5 tenths. So that would mean you add 1 to the units place. So the units place would be 3 + 2 + 1 = 6, as before.

So then, putting it all together, the result should be 326.5.

Wait, let me check with another method. Let's convert both numbers to fractions.

313.9 is equal to 313 and 9/10, which is 313 + 9/10.

12.6 is 12 and 6/10, which is 12 + 6/10.

Adding them together: (313 + 12) + (9/10 + 6/10) = 325 + 15/10.

Simplify 15/10 to 1.5, so 325 + 1.5 = 326.5. Yep, same answer.

Alternatively, maybe adding them as decimals directly. Let me think. If I ignore the decimals for a moment, 313.9 is 3139 tenths, and 12.6 is 126 tenths. Adding them together: 3139 + 126 = 3265 tenths, which is 326.5. Exactly. So that works.

Another way: think of 313.9 + 12.6. Let me add 12.6 to 313.9. Maybe break down 12.6 into 10 + 2.6. So first add 10 to 313.9: 313.9 + 10 = 323.9. Then add 2.6: 323.9 + 2.6. Again, breaking it down: 323.9 + 2 = 325.9, then +0.6 = 326.5. Same result.

Alternatively, adding 313.9 + 12.6. Think of 313.9 + 12.6 as 313 + 0.9 + 12 + 0.6 = (313 + 12) + (0.9 + 0.6) = 325 + 1.5 = 326.5.

All methods lead to 326.5. So I think that is correct.

Wait, but let me make sure there are no mistakes in my steps. Sometimes when you carry over numbers, you can make a mistake. Let me do the vertical addition again, carefully:

 313.9
+  12.6
--------
Starting from the tenths place: 9 + 6 = 15. Write down 5, carry over 1.

Units place: 3 + 2 = 5. Plus the carried over 1 makes 6. Write down 6.

Tens place: 1 + 1 = 2. Write down 2.

Hundreds place: 3 + 0 = 3. Write down 3.

So altogether, 326.5. Yes, that's correct. No decimal points in the hundreds place, so the decimal is placed after the 326. So 326.5.

Another check: Use estimation. 313.9 is approximately 314, and 12.6 is approximately 13. Adding 314 + 13 = 327. The exact answer is 326.5, which is close to 327, so that makes sense. The estimation is slightly higher because we rounded up both numbers. So 326.5 is reasonable.

Alternatively, if I subtract 0.5 from 327, I get 326.5, which matches. So all checks out.

Therefore, the result is 326.5.

**Final Answer**
The result of adding 313.9 and 12.6 is \boxed{326.5}.

To add the two decimals $313.9$ and $12.6$, we follow a standard procedure for decimal addition, aligning the decimal points and performing the addition column by column from right to left, including any necessary carryovers.

---

### Step 1: Align the Numbers
We align the numbers based on their decimal points:

```
  313.9
+  12.6
--------
```

---

### Step 2: Add Tenths Place
Add the digits in the tenths (decimal) place:

$$
9 + 6 = 15
$$

Write down **5** in the tenths place and **carry over 1** to the units place.

---

### Step 3: Add Units Place
Now, add the digits in the units place along with the carry-over:

$$
3 + 2 + 1 = 6
$$

Write down **6** in the units place.

---

### Step 4: Add Tens Place
Add the tens digits:

$$
1 + 1 = 2
$$

Write down **2** in the tens place.

---

### Step 5: Add Hundreds Place
Add the hundreds digits:

$$
3 + 0 = 3
$$

Write down **3** in the hundreds place.

---

### Final Sum
Putting it all together, we get:

$$
326.5
$$

---

### Verification
To verify, we can also add the whole numbers and the decimal parts separately:

- Whole numbers: $313 + 12 = 325$
- Decimal parts: $0.9 + 0.6 = 1.5$
- Total: $325 + 1.5 = 326.5$

This matches our earlier result, confirming the accuracy of the addition.

---

### Final Answer
$$
\boxed{326.5}
$$

Solution:
We have \[
\begin{array}{@{}c@{}c@{}c@{}c@{}c@{}c}
& & & 1 & \\
& 3 & 1 & 3. & 9 \\
+ & & 1 & 2. & 6
\\ \cline{1-5}
& 3 & 2 & 6. & 5 \\
\end{array}
\]Since $9+6=15$ in the right-most column is greater than 9, we record the 5 and "carry" the 10 to the next column as a 1 (shown over the second 3 in the addend 313.9). The answer is $\boxed{326.5}$.


Explanation:
Okay, let me try to figure out this problem. The question is asking for the nonzero value of N that makes the number 21420N divisible by 6. And N is the units digit of the number 21420N. Hmm, wait, actually, let me parse that again. Wait, the number is 21420N, and N is the units digit. So does that mean the number is 21420N where N is a single digit? Wait, maybe the number is 21420 followed by the digit N? For example, if N is 3, then the number is 214203? So the number is 21420N, where N is a digit from 0 to 9, and we need to find the nonzero N such that the entire number is divisible by 6.

Alright, so first, let me recall that a number is divisible by 6 if and only if it is divisible by both 2 and 3. So, I need to find N (nonzero) such that 21420N is divisible by 2 and 3.

Let me start with divisibility by 2. A number is divisible by 2 if its units digit is even. Since N is the units digit, that means N has to be even. So possible values for N are 2, 4, 6, or 8. But we also need to check divisibility by 3.

Divisibility by 3 requires that the sum of the digits is a multiple of 3. So let me compute the sum of the digits of 21420N. Let me write down the digits:

The number is 2, 1, 4, 2, 0, N. Wait, hold on, let me confirm. The original number is 21420N. If it's written as 21420N, then it's 21420 followed by the digit N. So that would be a 6-digit number? Wait, 21420 is a 5-digit number. So adding a digit N at the end makes it a 6-digit number. So the digits are 2, 1, 4, 2, 0, N. Therefore, the digits are 2, 1, 4, 2, 0, and N.

So the sum of the digits is 2 + 1 + 4 + 2 + 0 + N. Let me compute that. 2 + 1 is 3, plus 4 is 7, plus 2 is 9, plus 0 is still 9, so total sum is 9 + N. Therefore, the sum is 9 + N. For the number to be divisible by 3, 9 + N must be divisible by 3.

Since 9 is already divisible by 3 (because 9 ÷ 3 = 3), adding N must result in another multiple of 3. Therefore, N must be such that N is divisible by 3. Because 9 is a multiple of 3, adding N must also result in a multiple of 3. Therefore, N must be a multiple of 3. However, earlier we found that N must be even (for divisibility by 2). So N has to be both even and a multiple of 3.

So the possible values for N are digits from 0-9 that are both even and multiples of 3. Let me list the digits:

Multiples of 3 between 0-9: 0, 3, 6, 9.

But N has to be even, so from those, the even ones are 0 and 6. However, the problem says "nonzero value of N", so N can't be 0. Therefore, the only possible value is 6. Therefore, N = 6.

Wait, but let me confirm. Let me check if 214206 is divisible by 6. Let me check divisibility by 2: the last digit is 6, which is even, so that's good. Divisibility by 3: sum of digits is 9 + 6 = 15, which is divisible by 3. So yes, 15 ÷ 3 = 5. So that works. So N = 6. Is that the answer?

Wait, but let me check if there are other possibilities. Let me make sure I didn't make a mistake in the sum of digits.

Original number: 21420N. Let me write all digits:

First digit: 2, second: 1, third: 4, fourth: 2, fifth: 0, sixth: N. So digits are 2,1,4,2,0,N. Sum is 2 +1 +4 +2 +0 +N = 9 + N. That seems correct.

So if N is 6, sum is 15, which is divisible by 3. If N is 0, sum is 9, which is also divisible by 3. But since N is nonzero, only 6. Therefore, answer is 6.

But let me think again. Wait, the problem says "the units digit of the number 21420N". Wait, maybe I misinterpreted the number. Maybe the number is 21420N, where N is the units digit. But is that the same as 21420 followed by N? Or is it something else?

Wait, for example, if N is 6, then the number is 214206. But maybe the number is 21420 multiplied by N? But that would be written as 21420×N, but the problem says "the number 21420N". Usually, when a number is written with a letter like that, it's supposed to be a concatenated number, not multiplication. For example, 21420N is like a number where N is the last digit. So I think my original interpretation is correct. Therefore, the answer is 6.

But let me check if there's another way. Suppose someone might think that 21420N is a multiplication, but then N is the units digit. Wait, but if it is 21420 multiplied by N, then the units digit of the product would be something. But the problem says "the units digit of the number 21420N". So if 21420N is a number, then it's more likely the concatenated number. However, in that case, maybe the problem is written as 21420N, where N is a digit, making it a 6-digit number. Therefore, I think my original reasoning is correct.

But just to be thorough, let me check both interpretations.

First interpretation: 21420N is a 6-digit number with digits 2,1,4,2,0,N. Then N must be 6 as above.

Second interpretation: 21420 multiplied by N, and N is the units digit of that product. But then the problem says "the units digit of the number 21420N", which is a bit confusing. If it's 21420×N, then the units digit of that product is something, but the problem says "N is the units digit of the number 21420N". Therefore, this would mean that N is equal to the units digit of 21420×N. So in that case, we have to find N (nonzero digit) such that when you compute 21420×N, the units digit of that product is N.

But the problem says "N is the units digit of the number 21420N". So maybe it's a way of saying that the number is 21420N, where N is the units digit. Therefore, the number is 21420 followed by digit N, making it a 6-digit number.

But maybe the problem is in another language, and the translation is a bit off. However, given the standard way these problems are presented, it's more likely that it is a concatenated number. For example, in problems like "What is the units digit of 21420N", N is a digit. So here, it's saying that N is the units digit of the number 21420N, which would mean that the number is 21420N with N as the last digit. Therefore, my original interpretation is correct. So N is a digit, making the number 21420N, and the question is to find which non-zero N makes this number divisible by 6. As we saw, N must be 6.

But let me check with the second interpretation just to make sure.

If the second interpretation is correct: Let me suppose that the number is 21420 × N, and N is the units digit of that number. So we need to find N (nonzero digit) such that when you compute 21420 × N, the units digit is N. Then, the question is to find N such that:

units digit of (21420 × N) = N.

So, let me compute 21420 × N mod 10 = N.

Therefore, 21420 × N ≡ N mod 10.

Which implies that (21420 × N - N) ≡ 0 mod 10.

Which is N × (21420 - 1) ≡ 0 mod 10.

Therefore, N × 21419 ≡ 0 mod 10.

So, 21419 mod 10 is 9. So this is equivalent to N × 9 ≡ 0 mod 10.

Therefore, 9N ≡ 0 mod 10. Which implies that 9N is divisible by 10. Therefore, since 9 and 10 are coprime, N must be divisible by 10. But N is a digit from 1 to 9 (nonzero), so there is no solution. Therefore, this interpretation leads to no solution, which contradicts the problem's statement that there is a nonzero N. Therefore, this suggests that the second interpretation is wrong. Therefore, the first interpretation must be correct. Therefore, N is the units digit of the number 21420N, meaning it's a concatenated number.

Therefore, going back to the original problem, N is the units digit of the number 21420N, which is a 6-digit number, and we need to find the nonzero N such that the entire number is divisible by 6. Which, as we found earlier, N must be 6. Therefore, the answer should be 6.

But to make sure, let me verify with an example. Let me take N=6. Then the number is 214206. Divided by 6, what do we get?

First, check divisibility by 2: last digit is 6, which is even. Divisible by 2.

Divisibility by 3: sum of digits is 2 + 1 + 4 + 2 + 0 + 6 = 15. 15 is divisible by 3. Therefore, 214206 is divisible by 6.

Alternatively, divide it by 6. Let me perform the division.

Divide 214206 by 6:

6 × 35701 = 214206. Let me check:

6 × 35000 = 210000

6 × 700 = 4200 → 210000 + 4200 = 214200

Then 6 × 1 = 6 → 214200 + 6 = 214206. Yes, so 35701 × 6 = 214206. Therefore, correct. So N=6 works.

Is there another possible value? Let me check. Earlier reasoning said that N has to be even (for divisibility by 2) and N has to be 0 or 6 (for divisibility by 3). Since N is non-zero, only 6. But let me check for example N=3. Wait, N=3 is not even, so the number would end with 3, which is odd, hence not divisible by 2. Therefore, not divisible by 6. Similarly N=9, same problem. N=0 is excluded. So only N=6 works. Therefore, answer is 6.

Alternatively, let me check the sum of digits again. The digits are 2, 1, 4, 2, 0, N. So sum is 2+1+4+2+0+N=9 + N. For divisibility by 3, 9 + N must be divisible by 3. Since 9 is already divisible by 3, N must be a multiple of 3. Therefore N ∈ {0,3,6,9}. But N must be even (for divisibility by 2). So among those, only 0 and 6 are even. Since N ≠ 0, only 6. Therefore, N=6. So yes, that's solid.

Therefore, the answer is 6. Therefore, I think that is correct.

**Final Answer**
The nonzero value of $ N $ is \boxed{6}.

To solve the problem, we're given a number in the form $ 21420N $, where $ N $ is a **single digit** (0–9), and our goal is to find the **nonzero value of $ N $** such that the number is **divisible by 6**.

---

### Step 1: Understand the Requirements

A number is divisible by **6** if it is divisible by **both 2 and 3**.

---

### Step 2: Use the Divisibility Rules

- **Divisibility by 2**: A number is divisible by 2 if its **last digit is even**.
  - Therefore, $ N $ must be **even**, and the possible values for $ N $ are: **0, 2, 4, 6, 8**

- **Divisibility by 3**: A number is divisible by 3 if the **sum of its digits** is divisible by 3.
  - The digits of the number $ 21420N $ are: **2, 1, 4, 2, 0, N**
  - Sum of these digits: $ 2 + 1 + 4 + 2 + 0 + N = 9 + N $
  - So, to satisfy divisibility by 3, $ 9 + N $ must be divisible by 3.
    - Since 9 is already divisible by 3, **N must be divisible by 3** as well.
    - Possible values for $ N $: **0, 3, 6, 9**

---

### Step 3: Find the Common Value

We now look for values of $ N $ that are **even** (from Step 2) **and** **divisible by 3** (from Step 2).
- From the even digits: **0, 2, 4, 6, 8**
- From the multiples of 3: **0, 3, 6, 9**
- The **only common value** is: **6**

Note: We are told $ N $ must be **nonzero**, so we exclude 0.

---

### Step 4: Verify the Answer

Let’s check if $ N = 6 $ satisfies both conditions:

- **Divisibility by 2**: The number becomes $ 214206 $, whose last digit is 6 (**even**) — ✅
- **Divisibility by 3**: Sum of digits is $ 2 + 1 + 4 + 2 + 0 + 6 = 15 $, which is divisible by 3 — ✅
So, the full number **214206** is divisible by 6.

---

### Final Answer

$$
\boxed{6}
$$

[Assistant]
Explanation:
Okay, let me try to figure out this problem. The question is asking for the nonzero value of N that makes the number 21420N divisible by 6. And N is the units digit of the number 21420N. Hmm, wait, actually, let me parse that again. Wait, the number is 21420N, and N is the units digit. So does that mean the number is 21420N where N is a single digit? Wait, maybe the number is 21420 followed by the digit N? For example, if N is 3, then the number is 214203? So the number is 21420N, where N is a digit from 0 to 9, and we need to find the nonzero N such that the entire number is divisible by 6.

Alright, so first, let me recall that a number is divisible by 6 if and only if it is divisible by both 2 and 3. So, I need to find N (nonzero) such that 21420N is divisible by 2 and 3.

Let me start with divisibility by 2. A number is divisible by 2 if its units digit is even. Since N is the units digit, that means N has to be even. So possible values for N are 2, 4, 6, or 8. But we also need to check divisibility by 3.

Divisibility by 3 requires that the sum of the digits is a multiple of 3. So let me compute the sum of the digits of 21420N. Let me write down the digits:

The number is 2, 1, 4, 2, 0, N. Wait, hold on, let me confirm. The original number is 21420N. If it's written as 21420N, then it's 21420 followed by the digit N. So that would be a 6-digit number? Wait, 21420 is a 5-digit number. So adding a digit N at the end makes it a 6-digit number. So the digits are 2, 1, 4, 2, 0, N. Therefore, the digits are 2, 1, 4, 2, 0, and N.

So the sum of the digits is 2 + 1 + 4 + 2 + 0 + N. Let me compute that. 2 + 1 is 3, plus 4 is 7, plus 2 is 9, plus 0 is still 9, so total sum is 9 + N. Therefore, the sum is 9 + N. For the number to be divisible by 3, 9 + N must be divisible by 3.

Since 9 is already divisible by 3 (because 9 ÷ 3 = 3), adding N must result in another multiple of 3. Therefore, N must be such that N is divisible by 3. Because 9 is a multiple of 3, adding N must also result in a multiple of 3. Therefore, N must be a multiple of 3. However, earlier we found that N must be even (for divisibility by 2). So N has to be both even and a multiple of 3.

So the possible values for N are digits from 0-9 that are both even and multiples of 3. Let me list the digits:

Multiples of 3 between 0-9: 0, 3, 6, 9.

But N has to be even, so from those, the even ones are 0 and 6. However, the problem says "nonzero value of N", so N can't be 0. Therefore, the only possible value is 6. Therefore, N = 6.

Wait, but let me confirm. Let me check if 214206 is divisible by 6. Let me check divisibility by 2: the last digit is 6, which is even, so that's good. Divisibility by 3: sum of digits is 9 + 6 = 15, which is divisible by 3. So yes, 15 ÷ 3 = 5. So that works. So N = 6. Is that the answer?

Wait, but let me check if there are other possibilities. Let me make sure I didn't make a mistake in the sum of digits.

Original number: 21420N. Let me write all digits:

First digit: 2, second: 1, third: 4, fourth: 2, fifth: 0, sixth: N. So digits are 2,1,4,2,0,N. Sum is 2 +1 +4 +2 +0 +N = 9 + N. That seems correct.

So if N is 6, sum is 15, which is divisible by 3. If N is 0, sum is 9, which is also divisible by 3. But since N is nonzero, only 6. Therefore, answer is 6.

But let me think again. Wait, the problem says "the units digit of the number 21420N". Wait, maybe I misinterpreted the number. Maybe the number is 21420N, where N is the units digit. But is that the same as 21420 followed by N? Or is it something else?

Wait, for example, if N is 6, then the number is 214206. But maybe the number is 21420 multiplied by N? But that would be written as 21420×N, but the problem says "the number 21420N". Usually, when a number is written with a letter like that, it's supposed to be a concatenated number, not multiplication. For example, 21420N is like a number where N is the last digit. So I think my original interpretation is correct. Therefore, the answer is 6.

But let me check if there's another way. Suppose someone might think that 21420N is a multiplication, but then N is the units digit. Wait, but if it is 21420 multiplied by N, then the units digit of the product would be something. But the problem says "the units digit of the number 21420N". So if 21420N is a number, then it's more likely the concatenated number. However, in that case, maybe the problem is written as 21420N, where N is a digit, making it a 6-digit number. Therefore, I think my original reasoning is correct.

But just to be thorough, let me check both interpretations.

First interpretation: 21420N is a 6-digit number with digits 2,1,4,2,0,N. Then N must be 6 as above.

Second interpretation: 21420 multiplied by N, and N is the units digit of that product. But then the problem says "the units digit of the number 21420N", which is a bit confusing. If it's 21420×N, then the units digit of that product is something, but the problem says "N is the units digit of the number 21420N". Therefore, this would mean that N is equal to the units digit of 21420×N. So in that case, we have to find N (nonzero digit) such that when you compute 21420×N, the units digit of that product is N.

But the problem says "N is the units digit of the number 21420N". So maybe it's a way of saying that the number is 21420N, where N is the units digit. Therefore, the number is 21420 followed by digit N, making it a 6-digit number.

But maybe the problem is in another language, and the translation is a bit off. However, given the standard way these problems are presented, it's more likely that it is a concatenated number. For example, in problems like "What is the units digit of 21420N", N is a digit. So here, it's saying that N is the units digit of the number 21420N, which would mean that the number is 21420N with N as the last digit. Therefore, my original interpretation is correct. So N is a digit, making the number 21420N, and the question is to find which non-zero N makes this number divisible by 6. As we saw, N must be 6.

But let me check with the second interpretation just to make sure.

If the second interpretation is correct: Let me suppose that the number is 21420 × N, and N is the units digit of that number. So we need to find N (nonzero digit) such that when you compute 21420 × N, the units digit is N. Then, the question is to find N such that:

units digit of (21420 × N) = N.

So, let me compute 21420 × N mod 10 = N.

Therefore, 21420 × N ≡ N mod 10.

Which implies that (21420 × N - N) ≡ 0 mod 10.

Which is N × (21420 - 1) ≡ 0 mod 10.

Therefore, N × 21419 ≡ 0 mod 10.

So, 21419 mod 10 is 9. So this is equivalent to N × 9 ≡ 0 mod 10.

Therefore, 9N ≡ 0 mod 10. Which implies that 9N is divisible by 10. Therefore, since 9 and 10 are coprime, N must be divisible by 10. But N is a digit from 1 to 9 (nonzero), so there is no solution. Therefore, this interpretation leads to no solution, which contradicts the problem's statement that there is a nonzero N. Therefore, this suggests that the second interpretation is wrong. Therefore, the first interpretation must be correct. Therefore, N is the units digit of the number 21420N, meaning it's a concatenated number.

Therefore, going back to the original problem, N is the units digit of the number 21420N, which is a 6-digit number, and we need to find the nonzero N such that the entire number is divisible by 6. Which, as we found earlier, N must be 6. Therefore, the answer should be 6.

But to make sure, let me verify with an example. Let me take N=6. Then the number is 214206. Divided by 6, what do we get?

First, check divisibility by 2: last digit is 6, which is even. Divisible by 2.

Divisibility by 3: sum of digits is 2 + 1 + 4 + 2 + 0 + 6 = 15. 15 is divisible by 3. Therefore, 214206 is divisible by 6.

Alternatively, divide it by 6. Let me perform the division.

Divide 214206 by 6:

6 × 35701 = 214206. Let me check:

6 × 35000 = 210000

6 × 700 = 4200 → 210000 + 4200 = 214200

Then 6 × 1 = 6 → 214200 + 6 = 214206. Yes, so 35701 × 6 = 214206. Therefore, correct. So N=6 works.

Is there another possible value? Let me check. Earlier reasoning said that N has to be even (for divisibility by 2) and N has to be 0 or 6 (for divisibility by 3). Since N is non-zero, only 6. But let me check for example N=3. Wait, N=3 is not even, so the number would end with 3, which is odd, hence not divisible by 2. Therefore, not divisible by 6. Similarly N=9, same problem. N=0 is excluded. So only N=6 works. Therefore, answer is 6.

Alternatively, let me check the sum of digits again. The digits are 2, 1, 4, 2, 0, N. So sum is 2+1+4+2+0+N=9 + N. For divisibility by 3, 9 + N must be divisible by 3. Since 9 is already divisible by 3, N must be a multiple of 3. Therefore N ∈ {0,3,6,9}. But N must be even (for divisibility by 2). So among those, only 0 and 6 are even. Since N ≠ 0, only 6. Therefore, N=6. So yes, that's solid.

Therefore, the answer is 6. Therefore, I think that is correct.

**Final Answer**
The nonzero value of $ N $ is \boxed{6}.

To solve the problem, we're given a number in the form $ 21420N $, where $ N $ is a **single digit** (0–9), and our goal is to find the **nonzero value of $ N $** such that the number is **divisible by 6**.

---

### Step 1: Understand the Requirements

A number is divisible by **6** if it is divisible by **both 2 and 3**.

---

### Step 2: Use the Divisibility Rules

- **Divisibility by 2**: A number is divisible by 2 if its **last digit is even**.
  - Therefore, $ N $ must be **even**, and the possible values for $ N $ are: **0, 2, 4, 6, 8**

- **Divisibility by 3**: A number is divisible by 3 if the **sum of its digits** is divisible by 3.
  - The digits of the number $ 21420N $ are: **2, 1, 4, 2, 0, N**
  - Sum of these digits: $ 2 + 1 + 4 + 2 + 0 + N = 9 + N $
  - So, to satisfy divisibility by 3, $ 9 + N $ must be divisible by 3.
    - Since 9 is already divisible by 3, **N must be divisible by 3** as well.
    - Possible values for $ N $: **0, 3, 6, 9**

---

### Step 3: Find the Common Value

We now look for values of $ N $ that are **even** (from Step 2) **and** **divisible by 3** (from Step 2).
- From the even digits: **0, 2, 4, 6, 8**
- From the multiples of 3: **0, 3, 6, 9**
- The **only common value** is: **6**

Note: We are told $ N $ must be **nonzero**, so we exclude 0.

---

### Step 4: Verify the Answer

Let’s check if $ N = 6 $ satisfies both conditions:

- **Divisibility by 2**: The number becomes $ 214206 $, whose last digit is 6 (**even**) — ✅
- **Divisibility by 3**: Sum of digits is $ 2 + 1 + 4 + 2 + 0 + 6 = 15 $, which is divisible by 3 — ✅
So, the full number **214206** is divisible by 6.

---

### Final Answer

$$
\boxed{6}
$$

Solution: The number $21420N$ is divisible by $6$ if and only if it is even and the sum of its digits is divisible by $3$. So $N$ must be even, and $2 + 1 + 4 + 2 + 0 + N = 9 + N$ must be divisible by $3$.  Since $9$ is divisible by $3$, we see that $N$ must also be divisible by $3$. The only digit that works is $N = \boxed{6}$.


Explanation:
Okay, so I need to simplify the expression (10r³)(4r⁶) divided by 8r⁴. Let me think about how to approach this. First, I remember that when you have algebraic expressions like this, you can handle the coefficients (the numbers) and the variables separately. So maybe I should start by multiplying the coefficients in the numerator and then dealing with the r terms.

Alright, the numerator is (10r³)(4r⁶). Let me compute the coefficients first. 10 multiplied by 4... that's 40. So the coefficient part becomes 40. Now for the variables, when you multiply terms with the same base, you add the exponents. So r³ times r⁶ would be r raised to the power of 3 + 6, which is 9. So the numerator simplifies to 40r⁹.

Now the entire expression is that numerator divided by 8r⁴. So, 40r⁹ divided by 8r⁴. Let me separate the coefficients and the variables again. 40 divided by 8... that's 5. Then for the variables, it's r⁹ divided by r⁴. Again, when dividing like bases, you subtract the exponents. So 9 - 4 is 5. Therefore, r⁹/r⁴ is r⁵.

Putting it all together, the simplified expression should be 5r⁵. Let me check my steps again to make sure I didn't make a mistake. First, multiplying 10 and 4 gives 40. Then adding exponents 3 and 6 gives 9, so numerator is 40r⁹. Then dividing by 8 gives 5, and subtracting exponents 9 - 4 is 5. Yep, that seems right.

Wait, maybe I should verify the division part again. 40 divided by 8 is indeed 5. And for the exponents, r⁹ over r⁴ is r^(9-4) = r^5. That all checks out. I don't see any errors here. So the simplified form is 5r⁵.

Alternatively, maybe I can approach it step by step without combining the numerator first. Let me try that. Let me write the original expression:

(10r³ * 4r⁶) / (8r⁴)

I can rearrange the terms to group coefficients and variables:

(10 * 4 / 8) * (r³ * r⁶ / r⁴)

Calculating the coefficients: 10*4=40, then 40/8=5. For the variables: r³*r⁶=r^(3+6)=r^9, then divided by r^4 is r^(9-4)=r^5. So again, 5r^5. Yep, same result.

Another way to think about it: Maybe simplifying before multiplying? Let's see. The original numerator is 10*4=40, denominator is 8. So 40/8 is 5. Then the variables: r³*r⁶/r⁴. Which is same as r^(3+6-4)=r^5.

Alternatively, if I simplify the coefficients first. Let me see. 10 divided by 8 is 5/4, then multiplied by 4? Wait, no. Maybe that's complicating. Let me think.

Alternatively, could I cancel factors before multiplying? For example, 10 and 8 have a common factor. 10 is 2*5, 8 is 2³. So maybe simplifying fractions. Let me try that.

Original expression: (10 * 4 / 8) * r³ * r⁶ / r⁴. So coefficients: 10 * 4 / 8. Let me write that as (10/8)*4. Then simplifying 10/8 is 5/4, so 5/4 * 4 = 5. Yep, same as before.

Alternatively, breaking down 10*4 first: 10*4=40. Then 40/8=5. So that's straightforward.

So regardless of the method, I end up with 5r^5. Therefore, I think that's the correct simplified form.

Wait, but let me check if there's any chance of miscalculating exponents. For example, if I thought that multiplying exponents would be 3*6 instead of adding? But no, when you multiply same bases, you add exponents. When you divide, subtract. So for numerator, multiplying r³*r⁶, add exponents: 3+6=9. Then divide by r⁴, subtract 4: 9-4=5. Correct.

Another point to check: If the original problem was written with parentheses, maybe there was a different interpretation? Let me check the original problem again: it's (10r³)(4r⁶) over 8r⁴. So that's (10r³ * 4r⁶) divided by 8r⁴. So no ambiguity here. The numerator is two terms multiplied together, and denominator is 8r⁴.

Alternatively, if someone misreads it as 10r³ multiplied by (4r⁶ / 8r⁴), but that would be different. But the original problem is written as (10r³)(4r⁶) divided by 8r⁴. So the parentheses indicate that the numerator is the product of 10r³ and 4r⁶, so the order of operations is correct.

Another check: plugging in a value for r. Let's say r=2. Let's compute original expression and simplified expression.

Original expression: (10*(2)^3)*(4*(2)^6) / (8*(2)^4)

Calculate numerator: 10*8=80, 4*64=256. Multiply them: 80*256. Let me calculate that. 80*256: 80*200=16,000; 80*56=4,480. Total is 16,000 + 4,480 = 20,480.

Denominator: 8*(2)^4 = 8*16=128.

So the original expression is 20,480 / 128. Let me compute that. 128*160=20,480. So 20,480 divided by 128 is 160.

Now the simplified expression is 5r^5. Let's compute that with r=2: 5*(2)^5=5*32=160. Same result. So that checks out.

Another example with r=1. Original expression: (10*1)(4*1)/ (8*1)= 10*4 /8=40/8=5. Simplified expression: 5*(1)^5=5*1=5. Correct.

Another test with r=3. Original expression: (10*27)*(4*729)/(8*81). Let me compute step by step.

Numerator: 10*27=270; 4*729=2916. Multiply these: 270*2916. That's a big number. Maybe compute denominator first: 8*81=648. Then compute 270*2916 / 648. Let me simplify fractions first.

Note that 270 / 648: both divisible by 54? 270 ÷ 54=5, 648 ÷ 54=12. So 5/12. Then multiply by 2916. So 5/12 * 2916. Let me compute 2916 divided by 12. 12*243=2916. So 5*243=1215.

Simplified expression: 5*(3)^5. 3^5 is 243, times 5 is 1215. Which matches. So that works too.

Therefore, the simplification is correct.

Hence, after all these steps and checks, I can confidently say the simplified form is 5r^5.

**Final Answer**
The simplified expression is \boxed{5r^5}.

To simplify the expression
$$
\frac{(10r^3)(4r^6)}{8r^4},
$$
we break it down into manageable parts by considering both the **numerical coefficients** and the **variable parts** separately.

---

### Step 1: Combine the Numerator

Start by multiplying the coefficients and the powers of $ r $ in the numerator:

- **Coefficients:**
  $$
  10 \cdot 4 = 40
  $$

- **Variables:**
  When multiplying powers of the same base, add the exponents:
  $$
  r^3 \cdot r^6 = r^{3+6} = r^9
  $$

So the full numerator becomes:
$$
40r^9
$$

---

### Step 2: Divide by the Denominator

Now divide the result by the denominator $8r^4$:

- **Coefficients:**
  $$
  \frac{40}{8} = 5
  $$

- **Variables:**
  When dividing powers of the same base, subtract the exponents:
  $$
  \frac{r^9}{r^4} = r^{9-4} = r^5
  $$

Putting it all together:
$$
\frac{40r^9}{8r^4} = 5r^5
$$

---

### Final Result

After simplifying both the numerical part and the variable part, we find that the expression simplifies to:

$$
\boxed{5r^5}
$$

[Assistant]
Explanation:
Okay, so I need to simplify the expression (10r³)(4r⁶) divided by 8r⁴. Let me think about how to approach this. First, I remember that when you have algebraic expressions like this, you can handle the coefficients (the numbers) and the variables separately. So maybe I should start by multiplying the coefficients in the numerator and then dealing with the r terms.

Alright, the numerator is (10r³)(4r⁶). Let me compute the coefficients first. 10 multiplied by 4... that's 40. So the coefficient part becomes 40. Now for the variables, when you multiply terms with the same base, you add the exponents. So r³ times r⁶ would be r raised to the power of 3 + 6, which is 9. So the numerator simplifies to 40r⁹.

Now the entire expression is that numerator divided by 8r⁴. So, 40r⁹ divided by 8r⁴. Let me separate the coefficients and the variables again. 40 divided by 8... that's 5. Then for the variables, it's r⁹ divided by r⁴. Again, when dividing like bases, you subtract the exponents. So 9 - 4 is 5. Therefore, r⁹/r⁴ is r⁵.

Putting it all together, the simplified expression should be 5r⁵. Let me check my steps again to make sure I didn't make a mistake. First, multiplying 10 and 4 gives 40. Then adding exponents 3 and 6 gives 9, so numerator is 40r⁹. Then dividing by 8 gives 5, and subtracting exponents 9 - 4 is 5. Yep, that seems right.

Wait, maybe I should verify the division part again. 40 divided by 8 is indeed 5. And for the exponents, r⁹ over r⁴ is r^(9-4) = r^5. That all checks out. I don't see any errors here. So the simplified form is 5r⁵.

Alternatively, maybe I can approach it step by step without combining the numerator first. Let me try that. Let me write the original expression:

(10r³ * 4r⁶) / (8r⁴)

I can rearrange the terms to group coefficients and variables:

(10 * 4 / 8) * (r³ * r⁶ / r⁴)

Calculating the coefficients: 10*4=40, then 40/8=5. For the variables: r³*r⁶=r^(3+6)=r^9, then divided by r^4 is r^(9-4)=r^5. So again, 5r^5. Yep, same result.

Another way to think about it: Maybe simplifying before multiplying? Let's see. The original numerator is 10*4=40, denominator is 8. So 40/8 is 5. Then the variables: r³*r⁶/r⁴. Which is same as r^(3+6-4)=r^5.

Alternatively, if I simplify the coefficients first. Let me see. 10 divided by 8 is 5/4, then multiplied by 4? Wait, no. Maybe that's complicating. Let me think.

Alternatively, could I cancel factors before multiplying? For example, 10 and 8 have a common factor. 10 is 2*5, 8 is 2³. So maybe simplifying fractions. Let me try that.

Original expression: (10 * 4 / 8) * r³ * r⁶ / r⁴. So coefficients: 10 * 4 / 8. Let me write that as (10/8)*4. Then simplifying 10/8 is 5/4, so 5/4 * 4 = 5. Yep, same as before.

Alternatively, breaking down 10*4 first: 10*4=40. Then 40/8=5. So that's straightforward.

So regardless of the method, I end up with 5r^5. Therefore, I think that's the correct simplified form.

Wait, but let me check if there's any chance of miscalculating exponents. For example, if I thought that multiplying exponents would be 3*6 instead of adding? But no, when you multiply same bases, you add exponents. When you divide, subtract. So for numerator, multiplying r³*r⁶, add exponents: 3+6=9. Then divide by r⁴, subtract 4: 9-4=5. Correct.

Another point to check: If the original problem was written with parentheses, maybe there was a different interpretation? Let me check the original problem again: it's (10r³)(4r⁶) over 8r⁴. So that's (10r³ * 4r⁶) divided by 8r⁴. So no ambiguity here. The numerator is two terms multiplied together, and denominator is 8r⁴.

Alternatively, if someone misreads it as 10r³ multiplied by (4r⁶ / 8r⁴), but that would be different. But the original problem is written as (10r³)(4r⁶) divided by 8r⁴. So the parentheses indicate that the numerator is the product of 10r³ and 4r⁶, so the order of operations is correct.

Another check: plugging in a value for r. Let's say r=2. Let's compute original expression and simplified expression.

Original expression: (10*(2)^3)*(4*(2)^6) / (8*(2)^4)

Calculate numerator: 10*8=80, 4*64=256. Multiply them: 80*256. Let me calculate that. 80*256: 80*200=16,000; 80*56=4,480. Total is 16,000 + 4,480 = 20,480.

Denominator: 8*(2)^4 = 8*16=128.

So the original expression is 20,480 / 128. Let me compute that. 128*160=20,480. So 20,480 divided by 128 is 160.

Now the simplified expression is 5r^5. Let's compute that with r=2: 5*(2)^5=5*32=160. Same result. So that checks out.

Another example with r=1. Original expression: (10*1)(4*1)/ (8*1)= 10*4 /8=40/8=5. Simplified expression: 5*(1)^5=5*1=5. Correct.

Another test with r=3. Original expression: (10*27)*(4*729)/(8*81). Let me compute step by step.

Numerator: 10*27=270; 4*729=2916. Multiply these: 270*2916. That's a big number. Maybe compute denominator first: 8*81=648. Then compute 270*2916 / 648. Let me simplify fractions first.

Note that 270 / 648: both divisible by 54? 270 ÷ 54=5, 648 ÷ 54=12. So 5/12. Then multiply by 2916. So 5/12 * 2916. Let me compute 2916 divided by 12. 12*243=2916. So 5*243=1215.

Simplified expression: 5*(3)^5. 3^5 is 243, times 5 is 1215. Which matches. So that works too.

Therefore, the simplification is correct.

Hence, after all these steps and checks, I can confidently say the simplified form is 5r^5.

**Final Answer**
The simplified expression is \boxed{5r^5}.

To simplify the expression
$$
\frac{(10r^3)(4r^6)}{8r^4},
$$
we break it down into manageable parts by considering both the **numerical coefficients** and the **variable parts** separately.

---

### Step 1: Combine the Numerator

Start by multiplying the coefficients and the powers of $ r $ in the numerator:

- **Coefficients:**
  $$
  10 \cdot 4 = 40
  $$

- **Variables:**
  When multiplying powers of the same base, add the exponents:
  $$
  r^3 \cdot r^6 = r^{3+6} = r^9
  $$

So the full numerator becomes:
$$
40r^9
$$

---

### Step 2: Divide by the Denominator

Now divide the result by the denominator $8r^4$:

- **Coefficients:**
  $$
  \frac{40}{8} = 5
  $$

- **Variables:**
  When dividing powers of the same base, subtract the exponents:
  $$
  \frac{r^9}{r^4} = r^{9-4} = r^5
  $$

Putting it all together:
$$
\frac{40r^9}{8r^4} = 5r^5
$$

---

### Final Result

After simplifying both the numerical part and the variable part, we find that the expression simplifies to:

$$
\boxed{5r^5}
$$

Solution: We have  \[\frac{(10r^3)(4r^6)}{8r^4}= \frac{40r^{3+6}}{8r^4} = \frac{40}{8}r^{3+6-4} = \boxed{5r^5}.\]


Explanation:
Okay, let me try to solve this problem. So, we have three pencils and a jumbo eraser costing $1.24, and five pencils and a jumbo eraser costing $1.82. We need to find the cost of a pencil in cents. Alright, let me think about how to approach this.

First, I know this is a system of equations problem. There are two items: pencils and jumbo erasers. Each has a cost, and we have two different combinations of them with their total prices. So, if I can set up equations for each scenario, I can solve for the individual prices.

Let me denote the cost of one pencil as "p" dollars and the cost of a jumbo eraser as "e" dollars. Since the final answer needs to be in cents, maybe I should work in cents from the beginning to avoid dealing with decimals? Or maybe convert the given prices to cents first? Let me check.

The problem says the prices don't include tax, so that's fine. The given amounts are in dollars, but they want the answer in cents. Maybe converting them to cents will make the calculations easier. Let me try that.

First, let me convert $1.24 to cents. Since 1 dollar is 100 cents, $1.24 is 124 cents. Similarly, $1.82 is 182 cents. So, the equations would be:

3 pencils + 1 eraser = 124 cents.

5 pencils + 1 eraser = 182 cents.

So, if I let p be the cost of a pencil in cents, and e be the cost of an eraser in cents, then:

Equation 1: 3p + e = 124

Equation 2: 5p + e = 182

Now, I have two equations with two variables. To solve this system, I can use the elimination method. If I subtract Equation 1 from Equation 2, the e terms will cancel out.

Let me write that down:

Equation 2 minus Equation 1:

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

Simplify the left side:

5p + e - 3p - e = 2p

Right side: 182 - 124 = 58

So, 2p = 58

Therefore, p = 58 / 2 = 29

So, the cost of a pencil is 29 cents?

Wait, let me check if that makes sense. Let me verify by plugging back into the original equations.

If p = 29 cents, then:

First equation: 3p + e = 124 cents

So, 3*29 = 87 cents. Therefore, e = 124 - 87 = 37 cents.

So, the eraser would be 37 cents. Let me check the second equation:

5p + e = 5*29 + 37 = 145 + 37 = 182 cents. Which matches the given amount. Perfect.

So, that seems correct. Therefore, the cost of a pencil is 29 cents.

But let me make sure I didn't make any calculation errors. Let me redo the subtraction step:

Equation 2: 5p + e = 182

Equation 1: 3p + e = 124

Subtract Equation 1 from Equation 2:

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

Which is 2p = 58, so p = 29. Yep, that's right. Then e = 124 - 3*29.

Calculating 3*29: 30*3 = 90, minus 1*3 = 3, so 90 - 3 = 87. Then 124 - 87 is 37. Which is correct. Then 5*29 is 145, plus 37 is 182. Perfect.

So, the answer is 29 cents. Therefore, the cost of a pencil is 29 cents. Wait, but the problem says "In cents, what is the cost of a pencil?" So, since we already converted everything to cents, the answer is 29 cents. Therefore, I think that's the answer.

Alternatively, if I had kept the prices in dollars, would I get the same result?

Let me try that approach as a check. Let me let p be the price of a pencil in dollars, and e be the price of an eraser in 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

Which is 2p = 0.58

Therefore, p = 0.58 / 2 = 0.29 dollars. Which is 29 cents. Same result. Perfect. So that's consistent.

Therefore, the answer is 29 cents. So, I think that's correct.

But just to make sure there isn't a trick or something I missed. The problem says "no prices include tax," so that's just telling us that we don't have to worry about tax in the calculations. So, the given amounts are the actual total prices. So, the equations are straightforward.

Is there any chance that the eraser is priced differently? For example, maybe the eraser is a jumbo eraser, but maybe there's something else. But the problem states "three pencils and a jumbo eraser" and "five pencils and a jumbo eraser," so the eraser is the same in both cases. Therefore, the equations are set up correctly. So, the difference between the two purchases is just the number of pencils, which allows us to solve for the price of a pencil.

Therefore, I think 29 cents is correct. So, the answer is 29 cents. So, in boxed form, it would be \boxed{29}.

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

To determine the cost of a pencil, we can model the situation as a system of linear equations based on the given cost information.

---

### Step 1: Define Variables

Let:
- $ p $ be the cost of **one pencil** in cents
- $ e $ be the cost of **one jumbo eraser** in cents

---

### Step 2: Translate the Problem into Equations

The first statement gives us:
- "Three pencils and a jumbo eraser cost **$1.24**" → $ 3p + e = 124 $
- "Five pencils and a jumbo eraser cost **$1.82**" → $ 5p + e = 182 $

So, the two equations are:

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

---

### Step 3: Eliminate the Eraser Variable

Subtract equation (1) from equation (2):

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

Simplify:

$$
5p - 3p + e - e = 58 \Rightarrow 2p = 58
$$

Solve for $ p $:

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

---

### Step 4: Verify the Solution

Now, substitute $ p = 29 $ into equation (1) to find $ e $:

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

Check this with equation (2):

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

This matches the given value, so the solution is consistent.

---

### Final Answer

Thus, the cost of a pencil is:

$$
\boxed{29}
$$

[Assistant]
Explanation:
Okay, let me try to solve this problem. So, we have three pencils and a jumbo eraser costing $1.24, and five pencils and a jumbo eraser costing $1.82. We need to find the cost of a pencil in cents. Alright, let me think about how to approach this.

First, I know this is a system of equations problem. There are two items: pencils and jumbo erasers. Each has a cost, and we have two different combinations of them with their total prices. So, if I can set up equations for each scenario, I can solve for the individual prices.

Let me denote the cost of one pencil as "p" dollars and the cost of a jumbo eraser as "e" dollars. Since the final answer needs to be in cents, maybe I should work in cents from the beginning to avoid dealing with decimals? Or maybe convert the given prices to cents first? Let me check.

The problem says the prices don't include tax, so that's fine. The given amounts are in dollars, but they want the answer in cents. Maybe converting them to cents will make the calculations easier. Let me try that.

First, let me convert $1.24 to cents. Since 1 dollar is 100 cents, $1.24 is 124 cents. Similarly, $1.82 is 182 cents. So, the equations would be:

3 pencils + 1 eraser = 124 cents.

5 pencils + 1 eraser = 182 cents.

So, if I let p be the cost of a pencil in cents, and e be the cost of an eraser in cents, then:

Equation 1: 3p + e = 124

Equation 2: 5p + e = 182

Now, I have two equations with two variables. To solve this system, I can use the elimination method. If I subtract Equation 1 from Equation 2, the e terms will cancel out.

Let me write that down:

Equation 2 minus Equation 1:

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

Simplify the left side:

5p + e - 3p - e = 2p

Right side: 182 - 124 = 58

So, 2p = 58

Therefore, p = 58 / 2 = 29

So, the cost of a pencil is 29 cents?

Wait, let me check if that makes sense. Let me verify by plugging back into the original equations.

If p = 29 cents, then:

First equation: 3p + e = 124 cents

So, 3*29 = 87 cents. Therefore, e = 124 - 87 = 37 cents.

So, the eraser would be 37 cents. Let me check the second equation:

5p + e = 5*29 + 37 = 145 + 37 = 182 cents. Which matches the given amount. Perfect.

So, that seems correct. Therefore, the cost of a pencil is 29 cents.

But let me make sure I didn't make any calculation errors. Let me redo the subtraction step:

Equation 2: 5p + e = 182

Equation 1: 3p + e = 124

Subtract Equation 1 from Equation 2:

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

Which is 2p = 58, so p = 29. Yep, that's right. Then e = 124 - 3*29.

Calculating 3*29: 30*3 = 90, minus 1*3 = 3, so 90 - 3 = 87. Then 124 - 87 is 37. Which is correct. Then 5*29 is 145, plus 37 is 182. Perfect.

So, the answer is 29 cents. Therefore, the cost of a pencil is 29 cents. Wait, but the problem says "In cents, what is the cost of a pencil?" So, since we already converted everything to cents, the answer is 29 cents. Therefore, I think that's the answer.

Alternatively, if I had kept the prices in dollars, would I get the same result?

Let me try that approach as a check. Let me let p be the price of a pencil in dollars, and e be the price of an eraser in 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

Which is 2p = 0.58

Therefore, p = 0.58 / 2 = 0.29 dollars. Which is 29 cents. Same result. Perfect. So that's consistent.

Therefore, the answer is 29 cents. So, I think that's correct.

But just to make sure there isn't a trick or something I missed. The problem says "no prices include tax," so that's just telling us that we don't have to worry about tax in the calculations. So, the given amounts are the actual total prices. So, the equations are straightforward.

Is there any chance that the eraser is priced differently? For example, maybe the eraser is a jumbo eraser, but maybe there's something else. But the problem states "three pencils and a jumbo eraser" and "five pencils and a jumbo eraser," so the eraser is the same in both cases. Therefore, the equations are set up correctly. So, the difference between the two purchases is just the number of pencils, which allows us to solve for the price of a pencil.

Therefore, I think 29 cents is correct. So, the answer is 29 cents. So, in boxed form, it would be \boxed{29}.

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

To determine the cost of a pencil, we can model the situation as a system of linear equations based on the given cost information.

---

### Step 1: Define Variables

Let:
- $ p $ be the cost of **one pencil** in cents
- $ e $ be the cost of **one jumbo eraser** in cents

---

### Step 2: Translate the Problem into Equations

The first statement gives us:
- "Three pencils and a jumbo eraser cost **$1.24**" → $ 3p + e = 124 $
- "Five pencils and a jumbo eraser cost **$1.82**" → $ 5p + e = 182 $

So, the two equations are:

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

---

### Step 3: Eliminate the Eraser Variable

Subtract equation (1) from equation (2):

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

Simplify:

$$
5p - 3p + e - e = 58 \Rightarrow 2p = 58
$$

Solve for $ p $:

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

---

### Step 4: Verify the Solution

Now, substitute $ p = 29 $ into equation (1) to find $ e $:

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

Check this with equation (2):

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

This matches the given value, so the solution is consistent.

---

### Final Answer

Thus, 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... I remember that dividing by 2 is the same as finding half of something. So, 9 divided by 2 should be half of 9. Hmm, half of 9 is 4.5, right? Wait, let me make sure I'm not making a mistake here. Maybe I should do the division step by step.

Alright, let's start with the division. 9 divided by 2. Since 2 goes into 9 four times because 2 times 4 is 8. Then there's a remainder of 1, right? So, 2 goes into 9 four times with a remainder of 1. Then, if I add a decimal point and a zero, making it 10, 2 goes into 10 five times. So that would make it 4.5. Let me check that again.

So, 2 times 4 is 8, subtract 8 from 9, you get 1. Bring down a zero, making it 10. 2 goes into 10 exactly 5 times. So there's no remainder left. Therefore, the decimal is 4.5. Wait, that seems straightforward. Is there another way to verify this?

Maybe converting the fraction to a decimal by multiplying numerator and denominator to make the denominator a power of 10? Let me try that. The denominator is 2. If I multiply numerator and denominator by 5, that would make the denominator 10. So, 9 times 5 is 45, and 2 times 5 is 10. Therefore, 9/2 is equivalent to 45/10, which is 4.5. That matches what I got before.

Alternatively, I can think in terms of decimals. Since 1/2 is 0.5, then 9/2 is 9 times 0.5. Let me calculate that. 9 times 0.5 is the same as 9 divided by 2, which again is 4.5. So all these different methods are giving me the same answer.

I guess another way to think about it is using fractions. If I have 9 divided by 2, that's the same as 8/2 + 1/2, which is 4 + 0.5, so again 4.5.

Is there a chance I could have messed up the decimal places? Let me check the multiplication again. If I have 4.5 times 2, that should give me back 9. Let me do that multiplication. 4 times 2 is 8, and 0.5 times 2 is 1. So 8 + 1 is 9. Perfect, that's correct.

So all the checks I did confirm that 9 divided by 2 is 4.5. I don't think I made any mistakes here. It seems pretty straightforward. Maybe I can also think about it in terms of money? If I have $9 and I split it between two people, each person gets $4.50. That makes sense because 4 dollars times 2 is 8 dollars, and 0.50 times 2 is 1 dollar, so total is 9 dollars. Yep, that works.

Alternatively, if I use a calculator to verify, but since I don't have one here, I have to rely on my mental math. But all the methods I've used so far are consistent. So I think I can be confident that 9/2 as a decimal is 4.5.

Wait, let me just write it out long division style to make sure.

So 2 | 9.0

2 goes into 9 four times. 4 times 2 is 8. Subtract 8 from 9, you get 1. Bring down the 0, making it 10. 2 goes into 10 five times. Multiply 5 by 2 is 10. Subtract 10 from 10, you get 0. So the division ends here. So the result is 4.5. Yep, that's exactly what I did before.

No remainders, so it's a terminating decimal. So 4.5 is the exact decimal representation. No need for repeating decimals or anything.

Therefore, I think I can safely say that 9/2 as a decimal is 4.5.

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

To convert the fraction $\frac{9}{2}$ into a decimal, we can follow a few logical methods. Here's a step-by-step explanation:

---

### Step 1: Understand the fraction
The fraction $\frac{9}{2}$ represents 9 divided by 2.

---

### Step 2: Perform the division
Let’s perform the division manually:

1. **Divide 9 by 2:**
   - 2 fits into 9 **4 times** because $2 \times 4 = 8$.
   - Subtract 8 from 9, we get a **remainder of 1**.

2. **Continue the division:**
   - Add a **decimal point** to the quotient (now we're considering 9.0).
   - Bring down a **zero**, making it **10**.
   - 2 fits into 10 exactly **5 times** because $2 \times 5 = 10$.
   - No remainder remains.

So, putting it all together, we find that:

$$
\frac{9}{2} = 4.5
$$

---

### Step 3: Verification
We can verify the result in different ways:

- **Multiplication Check:** $4.5 \times 2 = 9$
- **Fraction Equivalency:** $\frac{9}{2} = \frac{45}{10} = 4.5$
- **Alternative Method:** Recognizing that $\frac{1}{2} = 0.5$, then $9 \times 0.5 = 4.5$

All these confirm that the result 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... I remember that dividing by 2 is the same as finding half of something. So, 9 divided by 2 should be half of 9. Hmm, half of 9 is 4.5, right? Wait, let me make sure I'm not making a mistake here. Maybe I should do the division step by step.

Alright, let's start with the division. 9 divided by 2. Since 2 goes into 9 four times because 2 times 4 is 8. Then there's a remainder of 1, right? So, 2 goes into 9 four times with a remainder of 1. Then, if I add a decimal point and a zero, making it 10, 2 goes into 10 five times. So that would make it 4.5. Let me check that again.

So, 2 times 4 is 8, subtract 8 from 9, you get 1. Bring down a zero, making it 10. 2 goes into 10 exactly 5 times. So there's no remainder left. Therefore, the decimal is 4.5. Wait, that seems straightforward. Is there another way to verify this?

Maybe converting the fraction to a decimal by multiplying numerator and denominator to make the denominator a power of 10? Let me try that. The denominator is 2. If I multiply numerator and denominator by 5, that would make the denominator 10. So, 9 times 5 is 45, and 2 times 5 is 10. Therefore, 9/2 is equivalent to 45/10, which is 4.5. That matches what I got before.

Alternatively, I can think in terms of decimals. Since 1/2 is 0.5, then 9/2 is 9 times 0.5. Let me calculate that. 9 times 0.5 is the same as 9 divided by 2, which again is 4.5. So all these different methods are giving me the same answer.

I guess another way to think about it is using fractions. If I have 9 divided by 2, that's the same as 8/2 + 1/2, which is 4 + 0.5, so again 4.5.

Is there a chance I could have messed up the decimal places? Let me check the multiplication again. If I have 4.5 times 2, that should give me back 9. Let me do that multiplication. 4 times 2 is 8, and 0.5 times 2 is 1. So 8 + 1 is 9. Perfect, that's correct.

So all the checks I did confirm that 9 divided by 2 is 4.5. I don't think I made any mistakes here. It seems pretty straightforward. Maybe I can also think about it in terms of money? If I have $9 and I split it between two people, each person gets $4.50. That makes sense because 4 dollars times 2 is 8 dollars, and 0.50 times 2 is 1 dollar, so total is 9 dollars. Yep, that works.

Alternatively, if I use a calculator to verify, but since I don't have one here, I have to rely on my mental math. But all the methods I've used so far are consistent. So I think I can be confident that 9/2 as a decimal is 4.5.

Wait, let me just write it out long division style to make sure.

So 2 | 9.0

2 goes into 9 four times. 4 times 2 is 8. Subtract 8 from 9, you get 1. Bring down the 0, making it 10. 2 goes into 10 five times. Multiply 5 by 2 is 10. Subtract 10 from 10, you get 0. So the division ends here. So the result is 4.5. Yep, that's exactly what I did before.

No remainders, so it's a terminating decimal. So 4.5 is the exact decimal representation. No need for repeating decimals or anything.

Therefore, I think I can safely say that 9/2 as a decimal is 4.5.

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

To convert the fraction $\frac{9}{2}$ into a decimal, we can follow a few logical methods. Here's a step-by-step explanation:

---

### Step 1: Understand the fraction
The fraction $\frac{9}{2}$ represents 9 divided by 2.

---

### Step 2: Perform the division
Let’s perform the division manually:

1. **Divide 9 by 2:**
   - 2 fits into 9 **4 times** because $2 \times 4 = 8$.
   - Subtract 8 from 9, we get a **remainder of 1**.

2. **Continue the division:**
   - Add a **decimal point** to the quotient (now we're considering 9.0).
   - Bring down a **zero**, making it **10**.
   - 2 fits into 10 exactly **5 times** because $2 \times 5 = 10$.
   - No remainder remains.

So, putting it all together, we find that:

$$
\frac{9}{2} = 4.5
$$

---

### Step 3: Verification
We can verify the result in different ways:

- **Multiplication Check:** $4.5 \times 2 = 9$
- **Fraction Equivalency:** $\frac{9}{2} = \frac{45}{10} = 4.5$
- **Alternative Method:** Recognizing that $\frac{1}{2} = 0.5$, then $9 \times 0.5 = 4.5$

All these confirm that the result 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 $\boxed{4.5}$.