Explanation:
Okay, so I need to find the value of 'a' that makes the function f(n) continuous. The function is defined piecewise: it's 4n + 3 when n is less than a, and 7n - 12 when n is greater than or equal to a. Hmm, continuity means that there are no jumps or breaks in the graph, right? So, at the point where the definition of the function changes, which is at n = a, the two pieces should meet. That means the value of the function from the left (when n approaches a from below) should be equal to the value from the right (when n approaches a from above).

Let me write that down. For continuity at n = a, the limit as n approaches a from the left should equal the limit as n approaches a from the right, and both should equal f(a). Since the function is defined differently on either side of a, I need to set the two expressions equal to each other at n = a.

So, if I plug n = a into both expressions, they should be the same. Let me check that.

First, when n is less than a, the function is 4n + 3. So, approaching a from the left, the value would be 4a + 3.

On the other hand, when n is greater than or equal to a, the function is 7n - 12. So, approaching a from the right, the value would be 7a - 12.

Since the function is continuous at a, these two expressions must be equal. Therefore, I can set up the equation:

4a + 3 = 7a - 12

Now, I need to solve for a. Let me subtract 4a from both sides:

3 = 3a - 12

Then, add 12 to both sides:

15 = 3a

Divide both sides by 3:

a = 5

Wait, let me check that again. If a is 5, then let's verify if the function is continuous there.

For n approaching 5 from the left, f(n) approaches 4*5 + 3 = 20 + 3 = 23.

For n approaching 5 from the right, f(n) approaches 7*5 - 12 = 35 - 12 = 23.

And f(5) itself is 7*5 - 12 = 23. So yes, both sides agree. Therefore, the function is continuous at a = 5.

But wait, let me make sure there isn't any other condition or something I missed. The problem says "the graph of y = f(n) is continuous." So, does that mean continuous everywhere? Well, since the function is defined as two linear functions, which are continuous everywhere on their own. The only potential point of discontinuity is at n = a, so by ensuring continuity at n = a, the entire function is continuous.

Therefore, the answer should be a = 5.

But let me think again. Is there a possibility that the function could have another point where it's discontinuous? For example, if there was a different definition for other values? But according to the problem, the function is defined as 4n + 3 when n < a and 7n - 12 when n ≥ a. So, there's no other points of definition. So, the only point where the function could be discontinuous is at n = a. Therefore, solving for a such that the left-hand limit equals the right-hand limit (and f(a)) is sufficient.

Therefore, my conclusion is that a = 5.

But just to be thorough, let me consider if there's any other way this could be approached. For example, maybe the problem is in terms of integers? Wait, the problem didn't specify that n is an integer. It just says n. So, n is a real number? Or is n an integer? Hmm, the problem didn't specify. But since it's talking about the graph of y = f(n), which is typically a function of a real variable. If n were an integer, the problem might have mentioned it, like "for integer values of n" or something. So, I think n is a real number here.

Therefore, the continuity condition is as I considered before. So, solving 4a + 3 = 7a - 12 gives a = 5.

Alternatively, if n were an integer, then the function would be defined for integer n, and the continuity would be about the function being continuous at integer points. But since the problem says "the graph of y = f(n) is continuous," which usually implies continuity for all real numbers, unless specified otherwise.

But even if n were an integer, the function would be continuous at each integer point. However, since the function is defined piecewise with different expressions for n < a and n ≥ a, the only point where the function could have a discontinuity is at n = a. Therefore, regardless of whether n is real or integer, the condition for continuity at n = a is the same.

Therefore, I think my answer is correct.

Wait, but if n is an integer, then the function is defined for integers. So, for example, if a is an integer, then for n approaching a from the left (i.e., n = a - 1), the function would be 4(a - 1) + 3, and for n = a, it's 7a - 12. So, in that case, the continuity would require that 4(a - 1) + 3 = 7a - 12. Let me check that.

If n is integer, then the left-hand limit as n approaches a from the left would be f(a - 1) = 4(a - 1) + 3 = 4a - 4 + 3 = 4a - 1. The right-hand limit is f(a) = 7a - 12. So, setting them equal: 4a - 1 = 7a - 12. Then, solving: -1 + 12 = 7a - 4a => 11 = 3a => a = 11/3 ≈ 3.666... But if a is supposed to be an integer, then this would not be possible. However, the problem didn't specify that a is an integer.

But the problem says "Find a if the graph of y = f(n) is continuous." If n is an integer, then the graph would consist of discrete points, but the term "continuous" is usually applied to functions defined on real numbers. So, if the problem is in the context of real numbers, then my original answer of a = 5 is correct. If it's in the context of integers, then the problem might be different, but since the problem didn't specify, I think the safe assumption is that n is a real number.

Therefore, the answer is a = 5.

Another way to check: if a = 5, then the function is 4n + 3 for n < 5 and 7n - 12 for n ≥ 5. Let me pick a value less than 5, say n = 4. Then f(4) = 4*4 + 3 = 19. For n = 5, f(5) = 7*5 - 12 = 35 - 12 = 23. If I check the value approaching 5 from the left, say n = 4.9, f(4.9) = 4*(4.9) + 3 = 19.6 + 3 = 22.6. Approaching from the right, n = 5.1, f(5.1) = 7*(5.1) - 12 = 35.7 - 12 = 23.7. Wait, but hold on, if a = 5, then at n = 5, the function is 7n - 12, which is 23. But approaching from the left, as n approaches 5, the limit is 4*5 + 3 = 23. So, the left limit is 23, the right limit is 7*5 - 12 = 23, and f(5) is 23. So, yes, continuous.

But when I checked n = 4.9, which is approaching 5 from the left, I get 22.6, which is approaching 23. Similarly, n = 5.1 gives 23.7, which is approaching 23? Wait, no. Wait, 7*5.1 is 35.7, minus 12 is 23.7. Wait, but as n approaches 5 from the right, the value approaches 7*5 - 12 = 23. So, 23.7 is when n is 5.1, which is 0.1 away from 5. So, as n approaches 5 from the right, the value approaches 23. Similarly, approaching from the left, as n approaches 5, the value approaches 23. Therefore, the function is continuous at 5.

So, my initial calculation was correct. Therefore, a = 5 is the correct answer.

I think I've checked it thoroughly. So, I think the answer is 5.

**Final Answer**
The value of $ a $ that makes the graph of $ y = f(n) $ continuous is \boxed{5}.

To determine the value of $ a $ that makes the function

$$
f(n) =
\begin{cases}
4n + 3 & \text{if } n < a, \\
7n - 12 & \text{if } n \ge a,
\end{cases}
$$

continuous, we must ensure that the function has no jumps or breaks at the point where the definition changes, i.e., at $ n = a $.

---

### Step 1: Understand the Concept of Continuity

A function $ f(n) $ is continuous at a point $ n = a $ if:

$$
\lim_{n \to a^-} f(n) = \lim_{n \to a^+} f(n) = f(a)
$$

In this case, the function is defined as two different linear expressions on either side of $ a $. Therefore, the only potential point of discontinuity is at $ n = a $, and we must ensure that the left-hand limit and the right-hand limit are equal to the function value at $ a $.

---

### Step 2: Evaluate the Left-Hand Limit

For $ n < a $, the function is $ f(n) = 4n + 3 $. So, the left-hand limit as $ n \to a^- $ is:

$$
\lim_{n \to a^-} f(n) = 4a + 3
$$

---

### Step 3: Evaluate the Right-Hand Limit

For $ n \ge a $, the function is $ f(n) = 7n - 12 $. So, the right-hand limit as $ n \to a^+ $ is:

$$
\lim_{n \to a^+} f(n) = 7a - 12
$$

---

### Step 4: Set the Limits Equal for Continuity

To ensure continuity at $ n = a $, we set the left-hand limit equal to the right-hand limit:

$$
4a + 3 = 7a - 12
$$

Solving for $ a $:

$$
4a + 3 = 7a - 12 \\
3 + 12 = 7a - 4a \\
15 = 3a \\
a = 5
$$

---

### Step 5: Verify the Continuity at $ a = 5 $

- Left-hand limit: $ 4(5) + 3 = 20 + 3 = 23 $
- Right-hand limit: $ 7(5) - 12 = 35 - 12 = 23 $
- Function value at $ a = 5 $: $ f(5) = 7(5) - 12 = 23 $

All three values are equal, confirming that the function is continuous at $ n = 5 $.

---

### Final Answer

$$
\boxed{5}
$$

[Assistant]
Explanation:
Okay, so I need to find the value of 'a' that makes the function f(n) continuous. The function is defined piecewise: it's 4n + 3 when n is less than a, and 7n - 12 when n is greater than or equal to a. Hmm, continuity means that there are no jumps or breaks in the graph, right? So, at the point where the definition of the function changes, which is at n = a, the two pieces should meet. That means the value of the function from the left (when n approaches a from below) should be equal to the value from the right (when n approaches a from above).

Let me write that down. For continuity at n = a, the limit as n approaches a from the left should equal the limit as n approaches a from the right, and both should equal f(a). Since the function is defined differently on either side of a, I need to set the two expressions equal to each other at n = a.

So, if I plug n = a into both expressions, they should be the same. Let me check that.

First, when n is less than a, the function is 4n + 3. So, approaching a from the left, the value would be 4a + 3.

On the other hand, when n is greater than or equal to a, the function is 7n - 12. So, approaching a from the right, the value would be 7a - 12.

Since the function is continuous at a, these two expressions must be equal. Therefore, I can set up the equation:

4a + 3 = 7a - 12

Now, I need to solve for a. Let me subtract 4a from both sides:

3 = 3a - 12

Then, add 12 to both sides:

15 = 3a

Divide both sides by 3:

a = 5

Wait, let me check that again. If a is 5, then let's verify if the function is continuous there.

For n approaching 5 from the left, f(n) approaches 4*5 + 3 = 20 + 3 = 23.

For n approaching 5 from the right, f(n) approaches 7*5 - 12 = 35 - 12 = 23.

And f(5) itself is 7*5 - 12 = 23. So yes, both sides agree. Therefore, the function is continuous at a = 5.

But wait, let me make sure there isn't any other condition or something I missed. The problem says "the graph of y = f(n) is continuous." So, does that mean continuous everywhere? Well, since the function is defined as two linear functions, which are continuous everywhere on their own. The only potential point of discontinuity is at n = a, so by ensuring continuity at n = a, the entire function is continuous.

Therefore, the answer should be a = 5.

But let me think again. Is there a possibility that the function could have another point where it's discontinuous? For example, if there was a different definition for other values? But according to the problem, the function is defined as 4n + 3 when n < a and 7n - 12 when n ≥ a. So, there's no other points of definition. So, the only point where the function could be discontinuous is at n = a. Therefore, solving for a such that the left-hand limit equals the right-hand limit (and f(a)) is sufficient.

Therefore, my conclusion is that a = 5.

But just to be thorough, let me consider if there's any other way this could be approached. For example, maybe the problem is in terms of integers? Wait, the problem didn't specify that n is an integer. It just says n. So, n is a real number? Or is n an integer? Hmm, the problem didn't specify. But since it's talking about the graph of y = f(n), which is typically a function of a real variable. If n were an integer, the problem might have mentioned it, like "for integer values of n" or something. So, I think n is a real number here.

Therefore, the continuity condition is as I considered before. So, solving 4a + 3 = 7a - 12 gives a = 5.

Alternatively, if n were an integer, then the function would be defined for integer n, and the continuity would be about the function being continuous at integer points. But since the problem says "the graph of y = f(n) is continuous," which usually implies continuity for all real numbers, unless specified otherwise.

But even if n were an integer, the function would be continuous at each integer point. However, since the function is defined piecewise with different expressions for n < a and n ≥ a, the only point where the function could have a discontinuity is at n = a. Therefore, regardless of whether n is real or integer, the condition for continuity at n = a is the same.

Therefore, I think my answer is correct.

Wait, but if n is an integer, then the function is defined for integers. So, for example, if a is an integer, then for n approaching a from the left (i.e., n = a - 1), the function would be 4(a - 1) + 3, and for n = a, it's 7a - 12. So, in that case, the continuity would require that 4(a - 1) + 3 = 7a - 12. Let me check that.

If n is integer, then the left-hand limit as n approaches a from the left would be f(a - 1) = 4(a - 1) + 3 = 4a - 4 + 3 = 4a - 1. The right-hand limit is f(a) = 7a - 12. So, setting them equal: 4a - 1 = 7a - 12. Then, solving: -1 + 12 = 7a - 4a => 11 = 3a => a = 11/3 ≈ 3.666... But if a is supposed to be an integer, then this would not be possible. However, the problem didn't specify that a is an integer.

But the problem says "Find a if the graph of y = f(n) is continuous." If n is an integer, then the graph would consist of discrete points, but the term "continuous" is usually applied to functions defined on real numbers. So, if the problem is in the context of real numbers, then my original answer of a = 5 is correct. If it's in the context of integers, then the problem might be different, but since the problem didn't specify, I think the safe assumption is that n is a real number.

Therefore, the answer is a = 5.

Another way to check: if a = 5, then the function is 4n + 3 for n < 5 and 7n - 12 for n ≥ 5. Let me pick a value less than 5, say n = 4. Then f(4) = 4*4 + 3 = 19. For n = 5, f(5) = 7*5 - 12 = 35 - 12 = 23. If I check the value approaching 5 from the left, say n = 4.9, f(4.9) = 4*(4.9) + 3 = 19.6 + 3 = 22.6. Approaching from the right, n = 5.1, f(5.1) = 7*(5.1) - 12 = 35.7 - 12 = 23.7. Wait, but hold on, if a = 5, then at n = 5, the function is 7n - 12, which is 23. But approaching from the left, as n approaches 5, the limit is 4*5 + 3 = 23. So, the left limit is 23, the right limit is 7*5 - 12 = 23, and f(5) is 23. So, yes, continuous.

But when I checked n = 4.9, which is approaching 5 from the left, I get 22.6, which is approaching 23. Similarly, n = 5.1 gives 23.7, which is approaching 23? Wait, no. Wait, 7*5.1 is 35.7, minus 12 is 23.7. Wait, but as n approaches 5 from the right, the value approaches 7*5 - 12 = 23. So, 23.7 is when n is 5.1, which is 0.1 away from 5. So, as n approaches 5 from the right, the value approaches 23. Similarly, approaching from the left, as n approaches 5, the value approaches 23. Therefore, the function is continuous at 5.

So, my initial calculation was correct. Therefore, a = 5 is the correct answer.

I think I've checked it thoroughly. So, I think the answer is 5.

**Final Answer**
The value of $ a $ that makes the graph of $ y = f(n) $ continuous is \boxed{5}.

To determine the value of $ a $ that makes the function

$$
f(n) =
\begin{cases}
4n + 3 & \text{if } n < a, \\
7n - 12 & \text{if } n \ge a,
\end{cases}
$$

continuous, we must ensure that the function has no jumps or breaks at the point where the definition changes, i.e., at $ n = a $.

---

### Step 1: Understand the Concept of Continuity

A function $ f(n) $ is continuous at a point $ n = a $ if:

$$
\lim_{n \to a^-} f(n) = \lim_{n \to a^+} f(n) = f(a)
$$

In this case, the function is defined as two different linear expressions on either side of $ a $. Therefore, the only potential point of discontinuity is at $ n = a $, and we must ensure that the left-hand limit and the right-hand limit are equal to the function value at $ a $.

---

### Step 2: Evaluate the Left-Hand Limit

For $ n < a $, the function is $ f(n) = 4n + 3 $. So, the left-hand limit as $ n \to a^- $ is:

$$
\lim_{n \to a^-} f(n) = 4a + 3
$$

---

### Step 3: Evaluate the Right-Hand Limit

For $ n \ge a $, the function is $ f(n) = 7n - 12 $. So, the right-hand limit as $ n \to a^+ $ is:

$$
\lim_{n \to a^+} f(n) = 7a - 12
$$

---

### Step 4: Set the Limits Equal for Continuity

To ensure continuity at $ n = a $, we set the left-hand limit equal to the right-hand limit:

$$
4a + 3 = 7a - 12
$$

Solving for $ a $:

$$
4a + 3 = 7a - 12 \\
3 + 12 = 7a - 4a \\
15 = 3a \\
a = 5
$$

---

### Step 5: Verify the Continuity at $ a = 5 $

- Left-hand limit: $ 4(5) + 3 = 20 + 3 = 23 $
- Right-hand limit: $ 7(5) - 12 = 35 - 12 = 23 $
- Function value at $ a = 5 $: $ f(5) = 7(5) - 12 = 23 $

All three values are equal, confirming that the function is continuous at $ n = 5 $.

---

### Final Answer

$$
\boxed{5}
$$

Solution: In order for the graph to be continuous, the two pieces of the function must meet at $n=a$. In order for this to happen, we know that $4a+3=7a-12$. Solving for $a$, we find that $a=\frac{15}{3}=\boxed{5}$.


Explanation:
Okay, so I need to compute the binomial coefficient $\dbinom{31}{28}$. Hmm, let me remember how binomial coefficients work. I think they're used in combinations, right? Like, the number of ways to choose 28 items out of 31 without considering the order. But wait, there's a formula for that. Let me recall... I think it's $\dbinom{n}{k} = \frac{n!}{k!(n - k)!}$. Yeah, that sounds right. So, applying that formula here, it should be $\frac{31!}{28!(31 - 28)!}$. Let me check the subtraction in the denominator: 31 minus 28 is 3. So, the denominator becomes 28! times 3!.

But wait, factorials can get really big. Maybe there's a way to simplify this before calculating? Let me think. Since 31! divided by 28! would be 31 × 30 × 29 × 28! divided by 28!... Oh, right! The 28! cancels out from numerator and denominator. So that simplifies to 31 × 30 × 29 divided by 3!.

So, let me write that down: $\dbinom{31}{28} = \frac{31 \times 30 \times 29}{3 \times 2 \times 1}$. Let me compute the numerator first. 31 multiplied by 30 is... 31*30. Let me do 30*30 which is 900, plus 1*30 which is 30, so total 930. Then multiply that by 29. Hmm, 930*29. Let me break that down. 930*30 would be 27,900, so subtract 930 to get 930*29. So 27,900 - 930 = 26,970. So the numerator is 26,970.

Now the denominator is 3! which is 6. So, divide 26,970 by 6. Let me do that division. 26,970 divided by 6. Let me see, 6 goes into 26 four times (4*6=24), remainder 2. Bring down the 9 to make 29. 6 into 29 is 4 times (4*6=24), remainder 5. Bring down the 7 to make 57. 6 into 57 is 9 times (9*6=54), remainder 3. Bring down the 0 to make 30. 6 into 30 is 5 times. So putting that all together: 4, 4, 9, 5. Wait, let me check again step by step.

Wait, maybe I should do it another way. Let me divide 26,970 by 6. Let me divide each part:

First, 26,970 divided by 6. Let me divide 26,970 by 6:

Divide 26,970 by 6:

6 × 4,000 = 24,000. Subtract 24,000 from 26,970: 2,970.

Now divide 2,970 by 6. 6 × 400 = 2,400. Subtract: 2,970 - 2,400 = 570.

Divide 570 by 6: 6 × 95 = 570. So total is 4,000 + 400 + 95 = 4,495.

So, 26,970 divided by 6 is 4,495. Therefore, $\dbinom{31}{28} = 4,495$.

Wait, but let me verify this because sometimes when I do division step by step, I might make a mistake. Let me check with another method. Let me compute 26,970 divided by 6.

Alternatively, 26,970 divided by 6. Let me divide 26,970 by 6:

First, divide 26,970 by 2: 13,485. Then divide by 3: 13,485 ÷ 3. Let me compute that. 3 × 4,000 = 12,000. Subtract 12,000 from 13,485: 1,485. 3 × 400 = 1,200. Subtract: 1,485 - 1,200 = 285. 3 × 95 = 285. So total is 4,000 + 400 + 95 = 4,495. Yep, same result. So that seems correct.

But just to be thorough, maybe I can check using another approach. Let me recall that $\dbinom{n}{k} = \dbinom{n}{n - k}$. So, $\dbinom{31}{28} = \dbinom{31}{3}$. Maybe computing $\dbinom{31}{3}$ is easier? Let me try that.

$\dbinom{31}{3} = \frac{31 \times 30 \times 29}{3 \times 2 \times 1}$, which is exactly the same as before. So that's 31*30*29 / 6. Which is the same calculation. So, same result. So that's reassuring.

Alternatively, maybe I can compute $\dbinom{31}{3}$ step by step. Let me compute 31 choose 3.

First, 31 × 30 = 930, as before. Then 930 × 29. Wait, that's the same numerator. Then divide by 6. So same thing. So, same answer.

Alternatively, maybe I can compute 31 × 30 × 29 divided by 6. Let me see if I can simplify before multiplying. For example, 31 divided by something? Well, 31 is a prime number, so maybe not. Let me see. Let me write the numerator as 31 × 30 × 29. The denominator is 6, which is 2 × 3. Let me see if I can cancel factors.

So, 30 in the numerator is 2 × 3 × 5. So, 30 divided by 6 is 5. So, if I divide 30 by 6, that would be 5. So, then the expression becomes 31 × 5 × 29. Let me check that.

Original expression: (31 × 30 × 29) / 6. Since 30 / 6 = 5, so yes, that simplifies to 31 × 5 × 29.

So, compute 31 × 5 first. 31 × 5 is 155. Then multiply by 29. So, 155 × 29. Let me compute that.

Breaking down 155 × 29: 155 × 30 = 4,650. Subtract 155 to get 155 × 29. So, 4,650 - 155 = 4,495.

Same result! So that's another way to compute it, and it gives the same answer. So that's good.

Therefore, regardless of the method, I get 4,495.

But just to be absolutely sure, maybe I can check with a calculator? Wait, since I don't have a calculator here, but maybe I can verify with smaller numbers. Let me test with a known value. For example, $\dbinom{5}{2} = 10$. Let me apply the same method.

$\dbinom{5}{2} = \frac{5!}{2!3!} = \frac{5 × 4}{2 × 1} = 10$. Correct.

Another example: $\dbinom{10}{3} = 120$. Let me compute that. 10 × 9 × 8 / 6 = 720 / 6 = 120. Correct.

So, the method works. Therefore, applying the same method to $\dbinom{31}{28}$ gives 4,495.

Alternatively, maybe I can use Pascal's triangle or some recursive formula? But that might take longer.

Alternatively, think about the multiplicative formula for combinations. The formula is $\dbinom{n}{k} = \frac{n(n - 1)(n - 2)...(n - k + 1)}{k!}$. So, for $\dbinom{31}{28}$, that would be $\frac{31 × 30 × 29 × ... × (31 - 28 + 1)}{28!}$. Wait, but 31 - 28 + 1 is 4. So, the numerator would be 31 × 30 × 29 × 28! divided by 28! which cancels out, giving 31 × 30 × 29 / 3! again. So same as before.

Therefore, regardless of the approach, the answer is 4,495.

Wait, but just to make sure, let me check if 31 choose 28 is indeed equal to 31 choose 3. Since 31 - 28 = 3. So, yes, $\dbinom{31}{28} = \dbinom{31}{3}$. Let me check with another example. For instance, $\dbinom{5}{2} = \dbinom{5}{3} = 10$. Correct. So, that's a valid property. Therefore, computing $\dbinom{31}{3}$ is equivalent.

So, if I compute $\dbinom{31}{3}$, which is 31 × 30 × 29 / 6, which we did earlier, and got 4,495.

Alternatively, maybe I can compute 31 × 30 first, which is 930, then multiply by 29. Let me do that again.

930 × 29. Let me think of 29 as 30 - 1. So, 930 × 30 = 27,900. Then subtract 930 × 1 = 930. So, 27,900 - 930 = 26,970. Then divide by 6. Which is 26,970 / 6 = 4,495. Yep.

Alternatively, maybe I can compute 26,970 divided by 6 by breaking it down. 26,970 divided by 6. Let me divide each digit:

6 into 2 is 0, 6 into 26 is 4 (4×6=24), remainder 2. Bring down 9: 29. 6 into 29 is 4 (4×6=24), remainder 5. Bring down 7: 57. 6 into 57 is 9 (9×6=54), remainder 3. Bring down 0: 30. 6 into 30 is 5. So the quotient is 4,495. Yep, same as before.

So, all methods lead to 4,495. Therefore, I think that's the correct answer.

But just to be 100% sure, maybe I can check with another approach. Let me think.

Suppose I compute $\dbinom{31}{28}$ using the multiplicative formula step by step.

The multiplicative formula is $\dbinom{n}{k} = \frac{n}{k} \times \frac{n - 1}{k - 1} \times \cdots \times \frac{n - k + 1}{1}$.

But for $\dbinom{31}{28}$, that would be $\frac{31}{28} \times \frac{30}{27} \times \frac{29}{26} \times \cdots$? Wait, no, that seems more complicated. Maybe not the best approach.

Alternatively, since $\dbinom{31}{28} = \dbinom{31}{3}$, let me compute $\dbinom{31}{3}$ using the multiplicative formula.

The formula is $\dbinom{n}{k} = \frac{n \times (n - 1) \times (n - 2)}{k \times (k - 1) \times (k - 2)}$.

So, for $\dbinom{31}{3}$, that would be $\frac{31 × 30 × 29}{3 × 2 × 1}$, which is the same as before. So, again, same calculation.

Alternatively, maybe compute step by step:

First term: 31 / 3 = 10.333... Hmm, but fractions might complicate.

Alternatively, compute 31 × 30 = 930, then 930 × 29 = 26,970, then divide by 6. Which is 4,495.

Alternatively, think of 31 × 30 × 29 divided by 6. Let me factor all numbers:

31 is prime. 30 is 2 × 3 × 5. 29 is prime. So numerator is 31 × 2 × 3 × 5 × 29. Denominator is 2 × 3. So cancel out 2 and 3, leaving 31 × 5 × 29. Which is 31 × 5 = 155, 155 × 29. Which is 155 × 30 - 155 = 4,650 - 155 = 4,495.

Same result.

Therefore, I think I can be confident that the answer is 4,495.

But just to make sure, let me check with another example. Let me compute $\dbinom{10}{3}$ using this method.

$\dbinom{10}{3} = \frac{10 × 9 × 8}{3 × 2 × 1} = \frac{720}{6} = 120$. Which is correct.

Another example: $\dbinom{5}{2} = \frac{5 × 4}{2 × 1} = 10$. Correct.

So, the method works. Therefore, applying it to $\dbinom{31}{28}$ gives 4,495.

Alternatively, maybe I can use the recursive formula: $\dbinom{n}{k} = \dbinom{n - 1}{k - 1} + \dbinom{n - 1}{k}$. But computing $\dbinom{31}{28}$ recursively would take a lot of steps, so maybe not practical here.

But just for fun, let me try to compute $\dbinom{31}{28}$ using recursion.

But since $\dbinom{31}{28} = \dbinom{30}{27} + \dbinom{30}{28}$. Then $\dbinom{30}{27} = \dbinom{29}{26} + \dbinom{29}{27}$, and so on. This seems too tedious.

Alternatively, maybe use known values. But without a table, it's not helpful.

Alternatively, think about the value of $\dbinom{31}{3}$. Let me think of 31 choose 3.

I know that $\dbinom{n}{3} = \frac{n(n - 1)(n - 2)}{6}$. So, plugging in n = 31, we get $\frac{31 × 30 × 29}{6}$. Which is exactly what we did.

Alternatively, maybe approximate? Well, 31 × 30 × 29 is roughly 30 × 30 × 30 = 27,000. Divided by 6 is 4,500. Which is very close to our exact answer of 4,495. So that approximation makes sense.

Therefore, given all these cross-checks and methods, I can be confident that the answer is 4,495.

**Final Answer**
\boxed{4495}

To compute the binomial coefficient $\dbinom{31}{28}$, we use the standard formula for combinations:

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

---

### Step 1: Apply the formula

We are given $n = 31$ and $k = 28$, so:

$$
\dbinom{31}{28} = \frac{31!}{28!(31 - 28)!} = \frac{31!}{28! \cdot 3!}
$$

---

### Step 2: Simplify the factorials

Note that:

$$
\frac{31!}{28!} = 31 \times 30 \times 29
$$

So the expression becomes:

$$
\dbinom{31}{28} = \frac{31 \times 30 \times 29}{3!} = \frac{31 \times 30 \times 29}{6}
$$

---

### Step 3: Compute the numerator

$$
31 \times 30 = 930
$$
$$
930 \times 29 = 26,970
$$

---

### Step 4: Divide by the denominator

$$
\frac{26,970}{6} = 4,495
$$

---

### Step 5: Final Answer

$$
\boxed{4495}
$$

[Assistant]
Explanation:
Okay, so I need to compute the binomial coefficient $\dbinom{31}{28}$. Hmm, let me remember how binomial coefficients work. I think they're used in combinations, right? Like, the number of ways to choose 28 items out of 31 without considering the order. But wait, there's a formula for that. Let me recall... I think it's $\dbinom{n}{k} = \frac{n!}{k!(n - k)!}$. Yeah, that sounds right. So, applying that formula here, it should be $\frac{31!}{28!(31 - 28)!}$. Let me check the subtraction in the denominator: 31 minus 28 is 3. So, the denominator becomes 28! times 3!.

But wait, factorials can get really big. Maybe there's a way to simplify this before calculating? Let me think. Since 31! divided by 28! would be 31 × 30 × 29 × 28! divided by 28!... Oh, right! The 28! cancels out from numerator and denominator. So that simplifies to 31 × 30 × 29 divided by 3!.

So, let me write that down: $\dbinom{31}{28} = \frac{31 \times 30 \times 29}{3 \times 2 \times 1}$. Let me compute the numerator first. 31 multiplied by 30 is... 31*30. Let me do 30*30 which is 900, plus 1*30 which is 30, so total 930. Then multiply that by 29. Hmm, 930*29. Let me break that down. 930*30 would be 27,900, so subtract 930 to get 930*29. So 27,900 - 930 = 26,970. So the numerator is 26,970.

Now the denominator is 3! which is 6. So, divide 26,970 by 6. Let me do that division. 26,970 divided by 6. Let me see, 6 goes into 26 four times (4*6=24), remainder 2. Bring down the 9 to make 29. 6 into 29 is 4 times (4*6=24), remainder 5. Bring down the 7 to make 57. 6 into 57 is 9 times (9*6=54), remainder 3. Bring down the 0 to make 30. 6 into 30 is 5 times. So putting that all together: 4, 4, 9, 5. Wait, let me check again step by step.

Wait, maybe I should do it another way. Let me divide 26,970 by 6. Let me divide each part:

First, 26,970 divided by 6. Let me divide 26,970 by 6:

Divide 26,970 by 6:

6 × 4,000 = 24,000. Subtract 24,000 from 26,970: 2,970.

Now divide 2,970 by 6. 6 × 400 = 2,400. Subtract: 2,970 - 2,400 = 570.

Divide 570 by 6: 6 × 95 = 570. So total is 4,000 + 400 + 95 = 4,495.

So, 26,970 divided by 6 is 4,495. Therefore, $\dbinom{31}{28} = 4,495$.

Wait, but let me verify this because sometimes when I do division step by step, I might make a mistake. Let me check with another method. Let me compute 26,970 divided by 6.

Alternatively, 26,970 divided by 6. Let me divide 26,970 by 6:

First, divide 26,970 by 2: 13,485. Then divide by 3: 13,485 ÷ 3. Let me compute that. 3 × 4,000 = 12,000. Subtract 12,000 from 13,485: 1,485. 3 × 400 = 1,200. Subtract: 1,485 - 1,200 = 285. 3 × 95 = 285. So total is 4,000 + 400 + 95 = 4,495. Yep, same result. So that seems correct.

But just to be thorough, maybe I can check using another approach. Let me recall that $\dbinom{n}{k} = \dbinom{n}{n - k}$. So, $\dbinom{31}{28} = \dbinom{31}{3}$. Maybe computing $\dbinom{31}{3}$ is easier? Let me try that.

$\dbinom{31}{3} = \frac{31 \times 30 \times 29}{3 \times 2 \times 1}$, which is exactly the same as before. So that's 31*30*29 / 6. Which is the same calculation. So, same result. So that's reassuring.

Alternatively, maybe I can compute $\dbinom{31}{3}$ step by step. Let me compute 31 choose 3.

First, 31 × 30 = 930, as before. Then 930 × 29. Wait, that's the same numerator. Then divide by 6. So same thing. So, same answer.

Alternatively, maybe I can compute 31 × 30 × 29 divided by 6. Let me see if I can simplify before multiplying. For example, 31 divided by something? Well, 31 is a prime number, so maybe not. Let me see. Let me write the numerator as 31 × 30 × 29. The denominator is 6, which is 2 × 3. Let me see if I can cancel factors.

So, 30 in the numerator is 2 × 3 × 5. So, 30 divided by 6 is 5. So, if I divide 30 by 6, that would be 5. So, then the expression becomes 31 × 5 × 29. Let me check that.

Original expression: (31 × 30 × 29) / 6. Since 30 / 6 = 5, so yes, that simplifies to 31 × 5 × 29.

So, compute 31 × 5 first. 31 × 5 is 155. Then multiply by 29. So, 155 × 29. Let me compute that.

Breaking down 155 × 29: 155 × 30 = 4,650. Subtract 155 to get 155 × 29. So, 4,650 - 155 = 4,495.

Same result! So that's another way to compute it, and it gives the same answer. So that's good.

Therefore, regardless of the method, I get 4,495.

But just to be absolutely sure, maybe I can check with a calculator? Wait, since I don't have a calculator here, but maybe I can verify with smaller numbers. Let me test with a known value. For example, $\dbinom{5}{2} = 10$. Let me apply the same method.

$\dbinom{5}{2} = \frac{5!}{2!3!} = \frac{5 × 4}{2 × 1} = 10$. Correct.

Another example: $\dbinom{10}{3} = 120$. Let me compute that. 10 × 9 × 8 / 6 = 720 / 6 = 120. Correct.

So, the method works. Therefore, applying the same method to $\dbinom{31}{28}$ gives 4,495.

Alternatively, maybe I can use Pascal's triangle or some recursive formula? But that might take longer.

Alternatively, think about the multiplicative formula for combinations. The formula is $\dbinom{n}{k} = \frac{n(n - 1)(n - 2)...(n - k + 1)}{k!}$. So, for $\dbinom{31}{28}$, that would be $\frac{31 × 30 × 29 × ... × (31 - 28 + 1)}{28!}$. Wait, but 31 - 28 + 1 is 4. So, the numerator would be 31 × 30 × 29 × 28! divided by 28! which cancels out, giving 31 × 30 × 29 / 3! again. So same as before.

Therefore, regardless of the approach, the answer is 4,495.

Wait, but just to make sure, let me check if 31 choose 28 is indeed equal to 31 choose 3. Since 31 - 28 = 3. So, yes, $\dbinom{31}{28} = \dbinom{31}{3}$. Let me check with another example. For instance, $\dbinom{5}{2} = \dbinom{5}{3} = 10$. Correct. So, that's a valid property. Therefore, computing $\dbinom{31}{3}$ is equivalent.

So, if I compute $\dbinom{31}{3}$, which is 31 × 30 × 29 / 6, which we did earlier, and got 4,495.

Alternatively, maybe I can compute 31 × 30 first, which is 930, then multiply by 29. Let me do that again.

930 × 29. Let me think of 29 as 30 - 1. So, 930 × 30 = 27,900. Then subtract 930 × 1 = 930. So, 27,900 - 930 = 26,970. Then divide by 6. Which is 26,970 / 6 = 4,495. Yep.

Alternatively, maybe I can compute 26,970 divided by 6 by breaking it down. 26,970 divided by 6. Let me divide each digit:

6 into 2 is 0, 6 into 26 is 4 (4×6=24), remainder 2. Bring down 9: 29. 6 into 29 is 4 (4×6=24), remainder 5. Bring down 7: 57. 6 into 57 is 9 (9×6=54), remainder 3. Bring down 0: 30. 6 into 30 is 5. So the quotient is 4,495. Yep, same as before.

So, all methods lead to 4,495. Therefore, I think that's the correct answer.

But just to be 100% sure, maybe I can check with another approach. Let me think.

Suppose I compute $\dbinom{31}{28}$ using the multiplicative formula step by step.

The multiplicative formula is $\dbinom{n}{k} = \frac{n}{k} \times \frac{n - 1}{k - 1} \times \cdots \times \frac{n - k + 1}{1}$.

But for $\dbinom{31}{28}$, that would be $\frac{31}{28} \times \frac{30}{27} \times \frac{29}{26} \times \cdots$? Wait, no, that seems more complicated. Maybe not the best approach.

Alternatively, since $\dbinom{31}{28} = \dbinom{31}{3}$, let me compute $\dbinom{31}{3}$ using the multiplicative formula.

The formula is $\dbinom{n}{k} = \frac{n \times (n - 1) \times (n - 2)}{k \times (k - 1) \times (k - 2)}$.

So, for $\dbinom{31}{3}$, that would be $\frac{31 × 30 × 29}{3 × 2 × 1}$, which is the same as before. So, again, same calculation.

Alternatively, maybe compute step by step:

First term: 31 / 3 = 10.333... Hmm, but fractions might complicate.

Alternatively, compute 31 × 30 = 930, then 930 × 29 = 26,970, then divide by 6. Which is 4,495.

Alternatively, think of 31 × 30 × 29 divided by 6. Let me factor all numbers:

31 is prime. 30 is 2 × 3 × 5. 29 is prime. So numerator is 31 × 2 × 3 × 5 × 29. Denominator is 2 × 3. So cancel out 2 and 3, leaving 31 × 5 × 29. Which is 31 × 5 = 155, 155 × 29. Which is 155 × 30 - 155 = 4,650 - 155 = 4,495.

Same result.

Therefore, I think I can be confident that the answer is 4,495.

But just to make sure, let me check with another example. Let me compute $\dbinom{10}{3}$ using this method.

$\dbinom{10}{3} = \frac{10 × 9 × 8}{3 × 2 × 1} = \frac{720}{6} = 120$. Which is correct.

Another example: $\dbinom{5}{2} = \frac{5 × 4}{2 × 1} = 10$. Correct.

So, the method works. Therefore, applying it to $\dbinom{31}{28}$ gives 4,495.

Alternatively, maybe I can use the recursive formula: $\dbinom{n}{k} = \dbinom{n - 1}{k - 1} + \dbinom{n - 1}{k}$. But computing $\dbinom{31}{28}$ recursively would take a lot of steps, so maybe not practical here.

But just for fun, let me try to compute $\dbinom{31}{28}$ using recursion.

But since $\dbinom{31}{28} = \dbinom{30}{27} + \dbinom{30}{28}$. Then $\dbinom{30}{27} = \dbinom{29}{26} + \dbinom{29}{27}$, and so on. This seems too tedious.

Alternatively, maybe use known values. But without a table, it's not helpful.

Alternatively, think about the value of $\dbinom{31}{3}$. Let me think of 31 choose 3.

I know that $\dbinom{n}{3} = \frac{n(n - 1)(n - 2)}{6}$. So, plugging in n = 31, we get $\frac{31 × 30 × 29}{6}$. Which is exactly what we did.

Alternatively, maybe approximate? Well, 31 × 30 × 29 is roughly 30 × 30 × 30 = 27,000. Divided by 6 is 4,500. Which is very close to our exact answer of 4,495. So that approximation makes sense.

Therefore, given all these cross-checks and methods, I can be confident that the answer is 4,495.

**Final Answer**
\boxed{4495}

To compute the binomial coefficient $\dbinom{31}{28}$, we use the standard formula for combinations:

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

---

### Step 1: Apply the formula

We are given $n = 31$ and $k = 28$, so:

$$
\dbinom{31}{28} = \frac{31!}{28!(31 - 28)!} = \frac{31!}{28! \cdot 3!}
$$

---

### Step 2: Simplify the factorials

Note that:

$$
\frac{31!}{28!} = 31 \times 30 \times 29
$$

So the expression becomes:

$$
\dbinom{31}{28} = \frac{31 \times 30 \times 29}{3!} = \frac{31 \times 30 \times 29}{6}
$$

---

### Step 3: Compute the numerator

$$
31 \times 30 = 930
$$
$$
930 \times 29 = 26,970
$$

---

### Step 4: Divide by the denominator

$$
\frac{26,970}{6} = 4,495
$$

---

### Step 5: Final Answer

$$
\boxed{4495}
$$

Solution:
\begin{align*}
\dbinom{31}{28} &= \dbinom{31}{3} \\
&= \dfrac{31!}{28!3!} \\
&= \dfrac{31\times 30\times 29}{3\times 2\times 1} \\
&= 31 \times \dfrac{30}{3\times 2\times 1} \times 29 \\
&= 31\times 5\times 29 \\
&= \boxed{4495}.
\end{align*}


Explanation:
Okay, so I need to find the units digit of 18 raised to the 6th power. Hmm, let me think. The units digit is the last digit of the number, right? So, if I can figure out what 18^6 ends with, that's the answer.

First, maybe I should recall that when you raise a number to a power, the units digit depends only on the units digit of the original number. Wait, is that true? Let me check. For example, if I have 18, the units digit is 8. So maybe I can just focus on 8^6 and find its units digit? That might be easier. Let me verify this idea with a smaller exponent. Let's take 18^1, which is 18, units digit 8. 18^2 is 324, units digit 4. If I do 8^2, that's 64, which also ends with 4. Okay, so that seems to work. Similarly, 18^3 would be 18*18*18. Let me compute 18^3. 18*18 is 324, then 324*18. Let me do that multiplication. 324*10 is 3240, 324*8 is 2592, so total is 3240 + 2592 = 5832. The units digit is 2. If I do 8^3, that's 512, which also ends with 2. So yeah, it seems like the units digit of 18^n is the same as the units digit of 8^n. So I can just compute 8^6 and find its units digit. That simplifies things!

Alright, so let me compute 8^6. Let me do this step by step. Let's start with 8^1 = 8. 8^2 = 8*8 = 64. 8^3 = 8*64. Let me compute that. 8*60 is 480, 8*4 is 32, so 480 + 32 = 512. 8^3 is 512. Then 8^4 is 8*512. Let me calculate that. 512*8. Let me break it down: 500*8 = 4000, 12*8 = 96, so total is 4000 + 96 = 4096. So 8^4 is 4096. The units digit here is 6.

Moving on to 8^5. That's 8^4 * 8 = 4096 * 8. Let me compute that. 4000*8 = 32,000. 96*8 = 768. Adding them together: 32,000 + 768 = 32,768. So 8^5 is 32,768. The units digit is 8.

Then 8^6 is 8^5 * 8 = 32,768 * 8. Let me calculate that. Let me break it down again. 32,768 * 8. Let me do 30,000 * 8 = 240,000. 2,000 * 8 = 16,000. 700 * 8 = 5,600. 60 * 8 = 480. 8 * 8 = 64. Now add all those up.

Wait, maybe that's too tedious. Alternatively, since I only care about the units digit, maybe there's a pattern here. Let me think.

Earlier, I saw that 8^1 ends with 8, 8^2 ends with 4, 8^3 ends with 2, 8^4 ends with 6, 8^5 ends with 8 again. Wait, so there's a cycle here. Let me list them out:

- 8^1: 8
- 8^2: 4
- 8^3: 2
- 8^4: 6
- 8^5: 8
- 8^6: 4

So the pattern repeats every 4 exponents. The cycle is 8, 4, 2, 6, then back to 8, 4, etc. So the cycle length is 4. Therefore, if I can figure out where 6 falls in this cycle, I can find the units digit.

Since the cycle is 4 numbers long, I can divide the exponent by 4 and find the remainder. For exponent 6, divide 6 by 4. 4 goes into 6 once with a remainder of 2. So the remainder is 2. Then, looking at the cycle:

Remainder 1: 8

Remainder 2: 4

Remainder 3: 2

Remainder 0: 6 (since 4 divides into 4 with remainder 0, which corresponds to the 4th position in the cycle)

Wait, so if the remainder is 0, it's the 4th position. So for exponent 6, remainder 2, so the units digit is 4? Let me check with my previous calculation. Earlier, I calculated 8^5 ends with 8, then 8^6 would be 8^5 * 8, so 8*8=64, so units digit 4. Yes, that matches. So according to this pattern, the units digit of 8^6 is 4. Therefore, the units digit of 18^6 is also 4.

But let me confirm this by actually computing 18^6. Wait, that might be a big number, but maybe I can compute it step by step, keeping track of the units digit each time. Let me try that.

Starting with 18^1: units digit 8.

18^2: 18*18. Units digit is 8*8=64, so 4.

18^3: 18^2 * 18. Units digit is 4*8=32, so 2.

18^4: 18^3 * 18. Units digit is 2*8=16, so 6.

18^5: 18^4 * 18. Units digit is 6*8=48, so 8.

18^6: 18^5 * 18. Units digit is 8*8=64, so 4.

Yes! So that's the same result. So the units digit is 4. Therefore, the answer is 4.

Alternatively, another way to think about it is using modular arithmetic. Since we only care about the units digit, we can compute 18^6 mod 10.

But 18 mod 10 is 8, so 18^6 mod 10 is equivalent to 8^6 mod 10. Which is exactly what I did before. So 8^6 mod 10.

Alternatively, since 8 and 10 are not coprime, Euler's theorem might not apply directly, but since we have a small exponent, it's easier to compute the pattern.

But maybe I can use Euler's theorem here. Wait, Euler's theorem says that if a and n are coprime, then a^φ(n) ≡ 1 mod n. For n=10, φ(10)=φ(2*5)=φ(2)*φ(5)=1*4=4. So φ(10)=4. However, 8 and 10 are not coprime, since gcd(8,10)=2. Therefore, Euler's theorem doesn't apply here.

But maybe I can still use the concept of cycles. Since we saw that the units digit cycles every 4 exponents. So for 8^k mod 10, the cycle length is 4. Therefore, 8^(k) mod 10 = 8^(k mod 4) mod 10, but with adjustment if k mod 4 is 0.

So for exponent 6, 6 divided by 4 is 1 with remainder 2. Therefore, 8^6 mod 10 = 8^2 mod 10 = 64 mod 10 = 4. Which matches our previous results.

Alternatively, if I wanted to use modular exponentiation, I could compute 8^6 mod 10 step by step:

Compute 8^1 mod 10 = 8

8^2 mod 10 = (8*8) mod 10 = 64 mod 10 = 4

8^3 mod 10 = (8^2 * 8) mod 10 = (4 * 8) mod 10 = 32 mod 10 = 2

8^4 mod 10 = (8^3 * 8) mod 10 = (2 * 8) mod 10 = 16 mod 10 = 6

8^5 mod 10 = (8^4 * 8) mod 10 = (6 * 8) mod 10 = 48 mod 10 = 8

8^6 mod 10 = (8^5 * 8) mod 10 = (8 * 8) mod 10 = 64 mod 10 = 4

Same result again. So regardless of the method, the units digit is 4.

Just to make sure I didn't make a mistake in my reasoning, let me check with another approach. Let's compute 18^6 directly, but only keep track of the units digit at each multiplication step.

Starting with 18^1: units digit 8.

18^2: 18 * 18. Let's compute 18*18. As before, 18*10=180, 18*8=144, so total is 180+144=324. Units digit 4.

18^3: 324 * 18. Let's compute 324*18. Again, breaking it down: 324*10=3240, 324*8=2592. Adding them gives 3240+2592=5832. Units digit 2.

18^4: 5832 * 18. Let me compute 5832*18. Let me do 5832*10=58320, 5832*8=46656. Adding them: 58320 + 46656. Let me add these numbers. 58320 + 46656. Let me add 58320 + 40000 = 98320, then +6656 = 98320 + 6656. 98320 + 6000 = 104320, then +656 = 104,976. So 18^4 is 104,976. Units digit 6.

18^5: 104,976 * 18. Let me compute this. Again, breaking it down: 104,976 * 10 = 1,049,760. 104,976 * 8 = let's calculate that. 100,000 * 8 = 800,000. 4,976 * 8. Let me compute 4,000*8=32,000. 976*8. 900*8=7,200. 76*8=608. So 7,200 + 608 = 7,808. Then 32,000 + 7,808 = 39,808. So total 800,000 + 39,808 = 839,808. Therefore, 104,976 * 8 = 839,808. Now add that to 1,049,760.

So 1,049,760 + 839,808. Let me add them:

1,049,760 + 800,000 = 1,849,760

Then +39,808 = 1,849,760 + 39,808. Let me compute 1,849,760 + 30,000 = 1,879,760. Then +9,808 = 1,879,760 + 9,808 = 1,889,568. Therefore, 18^5 is 1,889,568. Units digit is 8.

Then 18^6: 1,889,568 * 18. Let me compute this. Again, breaking it down:

First, 1,889,568 * 10 = 18,895,680

Then 1,889,568 * 8. Let me compute that.

Let me do 1,889,568 * 8:

Break it down:

1,000,000 * 8 = 8,000,000

800,000 * 8 = 6,400,000

80,000 * 8 = 640,000

9,000 * 8 = 72,000

500 * 8 = 4,000

60 * 8 = 480

8 * 8 = 64

Wait, maybe that's too detailed. Alternatively, let me compute 1,889,568 * 8 step by step:

Starting from the right:

8 * 8 = 64, write down 4, carryover 6.

8 * 6 = 48 + 6 = 54, write down 4, carryover 5.

8 * 5 = 40 + 5 = 45, write down 5, carryover 4.

8 * 9 = 72 + 4 = 76, write down 6, carryover 7.

8 * 8 = 64 + 7 = 71, write down 1, carryover 7.

8 * 8 = 64 + 7 = 71, write down 1, carryover 7.

8 * 1 = 8 + 7 = 15, write down 15.

Wait, this is getting complicated. Maybe I should use another approach. Alternatively, since I only need the units digit, maybe I can just track the units digit through each multiplication. Wait, but I already know that 18^5 ends with 8, so multiplying by 18 again, which ends with 8, so 8 * 8 = 64, so units digit 4. Therefore, 18^6 ends with 4.

But just to be thorough, let me compute the units digit of 1,889,568 * 18. The units digit of 1,889,568 is 8, and the units digit of 18 is 8. So 8 * 8 = 64, so the units digit is 4. Therefore, regardless of the rest of the number, the units digit is 4.

Therefore, all methods point to the units digit being 4.

Just to recap, the key idea here is that the units digit of a number raised to a power depends only on the units digit of the original number. So instead of dealing with 18^6, which is a huge number, I can focus on 8^6. Then, by recognizing the repeating pattern in the units digits of powers of 8 (which cycles every 4 exponents: 8, 4, 2, 6), I can determine that 8^6 has the same units digit as 8^(6 mod 4) = 8^2, which is 4.

Alternatively, using modular arithmetic, since we're only interested in the result modulo 10, we can compute 8^6 mod 10 step by step, which also gives 4.

Therefore, I can be confident that the units digit of 18^6 is 4.

**Final Answer**
The units digit of $18^6$ is \boxed{4}.

To determine the **units digit** of $ 18^6 $, we focus on the **last digit** of the number, which is influenced only by the **last digit of the base** (in this case, 8) and the **exponent**.

---

### Step 1: Understand the Pattern of Units Digits for Powers of 8

We observe that the units digit of powers of 8 follows a **cyclic pattern**:

$$
\begin{align*}
8^1 &= 8 \quad \text{(units digit: 8)} \\
8^2 &= 64 \quad \text{(units digit: 4)} \\
8^3 &= 512 \quad \text{(units digit: 2)} \\
8^4 &= 4096 \quad \text{(units digit: 6)} \\
8^5 &= 32768 \quad \text{(units digit: 8)} \\
8^6 &= 262144 \quad \text{(units digit: 4)}
\end{align*}
$$

So the **cycle of units digits** for powers of 8 is:
$$
8, 4, 2, 6
$$

This cycle repeats every **4** exponents.

---

### Step 2: Use the Pattern to Find the Units Digit of $ 8^6 $

Since the cycle length is 4, we can find the position of the exponent 6 in the cycle by computing:

$$
6 \mod 4 = 2
$$

This means the units digit of $ 8^6 $ is the **same as** the units digit of $ 8^2 $, which is **4**.

---

### Step 3: Confirm with Modular Arithmetic

We can also compute $ 8^6 \mod 10 $ directly:

$$
8^1 \mod 10 = 8 \\
8^2 \mod 10 = 64 \mod 10 = 4 \\
8^3 \mod 10 = 4 \cdot 8 = 32 \mod 10 = 2 \\
8^4 \mod 10 = 2 \cdot 8 = 16 \mod 10 = 6 \\
8^5 \mod 10 = 6 \cdot 8 = 48 \mod 10 = 8 \\
8^6 \mod 10 = 8 \cdot 8 = 64 \mod 10 = 4
$$

Again, we find that the units digit of $ 8^6 $ is **4**.

---

### Final Answer

Since the units digit of $ 18^6 $ is the same as the units digit of $ 8^6 $, we conclude:

$$
\boxed{4}
$$

[Assistant]
Explanation:
Okay, so I need to find the units digit of 18 raised to the 6th power. Hmm, let me think. The units digit is the last digit of the number, right? So, if I can figure out what 18^6 ends with, that's the answer.

First, maybe I should recall that when you raise a number to a power, the units digit depends only on the units digit of the original number. Wait, is that true? Let me check. For example, if I have 18, the units digit is 8. So maybe I can just focus on 8^6 and find its units digit? That might be easier. Let me verify this idea with a smaller exponent. Let's take 18^1, which is 18, units digit 8. 18^2 is 324, units digit 4. If I do 8^2, that's 64, which also ends with 4. Okay, so that seems to work. Similarly, 18^3 would be 18*18*18. Let me compute 18^3. 18*18 is 324, then 324*18. Let me do that multiplication. 324*10 is 3240, 324*8 is 2592, so total is 3240 + 2592 = 5832. The units digit is 2. If I do 8^3, that's 512, which also ends with 2. So yeah, it seems like the units digit of 18^n is the same as the units digit of 8^n. So I can just compute 8^6 and find its units digit. That simplifies things!

Alright, so let me compute 8^6. Let me do this step by step. Let's start with 8^1 = 8. 8^2 = 8*8 = 64. 8^3 = 8*64. Let me compute that. 8*60 is 480, 8*4 is 32, so 480 + 32 = 512. 8^3 is 512. Then 8^4 is 8*512. Let me calculate that. 512*8. Let me break it down: 500*8 = 4000, 12*8 = 96, so total is 4000 + 96 = 4096. So 8^4 is 4096. The units digit here is 6.

Moving on to 8^5. That's 8^4 * 8 = 4096 * 8. Let me compute that. 4000*8 = 32,000. 96*8 = 768. Adding them together: 32,000 + 768 = 32,768. So 8^5 is 32,768. The units digit is 8.

Then 8^6 is 8^5 * 8 = 32,768 * 8. Let me calculate that. Let me break it down again. 32,768 * 8. Let me do 30,000 * 8 = 240,000. 2,000 * 8 = 16,000. 700 * 8 = 5,600. 60 * 8 = 480. 8 * 8 = 64. Now add all those up.

Wait, maybe that's too tedious. Alternatively, since I only care about the units digit, maybe there's a pattern here. Let me think.

Earlier, I saw that 8^1 ends with 8, 8^2 ends with 4, 8^3 ends with 2, 8^4 ends with 6, 8^5 ends with 8 again. Wait, so there's a cycle here. Let me list them out:

- 8^1: 8
- 8^2: 4
- 8^3: 2
- 8^4: 6
- 8^5: 8
- 8^6: 4

So the pattern repeats every 4 exponents. The cycle is 8, 4, 2, 6, then back to 8, 4, etc. So the cycle length is 4. Therefore, if I can figure out where 6 falls in this cycle, I can find the units digit.

Since the cycle is 4 numbers long, I can divide the exponent by 4 and find the remainder. For exponent 6, divide 6 by 4. 4 goes into 6 once with a remainder of 2. So the remainder is 2. Then, looking at the cycle:

Remainder 1: 8

Remainder 2: 4

Remainder 3: 2

Remainder 0: 6 (since 4 divides into 4 with remainder 0, which corresponds to the 4th position in the cycle)

Wait, so if the remainder is 0, it's the 4th position. So for exponent 6, remainder 2, so the units digit is 4? Let me check with my previous calculation. Earlier, I calculated 8^5 ends with 8, then 8^6 would be 8^5 * 8, so 8*8=64, so units digit 4. Yes, that matches. So according to this pattern, the units digit of 8^6 is 4. Therefore, the units digit of 18^6 is also 4.

But let me confirm this by actually computing 18^6. Wait, that might be a big number, but maybe I can compute it step by step, keeping track of the units digit each time. Let me try that.

Starting with 18^1: units digit 8.

18^2: 18*18. Units digit is 8*8=64, so 4.

18^3: 18^2 * 18. Units digit is 4*8=32, so 2.

18^4: 18^3 * 18. Units digit is 2*8=16, so 6.

18^5: 18^4 * 18. Units digit is 6*8=48, so 8.

18^6: 18^5 * 18. Units digit is 8*8=64, so 4.

Yes! So that's the same result. So the units digit is 4. Therefore, the answer is 4.

Alternatively, another way to think about it is using modular arithmetic. Since we only care about the units digit, we can compute 18^6 mod 10.

But 18 mod 10 is 8, so 18^6 mod 10 is equivalent to 8^6 mod 10. Which is exactly what I did before. So 8^6 mod 10.

Alternatively, since 8 and 10 are not coprime, Euler's theorem might not apply directly, but since we have a small exponent, it's easier to compute the pattern.

But maybe I can use Euler's theorem here. Wait, Euler's theorem says that if a and n are coprime, then a^φ(n) ≡ 1 mod n. For n=10, φ(10)=φ(2*5)=φ(2)*φ(5)=1*4=4. So φ(10)=4. However, 8 and 10 are not coprime, since gcd(8,10)=2. Therefore, Euler's theorem doesn't apply here.

But maybe I can still use the concept of cycles. Since we saw that the units digit cycles every 4 exponents. So for 8^k mod 10, the cycle length is 4. Therefore, 8^(k) mod 10 = 8^(k mod 4) mod 10, but with adjustment if k mod 4 is 0.

So for exponent 6, 6 divided by 4 is 1 with remainder 2. Therefore, 8^6 mod 10 = 8^2 mod 10 = 64 mod 10 = 4. Which matches our previous results.

Alternatively, if I wanted to use modular exponentiation, I could compute 8^6 mod 10 step by step:

Compute 8^1 mod 10 = 8

8^2 mod 10 = (8*8) mod 10 = 64 mod 10 = 4

8^3 mod 10 = (8^2 * 8) mod 10 = (4 * 8) mod 10 = 32 mod 10 = 2

8^4 mod 10 = (8^3 * 8) mod 10 = (2 * 8) mod 10 = 16 mod 10 = 6

8^5 mod 10 = (8^4 * 8) mod 10 = (6 * 8) mod 10 = 48 mod 10 = 8

8^6 mod 10 = (8^5 * 8) mod 10 = (8 * 8) mod 10 = 64 mod 10 = 4

Same result again. So regardless of the method, the units digit is 4.

Just to make sure I didn't make a mistake in my reasoning, let me check with another approach. Let's compute 18^6 directly, but only keep track of the units digit at each multiplication step.

Starting with 18^1: units digit 8.

18^2: 18 * 18. Let's compute 18*18. As before, 18*10=180, 18*8=144, so total is 180+144=324. Units digit 4.

18^3: 324 * 18. Let's compute 324*18. Again, breaking it down: 324*10=3240, 324*8=2592. Adding them gives 3240+2592=5832. Units digit 2.

18^4: 5832 * 18. Let me compute 5832*18. Let me do 5832*10=58320, 5832*8=46656. Adding them: 58320 + 46656. Let me add these numbers. 58320 + 46656. Let me add 58320 + 40000 = 98320, then +6656 = 98320 + 6656. 98320 + 6000 = 104320, then +656 = 104,976. So 18^4 is 104,976. Units digit 6.

18^5: 104,976 * 18. Let me compute this. Again, breaking it down: 104,976 * 10 = 1,049,760. 104,976 * 8 = let's calculate that. 100,000 * 8 = 800,000. 4,976 * 8. Let me compute 4,000*8=32,000. 976*8. 900*8=7,200. 76*8=608. So 7,200 + 608 = 7,808. Then 32,000 + 7,808 = 39,808. So total 800,000 + 39,808 = 839,808. Therefore, 104,976 * 8 = 839,808. Now add that to 1,049,760.

So 1,049,760 + 839,808. Let me add them:

1,049,760 + 800,000 = 1,849,760

Then +39,808 = 1,849,760 + 39,808. Let me compute 1,849,760 + 30,000 = 1,879,760. Then +9,808 = 1,879,760 + 9,808 = 1,889,568. Therefore, 18^5 is 1,889,568. Units digit is 8.

Then 18^6: 1,889,568 * 18. Let me compute this. Again, breaking it down:

First, 1,889,568 * 10 = 18,895,680

Then 1,889,568 * 8. Let me compute that.

Let me do 1,889,568 * 8:

Break it down:

1,000,000 * 8 = 8,000,000

800,000 * 8 = 6,400,000

80,000 * 8 = 640,000

9,000 * 8 = 72,000

500 * 8 = 4,000

60 * 8 = 480

8 * 8 = 64

Wait, maybe that's too detailed. Alternatively, let me compute 1,889,568 * 8 step by step:

Starting from the right:

8 * 8 = 64, write down 4, carryover 6.

8 * 6 = 48 + 6 = 54, write down 4, carryover 5.

8 * 5 = 40 + 5 = 45, write down 5, carryover 4.

8 * 9 = 72 + 4 = 76, write down 6, carryover 7.

8 * 8 = 64 + 7 = 71, write down 1, carryover 7.

8 * 8 = 64 + 7 = 71, write down 1, carryover 7.

8 * 1 = 8 + 7 = 15, write down 15.

Wait, this is getting complicated. Maybe I should use another approach. Alternatively, since I only need the units digit, maybe I can just track the units digit through each multiplication. Wait, but I already know that 18^5 ends with 8, so multiplying by 18 again, which ends with 8, so 8 * 8 = 64, so units digit 4. Therefore, 18^6 ends with 4.

But just to be thorough, let me compute the units digit of 1,889,568 * 18. The units digit of 1,889,568 is 8, and the units digit of 18 is 8. So 8 * 8 = 64, so the units digit is 4. Therefore, regardless of the rest of the number, the units digit is 4.

Therefore, all methods point to the units digit being 4.

Just to recap, the key idea here is that the units digit of a number raised to a power depends only on the units digit of the original number. So instead of dealing with 18^6, which is a huge number, I can focus on 8^6. Then, by recognizing the repeating pattern in the units digits of powers of 8 (which cycles every 4 exponents: 8, 4, 2, 6), I can determine that 8^6 has the same units digit as 8^(6 mod 4) = 8^2, which is 4.

Alternatively, using modular arithmetic, since we're only interested in the result modulo 10, we can compute 8^6 mod 10 step by step, which also gives 4.

Therefore, I can be confident that the units digit of 18^6 is 4.

**Final Answer**
The units digit of $18^6$ is \boxed{4}.

To determine the **units digit** of $ 18^6 $, we focus on the **last digit** of the number, which is influenced only by the **last digit of the base** (in this case, 8) and the **exponent**.

---

### Step 1: Understand the Pattern of Units Digits for Powers of 8

We observe that the units digit of powers of 8 follows a **cyclic pattern**:

$$
\begin{align*}
8^1 &= 8 \quad \text{(units digit: 8)} \\
8^2 &= 64 \quad \text{(units digit: 4)} \\
8^3 &= 512 \quad \text{(units digit: 2)} \\
8^4 &= 4096 \quad \text{(units digit: 6)} \\
8^5 &= 32768 \quad \text{(units digit: 8)} \\
8^6 &= 262144 \quad \text{(units digit: 4)}
\end{align*}
$$

So the **cycle of units digits** for powers of 8 is:
$$
8, 4, 2, 6
$$

This cycle repeats every **4** exponents.

---

### Step 2: Use the Pattern to Find the Units Digit of $ 8^6 $

Since the cycle length is 4, we can find the position of the exponent 6 in the cycle by computing:

$$
6 \mod 4 = 2
$$

This means the units digit of $ 8^6 $ is the **same as** the units digit of $ 8^2 $, which is **4**.

---

### Step 3: Confirm with Modular Arithmetic

We can also compute $ 8^6 \mod 10 $ directly:

$$
8^1 \mod 10 = 8 \\
8^2 \mod 10 = 64 \mod 10 = 4 \\
8^3 \mod 10 = 4 \cdot 8 = 32 \mod 10 = 2 \\
8^4 \mod 10 = 2 \cdot 8 = 16 \mod 10 = 6 \\
8^5 \mod 10 = 6 \cdot 8 = 48 \mod 10 = 8 \\
8^6 \mod 10 = 8 \cdot 8 = 64 \mod 10 = 4
$$

Again, we find that the units digit of $ 8^6 $ is **4**.

---

### Final Answer

Since the units digit of $ 18^6 $ is the same as the units digit of $ 8^6 $, we conclude:

$$
\boxed{4}
$$

Solution: The units digit of $18^6$ is the same as in $8^6$.  There are several ways we could go about finding that units digit, but notice that $8^6 = 2^{18}$.  It's easy to find the pattern of units digits for powers of 2:  \begin{align*} 2^1 &= 2 \\ 2^2 &= 4 \\ 2^3 &= 8 \\ 2^4 &= 16 \\ 2^5 &= 32 \end{align*}Using this pattern, the units digit is found to be $\boxed{4}$.


Explanation:
Okay, so I need to figure out what kind of graph the equation $\left(\frac{x}{2} - 3\right)^2 + y^2 = 10$ represents. Let me start by recalling the standard forms of conic sections.

First, I remember that conic sections can be circles, ellipses, parabolas, hyperbolas, or even lines, points, or empty sets. The general idea is that the equation can be rewritten in a standard form to identify which one it is.

Looking at the given equation: $\left(\frac{x}{2} - 3\right)^2 + y^2 = 10$. Hmm, this looks similar to the standard form of a circle or an ellipse. Let me recall the standard forms.

A circle has the equation $(x - h)^2 + (y - k)^2 = r^2$, where $(h, k)$ is the center and $r$ is the radius. An ellipse has the equation $\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1$, where $a$ and $b$ are the semi-major and semi-minor axes.

But in the given equation, there's a $\frac{x}{2}$ term. Let me try to manipulate the equation to see if it can be written in a standard form.

First, let me rewrite the equation:

$$
\left(\frac{x}{2} - 3\right)^2 + y^2 = 10
$$

I notice that $\frac{x}{2} - 3$ can be written as $\frac{1}{2}(x - 6)$. Let me check that:

$$
\frac{1}{2}(x - 6) = \frac{x}{2} - \frac{6}{2} = \frac{x}{2} - 3
$$

Yes, that's correct. So substituting back into the equation:

$$
\left(\frac{1}{2}(x - 6)\right)^2 + y^2 = 10
$$

Now, squaring the $\frac{1}{2}$ term:

$$
\frac{1}{4}(x - 6)^2 + y^2 = 10
$$

Okay, so now the equation is $\frac{(x - 6)^2}{4} + y^2 = 10$. Wait, if I want to write this in the standard form of an ellipse, I need to have 1 on the right-hand side. Let me divide both sides by 10 to do that:

$$
\frac{(x - 6)^2}{4 \times 10} + \frac{y^2}{10} = 1
$$

Which simplifies to:

$$
\frac{(x - 6)^2}{40} + \frac{y^2}{10} = 1
$$

Now, comparing this with the standard form of an ellipse $\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1$, where $a$ and $b$ are the lengths of the semi-major and semi-minor axes.

In this case, the center of the ellipse is at $(h, k) = (6, 0)$. The denominators under the $x$ term is 40 and under the $y$ term is 10. Since 40 is larger than 10, the major axis is along the x-axis.

But wait, hold on. Let me check if this is indeed an ellipse. The standard form of an ellipse requires that both denominators are positive and the equation equals 1. Since both denominators are positive and we have 1 on the right-hand side, this is indeed an ellipse.

But let me make sure that I didn't make a mistake in my algebra. Let me go back through the steps again.

Original equation: $\left(\frac{x}{2} - 3\right)^2 + y^2 = 10$.

Expanding $\left(\frac{x}{2} - 3\right)^2$:

First, $\frac{x}{2} - 3$ squared is $\left(\frac{x}{2}\right)^2 - 2 \times \frac{x}{2} \times 3 + 3^2 = \frac{x^2}{4} - 3x + 9$.

So substituting back into the equation:

$$
\frac{x^2}{4} - 3x + 9 + y^2 = 10
$$

Simplify:

$$
\frac{x^2}{4} - 3x + y^2 + 9 - 10 = 0 \\
\frac{x^2}{4} - 3x + y^2 - 1 = 0
$$

Hmm, this seems different from the previous form. Wait, but earlier I had rewritten the original equation as $\frac{(x - 6)^2}{40} + \frac{y^2}{10} = 1$. Let me check if these are equivalent.

Let me take the equation $\frac{(x - 6)^2}{40} + \frac{y^2}{10} = 1$ and expand it:

First, expand $(x - 6)^2$:

$$
(x - 6)^2 = x^2 - 12x + 36
$$

So substituting back:

$$
\frac{x^2 - 12x + 36}{40} + \frac{y^2}{10} = 1 \\
\frac{x^2}{40} - \frac{12x}{40} + \frac{36}{40} + \frac{y^2}{10} = 1 \\
\frac{x^2}{40} - \frac{3x}{10} + \frac{9}{10} + \frac{y^2}{10} = 1
$$

Multiply all terms by 40 to eliminate denominators:

$$
x^2 - 12x + 36 + 4y^2 = 40 \\
x^2 - 12x + 4y^2 + 36 - 40 = 0 \\
x^2 - 12x + 4y^2 - 4 = 0
$$

Wait, but earlier when I expanded the original equation, I had:

$$
\frac{x^2}{4} - 3x + y^2 - 1 = 0
$$

Which is equivalent to multiplying by 4:

$$
x^2 - 12x + 4y^2 - 4 = 0
$$

Which matches the result from the other method. So that's consistent. Therefore, my previous steps are correct.

So, going back, the equation can be written as $\frac{(x - 6)^2}{40} + \frac{y^2}{10} = 1$, which is an ellipse.

But let me check if there's another way to interpret the original equation. For example, if there was a coefficient in front of the squared terms, but in this case, the equation is already in a form that resembles an ellipse.

Alternatively, if I consider the original equation $\left(\frac{x}{2} - 3\right)^2 + y^2 = 10$, maybe I can think of it as a circle that's been stretched or compressed.

A circle has the form $(x - h)^2 + (y - k)^2 = r^2$. If we have different coefficients in front of the squared terms, it becomes an ellipse.

In this case, the original equation is $\left(\frac{x}{2} - 3\right)^2 + y^2 = 10$. If I factor out the coefficients, it's similar to $\frac{(x - 6)^2}{4} + y^2 = 10$, which is the same as $\frac{(x - 6)^2}{40} + \frac{y^2}{10} = 1$ when divided by 10.

So, since both denominators are positive and the equation equals 1, this is indeed an ellipse.

But let me check if there's a possibility that this could be a circle. For a circle, the coefficients of the squared terms must be equal. In this case, the coefficient of $(x - 6)^2$ is 1/4 and the coefficient of $y^2$ is 1. Since they are not equal, it's not a circle. Therefore, it's an ellipse.

Alternatively, if I think about the original equation, if I had $\left(\frac{x}{2} - 3\right)^2 + y^2 = 10$, maybe I can think of it as a circle with radius $\sqrt{10}$, but stretched horizontally by a factor of 2.

Wait, stretching a circle horizontally by a factor of 2 would result in an ellipse. So, yes, that's another way to think about it.

Alternatively, if I consider the equation $\left(\frac{x}{2} - 3\right)^2 + y^2 = 10$, if I let $u = \frac{x}{2} - 3$, then the equation becomes $u^2 + y^2 = 10$, which is a circle with radius $\sqrt{10}$ in the $u$-$y$ plane. But since $u = \frac{x}{2} - 3$, this is a horizontal scaling and shifting of the coordinate system.

Therefore, the original equation is a circle in the $u$-$y$ plane, which translates back to the $x$-$y$ plane as an ellipse.

So, all these methods lead me to the conclusion that the graph is an ellipse.

But let me check if there's any other possibility. For example, could this equation represent a hyperbola? A hyperbola has the form $\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1$ or similar. But in our case, both terms are added, so it's not a hyperbola.

What about a parabola? A parabola has one squared term and a linear term. But here, both terms are squared, so it's not a parabola.

What about a line or two lines? If the equation could be factored into linear terms, but since both terms are squared and added, it's not a line or two lines.

A point would be if the equation was something like $(x - h)^2 + (y - k)^2 = 0$, which only has one solution. But here, the right-hand side is 10, which is positive, so it's not a point.

An empty set would be if the equation had a negative number on the right-hand side, but since 10 is positive, it's not empty.

Therefore, the only possibility left is an ellipse.

Wait, but let me check if the equation could be a circle. If I have $\left(\frac{x}{2} - 3\right)^2 + y^2 = 10$, if I wanted to write it as a circle, I would need to have the same coefficient in front of both squared terms. For example, if it was $\left(\frac{x}{2} - 3\right)^2 + \left(\frac{y}{1}\right)^2 = 10$, but since the coefficients are different, it's not a circle.

Alternatively, if I had $\left(\frac{x}{2} - 3\right)^2 + y^2 = 10$, and I wanted to write it as a circle, I would need to adjust the equation. For example, if I multiplied both sides by 4, I would get $(x - 6)^2 + 4y^2 = 40$, which is the same as $\frac{(x - 6)^2}{40} + \frac{y^2}{10} = 1$, which is an ellipse.

Therefore, it's definitely an ellipse.

Another way to check is to consider the general form of a conic section. The general second-degree equation is $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$.

Let me write the original equation in this form.

Original equation: $\left(\frac{x}{2} - 3\right)^2 + y^2 = 10$.

Expanding this:

First, expand $\left(\frac{x}{2} - 3\right)^2$:

$$
\frac{x^2}{4} - 3x + 9
$$

So the equation becomes:

$$
\frac{x^2}{4} - 3x + 9 + y^2 = 10
$$

Subtract 10 from both sides:

$$
\frac{x^2}{4} - 3x + y^2 - 1 = 0
$$

Multiply all terms by 4 to eliminate the fraction:

$$
x^2 - 12x + 4y^2 - 4 = 0
$$

So, in the general form, this is:

$A x^2 + B xy + C y^2 + D x + E y + F = 0$, where:

- $A = 1$
- $B = 0$ (since there's no $xy$ term)
- $C = 4$
- $D = -12$
- $E = 0$
- $F = -4$

Now, to determine the type of conic, we can use the discriminant $B^2 - 4AC$.

Calculating discriminant:

$B^2 - 4AC = 0^2 - 4(1)(4) = -16$

Since the discriminant is negative ($-16 < 0$), the conic is an ellipse or a circle.

But since $A$ and $C$ are not equal (1 ≠ 4), it's not a circle, so it must be an ellipse.

Therefore, this confirms that the equation represents an ellipse.

So, after going through multiple methods—rewriting into standard form, checking the discriminant, and considering the general form—it's clear that the graph is an ellipse.

I don't see any mistakes in my reasoning. All steps lead to the conclusion that it's an ellipse.

**Final Answer**
The graph of the equation is an \boxed{ellipse}.

To determine the type of graph represented by the equation:

$$
\left(\frac{x}{2} - 3\right)^2 + y^2 = 10
$$

we begin by analyzing its structure and comparing it to the standard forms of conic sections.

---

### Step 1: Recognize the Form

The equation contains two squared terms, one involving $ x $ and the other involving $ y $, and the right-hand side is a constant. This suggests the possibility of a **circle** or **ellipse**.

Let’s rewrite the equation in a more recognizable form.

---

### Step 2: Simplify the Equation

We start by rewriting the $ x $-term:

$$
\left(\frac{x}{2} - 3\right)^2 = \left(\frac{1}{2}(x - 6)\right)^2 = \frac{1}{4}(x - 6)^2
$$

So the equation becomes:

$$
\frac{1}{4}(x - 6)^2 + y^2 = 10
$$

To match the standard form of an ellipse, we divide both sides by 10:

$$
\frac{(x - 6)^2}{40} + \frac{y^2}{10} = 1
$$

---

### Step 3: Identify the Conic Section

This is now in the standard form of an **ellipse**:

$$
\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1
$$

Where:
- $ h = 6 $, $ k = 0 $ (center of the ellipse)
- $ a^2 = 40 $, $ b^2 = 10 $

Since $ a^2 > b^2 $, the major axis is along the $ x $-axis.

---

### Step 4: Confirm with the General Form

We can also write the equation in the general second-degree form:

$$
Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0
$$

Starting from the original equation:

$$
\left(\frac{x}{2} - 3\right)^2 + y^2 = 10
$$

Expanding:

$$
\frac{x^2}{4} - 3x + 9 + y^2 = 10
$$

Simplify:

$$
\frac{x^2}{4} - 3x + y^2 - 1 = 0
$$

Multiply through by 4 to eliminate the fraction:

$$
x^2 - 12x + 4y^2 - 4 = 0
$$

Now, in the general form:

- $ A = 1 $
- $ B = 0 $
- $ C = 4 $
- $ D = -12 $
- $ E = 0 $
- $ F = -4 $

Compute the **discriminant**:

$$
B^2 - 4AC = 0^2 - 4(1)(4) = -16
$$

Since the discriminant is **negative**, the conic is either an **ellipse** or a **circle**. Since $ A \neq C $, it is **not a circle**, so it must be an **ellipse**.

---

### Final Conclusion

All methods—standard form, general form, and discriminant—confirm that the given equation represents an **ellipse**.

$$
\boxed{ellipse}
$$

[Assistant]
Explanation:
Okay, so I need to figure out what kind of graph the equation $\left(\frac{x}{2} - 3\right)^2 + y^2 = 10$ represents. Let me start by recalling the standard forms of conic sections.

First, I remember that conic sections can be circles, ellipses, parabolas, hyperbolas, or even lines, points, or empty sets. The general idea is that the equation can be rewritten in a standard form to identify which one it is.

Looking at the given equation: $\left(\frac{x}{2} - 3\right)^2 + y^2 = 10$. Hmm, this looks similar to the standard form of a circle or an ellipse. Let me recall the standard forms.

A circle has the equation $(x - h)^2 + (y - k)^2 = r^2$, where $(h, k)$ is the center and $r$ is the radius. An ellipse has the equation $\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1$, where $a$ and $b$ are the semi-major and semi-minor axes.

But in the given equation, there's a $\frac{x}{2}$ term. Let me try to manipulate the equation to see if it can be written in a standard form.

First, let me rewrite the equation:

$$
\left(\frac{x}{2} - 3\right)^2 + y^2 = 10
$$

I notice that $\frac{x}{2} - 3$ can be written as $\frac{1}{2}(x - 6)$. Let me check that:

$$
\frac{1}{2}(x - 6) = \frac{x}{2} - \frac{6}{2} = \frac{x}{2} - 3
$$

Yes, that's correct. So substituting back into the equation:

$$
\left(\frac{1}{2}(x - 6)\right)^2 + y^2 = 10
$$

Now, squaring the $\frac{1}{2}$ term:

$$
\frac{1}{4}(x - 6)^2 + y^2 = 10
$$

Okay, so now the equation is $\frac{(x - 6)^2}{4} + y^2 = 10$. Wait, if I want to write this in the standard form of an ellipse, I need to have 1 on the right-hand side. Let me divide both sides by 10 to do that:

$$
\frac{(x - 6)^2}{4 \times 10} + \frac{y^2}{10} = 1
$$

Which simplifies to:

$$
\frac{(x - 6)^2}{40} + \frac{y^2}{10} = 1
$$

Now, comparing this with the standard form of an ellipse $\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1$, where $a$ and $b$ are the lengths of the semi-major and semi-minor axes.

In this case, the center of the ellipse is at $(h, k) = (6, 0)$. The denominators under the $x$ term is 40 and under the $y$ term is 10. Since 40 is larger than 10, the major axis is along the x-axis.

But wait, hold on. Let me check if this is indeed an ellipse. The standard form of an ellipse requires that both denominators are positive and the equation equals 1. Since both denominators are positive and we have 1 on the right-hand side, this is indeed an ellipse.

But let me make sure that I didn't make a mistake in my algebra. Let me go back through the steps again.

Original equation: $\left(\frac{x}{2} - 3\right)^2 + y^2 = 10$.

Expanding $\left(\frac{x}{2} - 3\right)^2$:

First, $\frac{x}{2} - 3$ squared is $\left(\frac{x}{2}\right)^2 - 2 \times \frac{x}{2} \times 3 + 3^2 = \frac{x^2}{4} - 3x + 9$.

So substituting back into the equation:

$$
\frac{x^2}{4} - 3x + 9 + y^2 = 10
$$

Simplify:

$$
\frac{x^2}{4} - 3x + y^2 + 9 - 10 = 0 \\
\frac{x^2}{4} - 3x + y^2 - 1 = 0
$$

Hmm, this seems different from the previous form. Wait, but earlier I had rewritten the original equation as $\frac{(x - 6)^2}{40} + \frac{y^2}{10} = 1$. Let me check if these are equivalent.

Let me take the equation $\frac{(x - 6)^2}{40} + \frac{y^2}{10} = 1$ and expand it:

First, expand $(x - 6)^2$:

$$
(x - 6)^2 = x^2 - 12x + 36
$$

So substituting back:

$$
\frac{x^2 - 12x + 36}{40} + \frac{y^2}{10} = 1 \\
\frac{x^2}{40} - \frac{12x}{40} + \frac{36}{40} + \frac{y^2}{10} = 1 \\
\frac{x^2}{40} - \frac{3x}{10} + \frac{9}{10} + \frac{y^2}{10} = 1
$$

Multiply all terms by 40 to eliminate denominators:

$$
x^2 - 12x + 36 + 4y^2 = 40 \\
x^2 - 12x + 4y^2 + 36 - 40 = 0 \\
x^2 - 12x + 4y^2 - 4 = 0
$$

Wait, but earlier when I expanded the original equation, I had:

$$
\frac{x^2}{4} - 3x + y^2 - 1 = 0
$$

Which is equivalent to multiplying by 4:

$$
x^2 - 12x + 4y^2 - 4 = 0
$$

Which matches the result from the other method. So that's consistent. Therefore, my previous steps are correct.

So, going back, the equation can be written as $\frac{(x - 6)^2}{40} + \frac{y^2}{10} = 1$, which is an ellipse.

But let me check if there's another way to interpret the original equation. For example, if there was a coefficient in front of the squared terms, but in this case, the equation is already in a form that resembles an ellipse.

Alternatively, if I consider the original equation $\left(\frac{x}{2} - 3\right)^2 + y^2 = 10$, maybe I can think of it as a circle that's been stretched or compressed.

A circle has the form $(x - h)^2 + (y - k)^2 = r^2$. If we have different coefficients in front of the squared terms, it becomes an ellipse.

In this case, the original equation is $\left(\frac{x}{2} - 3\right)^2 + y^2 = 10$. If I factor out the coefficients, it's similar to $\frac{(x - 6)^2}{4} + y^2 = 10$, which is the same as $\frac{(x - 6)^2}{40} + \frac{y^2}{10} = 1$ when divided by 10.

So, since both denominators are positive and the equation equals 1, this is indeed an ellipse.

But let me check if there's a possibility that this could be a circle. For a circle, the coefficients of the squared terms must be equal. In this case, the coefficient of $(x - 6)^2$ is 1/4 and the coefficient of $y^2$ is 1. Since they are not equal, it's not a circle. Therefore, it's an ellipse.

Alternatively, if I think about the original equation, if I had $\left(\frac{x}{2} - 3\right)^2 + y^2 = 10$, maybe I can think of it as a circle with radius $\sqrt{10}$, but stretched horizontally by a factor of 2.

Wait, stretching a circle horizontally by a factor of 2 would result in an ellipse. So, yes, that's another way to think about it.

Alternatively, if I consider the equation $\left(\frac{x}{2} - 3\right)^2 + y^2 = 10$, if I let $u = \frac{x}{2} - 3$, then the equation becomes $u^2 + y^2 = 10$, which is a circle with radius $\sqrt{10}$ in the $u$-$y$ plane. But since $u = \frac{x}{2} - 3$, this is a horizontal scaling and shifting of the coordinate system.

Therefore, the original equation is a circle in the $u$-$y$ plane, which translates back to the $x$-$y$ plane as an ellipse.

So, all these methods lead me to the conclusion that the graph is an ellipse.

But let me check if there's any other possibility. For example, could this equation represent a hyperbola? A hyperbola has the form $\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1$ or similar. But in our case, both terms are added, so it's not a hyperbola.

What about a parabola? A parabola has one squared term and a linear term. But here, both terms are squared, so it's not a parabola.

What about a line or two lines? If the equation could be factored into linear terms, but since both terms are squared and added, it's not a line or two lines.

A point would be if the equation was something like $(x - h)^2 + (y - k)^2 = 0$, which only has one solution. But here, the right-hand side is 10, which is positive, so it's not a point.

An empty set would be if the equation had a negative number on the right-hand side, but since 10 is positive, it's not empty.

Therefore, the only possibility left is an ellipse.

Wait, but let me check if the equation could be a circle. If I have $\left(\frac{x}{2} - 3\right)^2 + y^2 = 10$, if I wanted to write it as a circle, I would need to have the same coefficient in front of both squared terms. For example, if it was $\left(\frac{x}{2} - 3\right)^2 + \left(\frac{y}{1}\right)^2 = 10$, but since the coefficients are different, it's not a circle.

Alternatively, if I had $\left(\frac{x}{2} - 3\right)^2 + y^2 = 10$, and I wanted to write it as a circle, I would need to adjust the equation. For example, if I multiplied both sides by 4, I would get $(x - 6)^2 + 4y^2 = 40$, which is the same as $\frac{(x - 6)^2}{40} + \frac{y^2}{10} = 1$, which is an ellipse.

Therefore, it's definitely an ellipse.

Another way to check is to consider the general form of a conic section. The general second-degree equation is $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$.

Let me write the original equation in this form.

Original equation: $\left(\frac{x}{2} - 3\right)^2 + y^2 = 10$.

Expanding this:

First, expand $\left(\frac{x}{2} - 3\right)^2$:

$$
\frac{x^2}{4} - 3x + 9
$$

So the equation becomes:

$$
\frac{x^2}{4} - 3x + 9 + y^2 = 10
$$

Subtract 10 from both sides:

$$
\frac{x^2}{4} - 3x + y^2 - 1 = 0
$$

Multiply all terms by 4 to eliminate the fraction:

$$
x^2 - 12x + 4y^2 - 4 = 0
$$

So, in the general form, this is:

$A x^2 + B xy + C y^2 + D x + E y + F = 0$, where:

- $A = 1$
- $B = 0$ (since there's no $xy$ term)
- $C = 4$
- $D = -12$
- $E = 0$
- $F = -4$

Now, to determine the type of conic, we can use the discriminant $B^2 - 4AC$.

Calculating discriminant:

$B^2 - 4AC = 0^2 - 4(1)(4) = -16$

Since the discriminant is negative ($-16 < 0$), the conic is an ellipse or a circle.

But since $A$ and $C$ are not equal (1 ≠ 4), it's not a circle, so it must be an ellipse.

Therefore, this confirms that the equation represents an ellipse.

So, after going through multiple methods—rewriting into standard form, checking the discriminant, and considering the general form—it's clear that the graph is an ellipse.

I don't see any mistakes in my reasoning. All steps lead to the conclusion that it's an ellipse.

**Final Answer**
The graph of the equation is an \boxed{ellipse}.

To determine the type of graph represented by the equation:

$$
\left(\frac{x}{2} - 3\right)^2 + y^2 = 10
$$

we begin by analyzing its structure and comparing it to the standard forms of conic sections.

---

### Step 1: Recognize the Form

The equation contains two squared terms, one involving $ x $ and the other involving $ y $, and the right-hand side is a constant. This suggests the possibility of a **circle** or **ellipse**.

Let’s rewrite the equation in a more recognizable form.

---

### Step 2: Simplify the Equation

We start by rewriting the $ x $-term:

$$
\left(\frac{x}{2} - 3\right)^2 = \left(\frac{1}{2}(x - 6)\right)^2 = \frac{1}{4}(x - 6)^2
$$

So the equation becomes:

$$
\frac{1}{4}(x - 6)^2 + y^2 = 10
$$

To match the standard form of an ellipse, we divide both sides by 10:

$$
\frac{(x - 6)^2}{40} + \frac{y^2}{10} = 1
$$

---

### Step 3: Identify the Conic Section

This is now in the standard form of an **ellipse**:

$$
\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1
$$

Where:
- $ h = 6 $, $ k = 0 $ (center of the ellipse)
- $ a^2 = 40 $, $ b^2 = 10 $

Since $ a^2 > b^2 $, the major axis is along the $ x $-axis.

---

### Step 4: Confirm with the General Form

We can also write the equation in the general second-degree form:

$$
Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0
$$

Starting from the original equation:

$$
\left(\frac{x}{2} - 3\right)^2 + y^2 = 10
$$

Expanding:

$$
\frac{x^2}{4} - 3x + 9 + y^2 = 10
$$

Simplify:

$$
\frac{x^2}{4} - 3x + y^2 - 1 = 0
$$

Multiply through by 4 to eliminate the fraction:

$$
x^2 - 12x + 4y^2 - 4 = 0
$$

Now, in the general form:

- $ A = 1 $
- $ B = 0 $
- $ C = 4 $
- $ D = -12 $
- $ E = 0 $
- $ F = -4 $

Compute the **discriminant**:

$$
B^2 - 4AC = 0^2 - 4(1)(4) = -16
$$

Since the discriminant is **negative**, the conic is either an **ellipse** or a **circle**. Since $ A \neq C $, it is **not a circle**, so it must be an **ellipse**.

---

### Final Conclusion

All methods—standard form, general form, and discriminant—confirm that the given equation represents an **ellipse**.

$$
\boxed{ellipse}
$$

Solution: This looks like the equation of a circle, but we have replaced $x$ with $\frac x2$. So, we suspect this equation defines an $\boxed{\text{ellipse}}$. To verify this we write \[\left(\frac x2 - 3\right)^2 = \frac 14 \left( x - 6\right)^2,\]and we see that the equation \[ \frac{\left(x - 6 \right)^2}{4} + y^2 = 10 \]is the equation of an ellipse.


Explanation:
Okay, so I need to solve this equation: Three plus the reciprocal of a number equals 7 divided by that number. Let me try to write that down first. Let me call the number x. So, the equation would be 3 plus 1 over x equals 7 over x. Let me check that again. "Three plus the reciprocal of a number" – reciprocal of x is 1/x, so 3 + 1/x. And that equals 7 divided by that number, which is 7/x. So the equation is 3 + (1/x) = 7/x.

Alright, so I need to solve for x. Let me think about how to approach this. Since there are fractions here, maybe I can eliminate the denominators by multiplying both sides by x. That might make it easier. Let me try that. If I multiply both sides by x, then on the left side, I have x times 3 plus x times (1/x). On the right side, x times (7/x). Let me compute each term.

Left side: 3 times x is 3x, and then 1/x times x is 1. So altogether, left side becomes 3x + 1. Right side: 7/x times x is 7. So the equation simplifies to 3x + 1 = 7.

Okay, that seems simpler. Now I can solve for x. Subtract 1 from both sides: 3x = 7 - 1, which is 6. Then divide both sides by 3: x = 6 / 3 = 2. So x equals 2. Let me check if that works in the original equation.

Original equation: 3 + (1/x) = 7/x. Let me plug in x = 2. Left side: 3 + 1/2 = 3.5. Right side: 7/2 = 3.5. Yep, they are equal. So x = 2 is the solution.

Wait, but let me make sure I didn't make any mistakes in my steps. Let me go through again. Starting with 3 + 1/x = 7/x. Multiply both sides by x: 3x + 1 = 7. Then 3x = 6, so x = 2. Seems straightforward.

But maybe I should consider if there are any restrictions on x? Since there's a reciprocal, x can't be zero. So x ≠ 0. But in this case, our solution is x = 2, which is fine. So no issues there.

Alternatively, maybe I can solve it another way to verify. Let me try rearranging the original equation. Let's subtract 7/x from both sides to get 3 + 1/x - 7/x = 0. Combine like terms: 3 + (1 - 7)/x = 0. That simplifies to 3 - 6/x = 0. Then, 3 = 6/x. Multiply both sides by x: 3x = 6. Then x = 2. Same result.

Another way: Let me write the original equation again: 3 + 1/x = 7/x. Let me subtract 3 from both sides: 1/x = 7/x - 3. Then, let me combine the terms on the right side. To do that, maybe write 3 as 3x/x. So 7/x - 3x/x = (7 - 3x)/x. Therefore, 1/x = (7 - 3x)/x. Multiply both sides by x: 1 = 7 - 3x. Then, subtract 7: 1 - 7 = -3x => -6 = -3x => x = (-6)/(-3) = 2. Same answer again.

So all methods lead to x = 2. Therefore, I think that's the correct solution.

But just to be thorough, let me check if there's any other possible solution or if I missed something. For example, sometimes when you multiply both sides by a variable, you might introduce extraneous solutions, but in this case, since we only had one solution and it checked out in the original equation, there's no problem.

Alternatively, maybe there's a case where x is negative? Let me see. Suppose x is negative. Let me try plugging in x = -2. Left side: 3 + 1/(-2) = 3 - 0.5 = 2.5. Right side: 7/(-2) = -3.5. Not equal. So that doesn't work. How about x = 1? Left side: 3 + 1/1 = 4. Right side: 7/1 = 7. Not equal. x = 3? Left side: 3 + 1/3 ≈ 3.333. Right side: 7/3 ≈ 2.333. Not equal. So no, only x = 2 works.

Therefore, I think the answer is 2.

**Final Answer**
The number is \boxed{2}.

To solve the equation:

$$
3 + \frac{1}{x} = \frac{7}{x}
$$

we begin by identifying the variable and simplifying the expression. Let the unknown number be $ x $, and note that $ x \neq 0 $, since division by zero is undefined.

---

### Step 1: Eliminate the Denominators

To simplify the equation, we can eliminate the denominators by multiplying both sides of the equation by $ x $, assuming $ x \neq 0 $:

$$
x \left(3 + \frac{1}{x}\right) = x \cdot \frac{7}{x}
$$

Distribute $ x $ on the left-hand side:

$$
3x + 1 = 7
$$

---

### Step 2: Solve the Simplified Equation

Now solve for $ x $:

$$
3x + 1 = 7
$$

Subtract 1 from both sides:

$$
3x = 6
$$

Divide both sides by 3:

$$
x = 2
$$

---

### Step 3: Verify the Solution

Plug $ x = 2 $ back into the original equation to confirm:

- Left-hand side: $ 3 + \frac{1}{2} = 3.5 $
- Right-hand side: $ \frac{7}{2} = 3.5 $

Both sides are equal, so the solution is valid.

---

### Step 4: Consider Restrictions

Since the original equation involves a reciprocal $ \frac{1}{x} $, we must ensure $ x \neq 0 $. Our solution $ x = 2 $ satisfies this condition.

---

### Final Answer

$$
\boxed{2}
$$

[Assistant]
Explanation:
Okay, so I need to solve this equation: Three plus the reciprocal of a number equals 7 divided by that number. Let me try to write that down first. Let me call the number x. So, the equation would be 3 plus 1 over x equals 7 over x. Let me check that again. "Three plus the reciprocal of a number" – reciprocal of x is 1/x, so 3 + 1/x. And that equals 7 divided by that number, which is 7/x. So the equation is 3 + (1/x) = 7/x.

Alright, so I need to solve for x. Let me think about how to approach this. Since there are fractions here, maybe I can eliminate the denominators by multiplying both sides by x. That might make it easier. Let me try that. If I multiply both sides by x, then on the left side, I have x times 3 plus x times (1/x). On the right side, x times (7/x). Let me compute each term.

Left side: 3 times x is 3x, and then 1/x times x is 1. So altogether, left side becomes 3x + 1. Right side: 7/x times x is 7. So the equation simplifies to 3x + 1 = 7.

Okay, that seems simpler. Now I can solve for x. Subtract 1 from both sides: 3x = 7 - 1, which is 6. Then divide both sides by 3: x = 6 / 3 = 2. So x equals 2. Let me check if that works in the original equation.

Original equation: 3 + (1/x) = 7/x. Let me plug in x = 2. Left side: 3 + 1/2 = 3.5. Right side: 7/2 = 3.5. Yep, they are equal. So x = 2 is the solution.

Wait, but let me make sure I didn't make any mistakes in my steps. Let me go through again. Starting with 3 + 1/x = 7/x. Multiply both sides by x: 3x + 1 = 7. Then 3x = 6, so x = 2. Seems straightforward.

But maybe I should consider if there are any restrictions on x? Since there's a reciprocal, x can't be zero. So x ≠ 0. But in this case, our solution is x = 2, which is fine. So no issues there.

Alternatively, maybe I can solve it another way to verify. Let me try rearranging the original equation. Let's subtract 7/x from both sides to get 3 + 1/x - 7/x = 0. Combine like terms: 3 + (1 - 7)/x = 0. That simplifies to 3 - 6/x = 0. Then, 3 = 6/x. Multiply both sides by x: 3x = 6. Then x = 2. Same result.

Another way: Let me write the original equation again: 3 + 1/x = 7/x. Let me subtract 3 from both sides: 1/x = 7/x - 3. Then, let me combine the terms on the right side. To do that, maybe write 3 as 3x/x. So 7/x - 3x/x = (7 - 3x)/x. Therefore, 1/x = (7 - 3x)/x. Multiply both sides by x: 1 = 7 - 3x. Then, subtract 7: 1 - 7 = -3x => -6 = -3x => x = (-6)/(-3) = 2. Same answer again.

So all methods lead to x = 2. Therefore, I think that's the correct solution.

But just to be thorough, let me check if there's any other possible solution or if I missed something. For example, sometimes when you multiply both sides by a variable, you might introduce extraneous solutions, but in this case, since we only had one solution and it checked out in the original equation, there's no problem.

Alternatively, maybe there's a case where x is negative? Let me see. Suppose x is negative. Let me try plugging in x = -2. Left side: 3 + 1/(-2) = 3 - 0.5 = 2.5. Right side: 7/(-2) = -3.5. Not equal. So that doesn't work. How about x = 1? Left side: 3 + 1/1 = 4. Right side: 7/1 = 7. Not equal. x = 3? Left side: 3 + 1/3 ≈ 3.333. Right side: 7/3 ≈ 2.333. Not equal. So no, only x = 2 works.

Therefore, I think the answer is 2.

**Final Answer**
The number is \boxed{2}.

To solve the equation:

$$
3 + \frac{1}{x} = \frac{7}{x}
$$

we begin by identifying the variable and simplifying the expression. Let the unknown number be $ x $, and note that $ x \neq 0 $, since division by zero is undefined.

---

### Step 1: Eliminate the Denominators

To simplify the equation, we can eliminate the denominators by multiplying both sides of the equation by $ x $, assuming $ x \neq 0 $:

$$
x \left(3 + \frac{1}{x}\right) = x \cdot \frac{7}{x}
$$

Distribute $ x $ on the left-hand side:

$$
3x + 1 = 7
$$

---

### Step 2: Solve the Simplified Equation

Now solve for $ x $:

$$
3x + 1 = 7
$$

Subtract 1 from both sides:

$$
3x = 6
$$

Divide both sides by 3:

$$
x = 2
$$

---

### Step 3: Verify the Solution

Plug $ x = 2 $ back into the original equation to confirm:

- Left-hand side: $ 3 + \frac{1}{2} = 3.5 $
- Right-hand side: $ \frac{7}{2} = 3.5 $

Both sides are equal, so the solution is valid.

---

### Step 4: Consider Restrictions

Since the original equation involves a reciprocal $ \frac{1}{x} $, we must ensure $ x \neq 0 $. Our solution $ x = 2 $ satisfies this condition.

---

### Final Answer

$$
\boxed{2}
$$

Solution: Let $x$ be the number.  Converting the words in the problem into an equation gives us $3+\dfrac{1}{x} = \dfrac{7}{x}$.  Subtracting $\dfrac{1}{x}$ from both sides gives $3 = \dfrac{6}{x}$. Multiplying both sides of this equation by $x$ gives $3x =6$, and dividing both sides of this equation by 3 gives $x = \boxed{2}$.