Explanation:
Okay, so I need to compute the binomial coefficient $\dbinom{31}{28}$. Hmm, let me remember how binomial coefficients work. From what I recall, the formula for $\dbinom{n}{k}$ is $\frac{n!}{k!(n - k)!}$. Right? So, in this case, $n$ is 31 and $k$ is 28. Let me write that down:

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

Wait, so simplifying the denominator first might make this easier. Let me check if there's a way to simplify the factorials before calculating them all out. Because calculating 31! directly would be a huge number, and it might not be necessary. Let me think.

Since $31!$ divided by $28!$ would leave me with $31 \times 30 \times 29 \times 28! / 28!$, right? So the $28!$ cancels out. That simplifies the equation to:

$$
\frac{31 \times 30 \times 29}{3!}
$$

Okay, so then $3!$ is 6. So now the problem becomes $\frac{31 \times 30 \times 29}{6}$. Let me compute the numerator first.

Calculating $31 \times 30$. Let me do that step by step. 31 times 30 is like 31 times 3 times 10. 31 times 3 is 93, so 93 times 10 is 930. So that's 930. Then multiply by 29. Hmm, 930 times 29. Let me break that down.

29 can be thought of as 30 - 1. So 930 times 30 is 27,900, and then subtract 930. So 27,900 - 930 = 26,970. Wait, is that right? Let me check again:

First, 930 * 29:

Multiply 930 by 20: 18,600

Multiply 930 by 9: 8,370

Add them together: 18,600 + 8,370. Let's add 18,600 + 8,000 = 26,600, then +370 = 26,970. Yeah, same result. So the numerator is 26,970. Now divide that by 6.

So 26,970 divided by 6. Let me do that division step by step. Let me see how many times 6 goes into 26,970.

First, divide 26,970 by 6. Maybe I can break this down. Let me think of 26,970 divided by 6. Let me divide 26,970 by 6.

Well, 6 times 4,000 is 24,000. Subtract 24,000 from 26,970, that leaves 2,970. Then, 6 times 400 is 2,400. Subtract that from 2,970: 2,970 - 2,400 = 570. Then 6 times 95 is 570. So total is 4,000 + 400 + 95 = 4,495. Wait, let me check:

6 * 4,495 = ?

Well 4,495 * 6:

First, 4,000 * 6 = 24,000

400 * 6 = 2,400

95 * 6 = 570

So 24,000 + 2,400 is 26,400, plus 570 is 26,970. Yes, correct! So 26,970 divided by 6 is indeed 4,495. Therefore, $\dbinom{31}{28}$ equals 4,495.

But wait, let me make sure I didn't make a mistake in simplifying earlier. Let me verify the steps again. Original calculation:

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

Yes, that seems correct. So then computing numerator as 31 * 30 * 29. Which is 31 * 30 = 930, then 930 * 29 = 26,970. Divided by 6 gives 4,495.

Alternatively, I can check if there's another way to compute the binomial coefficient. For instance, sometimes it's easier to compute $\dbinom{n}{k}$ as $\dbinom{n}{n - k}$. Since $\dbinom{31}{28} = \dbinom{31}{3}$, right? Because 31 - 28 = 3. That might be easier to compute. Let me try that way to verify.

So $\dbinom{31}{3}$ is $\frac{31 \times 30 \times 29}{3 \times 2 \times 1}$ which is the same as above. So that's 31*30*29 / 6, which is exactly what I calculated. So that's the same result. Therefore, 4,495 should be the correct answer. But let me confirm with another method just to be absolutely sure.

Alternatively, I can compute the value step by step. Let me compute 31 divided by 3 first? Wait, but it's (31*30*29)/6. Maybe simplifying fractions before multiplying. Let me try that.

So 31*30*29 divided by 6. Let me note that 30 divided by 6 is 5. So maybe rewrite the equation as 31 * (30 / 6) * 29. Which is 31 * 5 * 29. Then compute that. 31 * 5 is 155, and then 155 * 29. Let me check this way.

So 155 * 29. Let me calculate that. 155 * 30 would be 4,650, so subtract 155 to get 155 * 29. So 4,650 - 155 = 4,495. Yep, same answer. So that's another way to compute it: simplifying the division before multiplying.

So that gives me more confidence. So whether I do the numerator first and then divide or simplify the fraction first, I get 4,495. Therefore, I think that is the correct value.

Wait, but just to be thorough, maybe I can check with smaller numbers. Let me test if the formula works for a smaller binomial coefficient. For example, $\dbinom{5}{2}$ should be 10. Let me use the same method. $\frac{5!}{2!3!} = \frac{5*4*3!}{2!3!} = \frac{5*4}{2} = 10$. Correct. Similarly, if I use the other approach, $\dbinom{5}{3} = \frac{5*4*3}{3*2*1} = \frac{60}{6} = 10. Same result.

So that seems to check out. Therefore, my method is valid. Therefore, my answer of 4,495 is likely correct.

Alternatively, maybe I can use a calculator to check? Well, since I don't have a calculator here, but let me think of another way. Maybe breaking down the numbers.

Let me compute 31 * 30 * 29 / 6. Let me do 31 * 30 first. 31*30=930. Then, divide by 6: 930 / 6 = 155. Then, multiply by 29: 155*29. As above, 155*30=4650, minus 155 is 4495. Same result.

Alternatively, compute 31 * (30 / 6) * 29 = 31 * 5 * 29. 31*5=155. 155*29: Let me compute 155*29. Let's think of 29 as 20+9. 155*20=3,100. 155*9=1,395. Then add them together: 3,100 + 1,395 = 4,495. Yep, same.

So I think that's solid. Therefore, $\dbinom{31}{28} = 4,495$. Wait, but let me check once more with another approach. Maybe using Pascal's triangle or some recursive formula? Although for such a large number, that might not be practical.

Alternatively, I can recall that $\dbinom{n}{k}$ is the number of combinations of n things taken k at a time. So 31 choose 28 is the same as 31 choose 3, which is the number of ways to choose 3 items from 31, which is 4,495. Hmm, if I think of it in terms of combinations, maybe I can think of it as follows: choosing 3 items out of 31. For example, the first item can be any of 31, the second any of 30, the third any of 29. But since order doesn't matter, we have to divide by the number of orderings, which is 3!. So that's exactly 31*30*29 / 6, which again gives 4,495.

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

Just to be absolutely sure, maybe I can compute $\dbinom{31}{3}$ using another method. Let me remember that $\dbinom{n}{3} = \frac{n(n-1)(n-2)}{6}$. So substituting n=31:

$$
\frac{31 \times 30 \times 29}{6}
$$

Which is exactly what I did. So if I compute 31×30×29 first, which is 31×(30×29). Let me compute 30×29 first: 30×29 = 870. Then 31×870. Hmm, 31×870. Let me calculate that.

Breaking it down: 30×870 + 1×870 = 26,100 + 870 = 26,970. Then divide by 6 gives 26,970 / 6 = 4,495. Yep, same result again. So that confirms it once more.

So after multiple methods and checks, I can't find any mistake in my calculations. Therefore, I conclude that $\dbinom{31}{28}$ is indeed 4,495.

**Final Answer**
The value of $\dbinom{31}{28}$ is \boxed{4495}.

To compute the binomial coefficient $\dbinom{31}{28}$, we begin with the definition of the binomial coefficient:

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

In this case, $n = 31$ and $k = 28$, so we have:

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

Instead of directly computing large factorials, we notice that $31! = 31 \cdot 30 \cdot 29 \cdot 28!$, so we can simplify:

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

---

### Step 1: Multiply the numerator

Compute the numerator:

$$
31 \times 30 = 930 \quad \text{and} \quad 930 \times 29 = 26,970
$$

---

### Step 2: Divide by the denominator

Now divide by 6:

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

---

### Final Verification

Alternatively, we can switch to the symmetry property of binomial coefficients:

$$
\dbinom{31}{28} = \dbinom{31}{3}
$$

$$
\dbinom{31}{3} = \frac{31 \cdot 30 \cdot 29}{6} = 4,495
$$

Both approaches confirm the result.

---

### 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. From what I recall, the formula for $\dbinom{n}{k}$ is $\frac{n!}{k!(n - k)!}$. Right? So, in this case, $n$ is 31 and $k$ is 28. Let me write that down:

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

Wait, so simplifying the denominator first might make this easier. Let me check if there's a way to simplify the factorials before calculating them all out. Because calculating 31! directly would be a huge number, and it might not be necessary. Let me think.

Since $31!$ divided by $28!$ would leave me with $31 \times 30 \times 29 \times 28! / 28!$, right? So the $28!$ cancels out. That simplifies the equation to:

$$
\frac{31 \times 30 \times 29}{3!}
$$

Okay, so then $3!$ is 6. So now the problem becomes $\frac{31 \times 30 \times 29}{6}$. Let me compute the numerator first.

Calculating $31 \times 30$. Let me do that step by step. 31 times 30 is like 31 times 3 times 10. 31 times 3 is 93, so 93 times 10 is 930. So that's 930. Then multiply by 29. Hmm, 930 times 29. Let me break that down.

29 can be thought of as 30 - 1. So 930 times 30 is 27,900, and then subtract 930. So 27,900 - 930 = 26,970. Wait, is that right? Let me check again:

First, 930 * 29:

Multiply 930 by 20: 18,600

Multiply 930 by 9: 8,370

Add them together: 18,600 + 8,370. Let's add 18,600 + 8,000 = 26,600, then +370 = 26,970. Yeah, same result. So the numerator is 26,970. Now divide that by 6.

So 26,970 divided by 6. Let me do that division step by step. Let me see how many times 6 goes into 26,970.

First, divide 26,970 by 6. Maybe I can break this down. Let me think of 26,970 divided by 6. Let me divide 26,970 by 6.

Well, 6 times 4,000 is 24,000. Subtract 24,000 from 26,970, that leaves 2,970. Then, 6 times 400 is 2,400. Subtract that from 2,970: 2,970 - 2,400 = 570. Then 6 times 95 is 570. So total is 4,000 + 400 + 95 = 4,495. Wait, let me check:

6 * 4,495 = ?

Well 4,495 * 6:

First, 4,000 * 6 = 24,000

400 * 6 = 2,400

95 * 6 = 570

So 24,000 + 2,400 is 26,400, plus 570 is 26,970. Yes, correct! So 26,970 divided by 6 is indeed 4,495. Therefore, $\dbinom{31}{28}$ equals 4,495.

But wait, let me make sure I didn't make a mistake in simplifying earlier. Let me verify the steps again. Original calculation:

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

Yes, that seems correct. So then computing numerator as 31 * 30 * 29. Which is 31 * 30 = 930, then 930 * 29 = 26,970. Divided by 6 gives 4,495.

Alternatively, I can check if there's another way to compute the binomial coefficient. For instance, sometimes it's easier to compute $\dbinom{n}{k}$ as $\dbinom{n}{n - k}$. Since $\dbinom{31}{28} = \dbinom{31}{3}$, right? Because 31 - 28 = 3. That might be easier to compute. Let me try that way to verify.

So $\dbinom{31}{3}$ is $\frac{31 \times 30 \times 29}{3 \times 2 \times 1}$ which is the same as above. So that's 31*30*29 / 6, which is exactly what I calculated. So that's the same result. Therefore, 4,495 should be the correct answer. But let me confirm with another method just to be absolutely sure.

Alternatively, I can compute the value step by step. Let me compute 31 divided by 3 first? Wait, but it's (31*30*29)/6. Maybe simplifying fractions before multiplying. Let me try that.

So 31*30*29 divided by 6. Let me note that 30 divided by 6 is 5. So maybe rewrite the equation as 31 * (30 / 6) * 29. Which is 31 * 5 * 29. Then compute that. 31 * 5 is 155, and then 155 * 29. Let me check this way.

So 155 * 29. Let me calculate that. 155 * 30 would be 4,650, so subtract 155 to get 155 * 29. So 4,650 - 155 = 4,495. Yep, same answer. So that's another way to compute it: simplifying the division before multiplying.

So that gives me more confidence. So whether I do the numerator first and then divide or simplify the fraction first, I get 4,495. Therefore, I think that is the correct value.

Wait, but just to be thorough, maybe I can check with smaller numbers. Let me test if the formula works for a smaller binomial coefficient. For example, $\dbinom{5}{2}$ should be 10. Let me use the same method. $\frac{5!}{2!3!} = \frac{5*4*3!}{2!3!} = \frac{5*4}{2} = 10$. Correct. Similarly, if I use the other approach, $\dbinom{5}{3} = \frac{5*4*3}{3*2*1} = \frac{60}{6} = 10. Same result.

So that seems to check out. Therefore, my method is valid. Therefore, my answer of 4,495 is likely correct.

Alternatively, maybe I can use a calculator to check? Well, since I don't have a calculator here, but let me think of another way. Maybe breaking down the numbers.

Let me compute 31 * 30 * 29 / 6. Let me do 31 * 30 first. 31*30=930. Then, divide by 6: 930 / 6 = 155. Then, multiply by 29: 155*29. As above, 155*30=4650, minus 155 is 4495. Same result.

Alternatively, compute 31 * (30 / 6) * 29 = 31 * 5 * 29. 31*5=155. 155*29: Let me compute 155*29. Let's think of 29 as 20+9. 155*20=3,100. 155*9=1,395. Then add them together: 3,100 + 1,395 = 4,495. Yep, same.

So I think that's solid. Therefore, $\dbinom{31}{28} = 4,495$. Wait, but let me check once more with another approach. Maybe using Pascal's triangle or some recursive formula? Although for such a large number, that might not be practical.

Alternatively, I can recall that $\dbinom{n}{k}$ is the number of combinations of n things taken k at a time. So 31 choose 28 is the same as 31 choose 3, which is the number of ways to choose 3 items from 31, which is 4,495. Hmm, if I think of it in terms of combinations, maybe I can think of it as follows: choosing 3 items out of 31. For example, the first item can be any of 31, the second any of 30, the third any of 29. But since order doesn't matter, we have to divide by the number of orderings, which is 3!. So that's exactly 31*30*29 / 6, which again gives 4,495.

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

Just to be absolutely sure, maybe I can compute $\dbinom{31}{3}$ using another method. Let me remember that $\dbinom{n}{3} = \frac{n(n-1)(n-2)}{6}$. So substituting n=31:

$$
\frac{31 \times 30 \times 29}{6}
$$

Which is exactly what I did. So if I compute 31×30×29 first, which is 31×(30×29). Let me compute 30×29 first: 30×29 = 870. Then 31×870. Hmm, 31×870. Let me calculate that.

Breaking it down: 30×870 + 1×870 = 26,100 + 870 = 26,970. Then divide by 6 gives 26,970 / 6 = 4,495. Yep, same result again. So that confirms it once more.

So after multiple methods and checks, I can't find any mistake in my calculations. Therefore, I conclude that $\dbinom{31}{28}$ is indeed 4,495.

**Final Answer**
The value of $\dbinom{31}{28}$ is \boxed{4495}.

To compute the binomial coefficient $\dbinom{31}{28}$, we begin with the definition of the binomial coefficient:

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

In this case, $n = 31$ and $k = 28$, so we have:

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

Instead of directly computing large factorials, we notice that $31! = 31 \cdot 30 \cdot 29 \cdot 28!$, so we can simplify:

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

---

### Step 1: Multiply the numerator

Compute the numerator:

$$
31 \times 30 = 930 \quad \text{and} \quad 930 \times 29 = 26,970
$$

---

### Step 2: Divide by the denominator

Now divide by 6:

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

---

### Final Verification

Alternatively, we can switch to the symmetry property of binomial coefficients:

$$
\dbinom{31}{28} = \dbinom{31}{3}
$$

$$
\dbinom{31}{3} = \frac{31 \cdot 30 \cdot 29}{6} = 4,495
$$

Both approaches confirm the result.

---

### 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 power of 6. 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. But calculating 18^6 directly might be a bit tedious. Let me see if there's a smarter way to do this without computing the entire number.

I remember that the units digit of a number raised to a power often follows a pattern. Maybe I can focus just on the units digit of the base, which is 8 in this case, and see how 8^6 behaves. Because when you multiply numbers, the units digit only depends on the units digits of the numbers being multiplied. So, 18^6 will have the same units digit as 8^6. Is that right?

Let me check with a smaller exponent to verify. For example, 18^1 is 18, units digit 8. 18^2 is 324, units digit 4. If I compute 8^2, that's 64, units digit 4. Same result. Okay, that works. Let me try another one. 18^3 is 18*18*18. 18*18 is 324, then 324*18. Let me just check the units digit again. 324 has units digit 4, 18 has units digit 8, so 4*8 = 32, so the units digit is 2. If I calculate 8^3, that's 512, which also has units digit 2. Yep, same result. So it seems like focusing on the units digit of the base is sufficient. Therefore, I can just compute 8^6 and find its units digit.

Alright, so let me compute 8^1 through 8^6 and note the units digits. Let me list them out:

- 8^1 = 8. Units digit: 8
- 8^2 = 64. Units digit: 4
- 8^3 = 8*64 = 512. Units digit: 2
- 8^4 = 8*512 = 4096. Units digit: 6
- 8^5 = 8*4096 = 32768. Units digit: 8
- 8^6 = 8*32768 = 262144. Units digit: 4

Wait, so according to this, the units digit of 8^6 is 4? Let me check if there's a pattern here. Let me write down the units digits for each exponent:

Exponent: 1, Units digit: 8

Exponent: 2, Units digit: 4

Exponent: 3, Units digit: 2

Exponent: 4, Units digit: 6

Exponent: 5, Units digit: 8

Exponent: 6, Units digit: 4

So, the pattern seems to be 8, 4, 2, 6, 8, 4... So it cycles every 4 exponents? Let me confirm. From exponent 1 to 5, the units digits go 8, 4, 2, 6, 8... Yes, so the cycle is 8, 4, 2, 6 and then repeats. So the cycle length is 4.

Therefore, if we can find the position of the exponent 6 in this cycle, we can find the units digit. Since the cycle repeats every 4, we can divide the exponent by 4 and find the remainder. Let me recall that.

For exponents:

If exponent mod 4 is 1, then units digit is 8

If exponent mod 4 is 2, units digit is 4

If exponent mod 4 is 3, units digit is 2

If exponent mod 4 is 0 (i.e., divisible by 4), units digit is 6

So let's apply that to exponent 6.

Compute 6 divided by 4: 4*1=4, remainder 2. So the remainder is 2. Therefore, the units digit should be 4. Which matches what I found earlier when calculating 8^6 directly. So that's consistent.

Therefore, the units digit of 18^6 is 4. Wait, but let me check again just to be sure. Maybe I made a mistake in my calculation.

Alternatively, another way to compute 8^6 is by breaking it down:

8^1 = 8

8^2 = 64

8^3 = 8^2 * 8 = 64 * 8. Let's compute 64*8: 60*8=480, 4*8=32, so 480+32=512. Units digit 2. Correct.

8^4 = 8^3 * 8 = 512 * 8. Let's do 500*8=4000, 12*8=96, so total is 4000+96=4096. Units digit 6. Correct.

8^5 = 4096 * 8. 4000*8=32000, 96*8=768, so total is 32000+768=32768. Units digit 8. Correct.

8^6 = 32768 * 8. Let me compute that. 32000*8=256000, 768*8=6144. Then total is 256000 + 6144 = 262144. Units digit is 4. Yep, same result. So that's correct. So the units digit is 4. Therefore, the answer should be 4.

But just to ensure that my cycle idea is correct, let me check exponent 5: remainder when divided by 4 is 1 (since 5 divided by 4 is 1 with remainder 1). According to the cycle, remainder 1 should correspond to units digit 8. Which matches 8^5's units digit. Similarly, exponent 4: remainder 0, units digit 6. Correct. Exponent 3: remainder 3, units digit 2. Correct. So the pattern holds. Therefore, exponent 6: remainder 2, units digit 4. So that's correct.

Alternatively, another way to think about it is using modular arithmetic. Since we're only interested in the units digit, we can compute 18^6 mod 10. Since 18 ≡ 8 mod 10, so 18^6 ≡ 8^6 mod 10. Then, compute 8^6 mod 10. Which is the same as before. So again, we can find 8^6 mod 10.

Alternatively, maybe we can find a pattern or use Euler's theorem? Wait, Euler's theorem says that if a and n are coprime, then a^φ(n) ≡ 1 mod n. Here n is 10. φ(10) is 4, since there are 4 numbers less than 10 that are coprime to 10: 1, 3, 7, 9. So φ(10)=4. So if 8 and 10 are coprime? Wait, 8 and 10 share a common factor of 2, so they are not coprime. Therefore, Euler's theorem does not apply here. So that approach might not work. But maybe we can still use the concept of cycles.

But we already saw that the units digit cycles every 4 exponents. So for exponents of 8, the cycle length is 4. So even though Euler's theorem doesn't apply here, the cycle still exists. So, since 8^4 ≡ 6 mod 10, and then 8^5 = 8^4 * 8 ≡ 6*8=48≡8 mod 10, which is the same as 8^1. Then 8^6 = 8^5 * 8 ≡ 8*8=64≡4 mod 10. So that's another way to verify it. So regardless of the method, I get that the units digit is 4.

Alternatively, maybe I can use exponentiation by squaring or something? Let me try that. Let's see:

Compute 8^6. Let's write 6 in binary? Not sure if that helps. Alternatively, break down the exponent. For example, 8^6 = (8^3)^2. We already know 8^3 is 512. Then, 512 squared. Let me compute 512^2. Well, 512 is 500 + 12. So (500 + 12)^2 = 500^2 + 2*500*12 + 12^2 = 250000 + 12000 + 144 = 250000 + 12000 = 262000 + 144 = 262144. Which matches the earlier result. Units digit is 4. So again, same answer.

Alternatively, since I know that 8^4 is 4096, which ends with 6. So 8^4 ≡ 6 mod 10. Then 8^5 = 8^4 * 8 ≡ 6 * 8 = 48 ≡ 8 mod 10. Then 8^6 = 8^5 * 8 ≡ 8 * 8 = 64 ≡ 4 mod 10. Yep. So that's consistent as well.

So all methods lead to the conclusion that the units digit is 4. Therefore, I think that's the correct answer. But just to be absolutely thorough, let me check with another approach.

Suppose I didn't notice the pattern and tried to compute 18^6 step by step, keeping track only of the units digit:

Start with 18^1. Units digit: 8.

Multiply by 18 again for 18^2: 8 * 8 = 64. Units digit: 4.

Multiply by 18 again for 18^3: 4 * 8 = 32. Units digit: 2.

Multiply by 18 again for 18^4: 2 * 8 = 16. Units digit: 6.

Multiply by 18 again for 18^5: 6 * 8 = 48. Units digit: 8.

Multiply by 18 again for 18^6: 8 * 8 = 64. Units digit: 4.

Same result! So that's another way to think about it. Each time, just taking the previous units digit and multiplying by 8, then taking the units digit of that. So this method also gives 4.

Therefore, I feel confident that the units digit of 18^6 is indeed 4. I don't think I made any mistakes here. All different methods of approaching the problem—direct computation, modular arithmetic, recognizing cycles, and iterative multiplication—lead to the same answer. So unless there's some hidden trick or something I'm missing, which I don't think so, 4 should be the correct units digit.

Just to ensure that maybe I misread the question? The question was about 18^6. Yes. The units digit of 18^6. So, no, that's not a typo. And my calculations all point to 4. So, yeah.

Another check: Let me compute 18^2 = 324. 324 * 18 = 5832. 5832 * 18 = 104,976. Wait, wait, that's 18^4. Wait, no. Let me actually compute step by step:

18^1 = 18.

18^2 = 18 * 18 = 324.

18^3 = 324 * 18. Let me compute 324*10=3240, 324*8=2592. Then 3240 + 2592 = 5832. So 18^3 = 5832. Units digit is 2.

18^4 = 5832 * 18. Let me compute 5832*10=58320, 5832*8=46656. Adding them together: 58320 + 46656. Let me add:

58320 + 46656:

58320 + 40000 = 98320

98320 + 6656 = 104,976. So 18^4 is 104,976. Units digit is 6. Correct.

18^5 = 104,976 * 18. Let me compute this:

First, 104,976 * 10 = 1,049,760

104,976 * 8 = 839,808

Adding them: 1,049,760 + 839,808. Let's do this:

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

1,849,760 + 39,808 = 1,889,568. So 18^5 is 1,889,568. Units digit is 8.

Then, 18^6 = 1,889,568 * 18. Let me compute that:

Again, break into 1,889,568 * 10 = 18,895,680

1,889,568 * 8 = Let's compute step by step.

First, 1,889,568 * 2 = 3,779,136

Then multiply by 4: 3,779,136 * 2 = 7,558,272

Wait, no. Wait, 1,889,568 * 8 is equal to 1,889,568 * 2 * 2 * 2.

First, 1,889,568 * 2 = 3,779,136

Then, 3,779,136 * 2 = 7,558,272

Then, 7,558,272 * 2 = 15,116,544. Therefore, 1,889,568 * 8 = 15,116,544.

Now add that to the previous result of 18,895,680 + 15,116,544.

Let's compute:

18,895,680 + 15,116,544:

18,895,680 + 15,000,000 = 33,895,680

Then add 116,544: 33,895,680 + 116,544 = 34,012,224.

So 18^6 is 34,012,224. Let me check the units digit here. The number is 34,012,224. The last digit is 4. Yes! So that's the units digit. So, this direct computation also confirms it.

So, even though computing 18^6 directly is a bit of a pain, it's still manageable, and the units digit is indeed 4. Therefore, I can be absolutely sure that the units digit of 18^6 is 4.

I think I've verified it in multiple ways: by recognizing the cycle of units digits for powers of 8, by using modular arithmetic, by breaking down the exponentiation step by step and only keeping track of the units digit, and even by computing the entire number and checking the last digit. All methods lead to the same answer. I don't see any flaws in my reasoning, so I feel confident that 4 is the correct units digit.

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

To find the **units digit of** $ 18^6 $, we can utilize the fact that **only the units digit of the base affects the units digit of the result** when raising to a power. This is because multiplying any number by another only affects the units digit in a predictable way based on the units digits of the operands.

---

### Step 1: Reduce the Base

We observe that:
$$
18 \equiv 8 \pmod{10}
$$

Therefore:
$$
18^6 \equiv 8^6 \pmod{10}
$$

This simplifies the problem to finding the **units digit of $ 8^6 $**.

---

### Step 2: Identify the Pattern in the Units Digits of Powers of 8

Let's compute the units digit of successive powers of 8:

- $ 8^1 = 8 $ → units digit: **8**
- $ 8^2 = 64 $ → units digit: **4**
- $ 8^3 = 512 $ → units digit: **2**
- $ 8^4 = 4096 $ → units digit: **6**
- $ 8^5 = 32768 $ → units digit: **8**
- $ 8^6 = 262144 $ → units digit: **4**

The units digits of the powers of 8 follow a **repeating cycle** of length **4**:
$$
\text{Cycle: } 8, 4, 2, 6
$$

---

### Step 3: Use the Cycle to Find the Units Digit of $ 8^6 $

To find the position of $ 8^6 $ in the cycle, compute the exponent modulo 4:

$$
6 \mod 4 = 2
$$

This corresponds to the **second** digit in the cycle. Referring back to the pattern:
- Position 1: 8
- Position 2: 4

Thus, the **units digit of $ 8^6 $** is **4**.

---

### Step 4: Final Answer

Since:
$$
18^6 \equiv 8^6 \pmod{10}
$$

And the units digit of $ 8^6 $ is **4**, the units digit of $ 18^6 $ is also:

$$
\boxed{4}
$$

[Assistant]
Explanation:
Okay, so I need to find the units digit of 18 raised to the power of 6. 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. But calculating 18^6 directly might be a bit tedious. Let me see if there's a smarter way to do this without computing the entire number.

I remember that the units digit of a number raised to a power often follows a pattern. Maybe I can focus just on the units digit of the base, which is 8 in this case, and see how 8^6 behaves. Because when you multiply numbers, the units digit only depends on the units digits of the numbers being multiplied. So, 18^6 will have the same units digit as 8^6. Is that right?

Let me check with a smaller exponent to verify. For example, 18^1 is 18, units digit 8. 18^2 is 324, units digit 4. If I compute 8^2, that's 64, units digit 4. Same result. Okay, that works. Let me try another one. 18^3 is 18*18*18. 18*18 is 324, then 324*18. Let me just check the units digit again. 324 has units digit 4, 18 has units digit 8, so 4*8 = 32, so the units digit is 2. If I calculate 8^3, that's 512, which also has units digit 2. Yep, same result. So it seems like focusing on the units digit of the base is sufficient. Therefore, I can just compute 8^6 and find its units digit.

Alright, so let me compute 8^1 through 8^6 and note the units digits. Let me list them out:

- 8^1 = 8. Units digit: 8
- 8^2 = 64. Units digit: 4
- 8^3 = 8*64 = 512. Units digit: 2
- 8^4 = 8*512 = 4096. Units digit: 6
- 8^5 = 8*4096 = 32768. Units digit: 8
- 8^6 = 8*32768 = 262144. Units digit: 4

Wait, so according to this, the units digit of 8^6 is 4? Let me check if there's a pattern here. Let me write down the units digits for each exponent:

Exponent: 1, Units digit: 8

Exponent: 2, Units digit: 4

Exponent: 3, Units digit: 2

Exponent: 4, Units digit: 6

Exponent: 5, Units digit: 8

Exponent: 6, Units digit: 4

So, the pattern seems to be 8, 4, 2, 6, 8, 4... So it cycles every 4 exponents? Let me confirm. From exponent 1 to 5, the units digits go 8, 4, 2, 6, 8... Yes, so the cycle is 8, 4, 2, 6 and then repeats. So the cycle length is 4.

Therefore, if we can find the position of the exponent 6 in this cycle, we can find the units digit. Since the cycle repeats every 4, we can divide the exponent by 4 and find the remainder. Let me recall that.

For exponents:

If exponent mod 4 is 1, then units digit is 8

If exponent mod 4 is 2, units digit is 4

If exponent mod 4 is 3, units digit is 2

If exponent mod 4 is 0 (i.e., divisible by 4), units digit is 6

So let's apply that to exponent 6.

Compute 6 divided by 4: 4*1=4, remainder 2. So the remainder is 2. Therefore, the units digit should be 4. Which matches what I found earlier when calculating 8^6 directly. So that's consistent.

Therefore, the units digit of 18^6 is 4. Wait, but let me check again just to be sure. Maybe I made a mistake in my calculation.

Alternatively, another way to compute 8^6 is by breaking it down:

8^1 = 8

8^2 = 64

8^3 = 8^2 * 8 = 64 * 8. Let's compute 64*8: 60*8=480, 4*8=32, so 480+32=512. Units digit 2. Correct.

8^4 = 8^3 * 8 = 512 * 8. Let's do 500*8=4000, 12*8=96, so total is 4000+96=4096. Units digit 6. Correct.

8^5 = 4096 * 8. 4000*8=32000, 96*8=768, so total is 32000+768=32768. Units digit 8. Correct.

8^6 = 32768 * 8. Let me compute that. 32000*8=256000, 768*8=6144. Then total is 256000 + 6144 = 262144. Units digit is 4. Yep, same result. So that's correct. So the units digit is 4. Therefore, the answer should be 4.

But just to ensure that my cycle idea is correct, let me check exponent 5: remainder when divided by 4 is 1 (since 5 divided by 4 is 1 with remainder 1). According to the cycle, remainder 1 should correspond to units digit 8. Which matches 8^5's units digit. Similarly, exponent 4: remainder 0, units digit 6. Correct. Exponent 3: remainder 3, units digit 2. Correct. So the pattern holds. Therefore, exponent 6: remainder 2, units digit 4. So that's correct.

Alternatively, another way to think about it is using modular arithmetic. Since we're only interested in the units digit, we can compute 18^6 mod 10. Since 18 ≡ 8 mod 10, so 18^6 ≡ 8^6 mod 10. Then, compute 8^6 mod 10. Which is the same as before. So again, we can find 8^6 mod 10.

Alternatively, maybe we can find a pattern or use Euler's theorem? Wait, Euler's theorem says that if a and n are coprime, then a^φ(n) ≡ 1 mod n. Here n is 10. φ(10) is 4, since there are 4 numbers less than 10 that are coprime to 10: 1, 3, 7, 9. So φ(10)=4. So if 8 and 10 are coprime? Wait, 8 and 10 share a common factor of 2, so they are not coprime. Therefore, Euler's theorem does not apply here. So that approach might not work. But maybe we can still use the concept of cycles.

But we already saw that the units digit cycles every 4 exponents. So for exponents of 8, the cycle length is 4. So even though Euler's theorem doesn't apply here, the cycle still exists. So, since 8^4 ≡ 6 mod 10, and then 8^5 = 8^4 * 8 ≡ 6*8=48≡8 mod 10, which is the same as 8^1. Then 8^6 = 8^5 * 8 ≡ 8*8=64≡4 mod 10. So that's another way to verify it. So regardless of the method, I get that the units digit is 4.

Alternatively, maybe I can use exponentiation by squaring or something? Let me try that. Let's see:

Compute 8^6. Let's write 6 in binary? Not sure if that helps. Alternatively, break down the exponent. For example, 8^6 = (8^3)^2. We already know 8^3 is 512. Then, 512 squared. Let me compute 512^2. Well, 512 is 500 + 12. So (500 + 12)^2 = 500^2 + 2*500*12 + 12^2 = 250000 + 12000 + 144 = 250000 + 12000 = 262000 + 144 = 262144. Which matches the earlier result. Units digit is 4. So again, same answer.

Alternatively, since I know that 8^4 is 4096, which ends with 6. So 8^4 ≡ 6 mod 10. Then 8^5 = 8^4 * 8 ≡ 6 * 8 = 48 ≡ 8 mod 10. Then 8^6 = 8^5 * 8 ≡ 8 * 8 = 64 ≡ 4 mod 10. Yep. So that's consistent as well.

So all methods lead to the conclusion that the units digit is 4. Therefore, I think that's the correct answer. But just to be absolutely thorough, let me check with another approach.

Suppose I didn't notice the pattern and tried to compute 18^6 step by step, keeping track only of the units digit:

Start with 18^1. Units digit: 8.

Multiply by 18 again for 18^2: 8 * 8 = 64. Units digit: 4.

Multiply by 18 again for 18^3: 4 * 8 = 32. Units digit: 2.

Multiply by 18 again for 18^4: 2 * 8 = 16. Units digit: 6.

Multiply by 18 again for 18^5: 6 * 8 = 48. Units digit: 8.

Multiply by 18 again for 18^6: 8 * 8 = 64. Units digit: 4.

Same result! So that's another way to think about it. Each time, just taking the previous units digit and multiplying by 8, then taking the units digit of that. So this method also gives 4.

Therefore, I feel confident that the units digit of 18^6 is indeed 4. I don't think I made any mistakes here. All different methods of approaching the problem—direct computation, modular arithmetic, recognizing cycles, and iterative multiplication—lead to the same answer. So unless there's some hidden trick or something I'm missing, which I don't think so, 4 should be the correct units digit.

Just to ensure that maybe I misread the question? The question was about 18^6. Yes. The units digit of 18^6. So, no, that's not a typo. And my calculations all point to 4. So, yeah.

Another check: Let me compute 18^2 = 324. 324 * 18 = 5832. 5832 * 18 = 104,976. Wait, wait, that's 18^4. Wait, no. Let me actually compute step by step:

18^1 = 18.

18^2 = 18 * 18 = 324.

18^3 = 324 * 18. Let me compute 324*10=3240, 324*8=2592. Then 3240 + 2592 = 5832. So 18^3 = 5832. Units digit is 2.

18^4 = 5832 * 18. Let me compute 5832*10=58320, 5832*8=46656. Adding them together: 58320 + 46656. Let me add:

58320 + 46656:

58320 + 40000 = 98320

98320 + 6656 = 104,976. So 18^4 is 104,976. Units digit is 6. Correct.

18^5 = 104,976 * 18. Let me compute this:

First, 104,976 * 10 = 1,049,760

104,976 * 8 = 839,808

Adding them: 1,049,760 + 839,808. Let's do this:

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

1,849,760 + 39,808 = 1,889,568. So 18^5 is 1,889,568. Units digit is 8.

Then, 18^6 = 1,889,568 * 18. Let me compute that:

Again, break into 1,889,568 * 10 = 18,895,680

1,889,568 * 8 = Let's compute step by step.

First, 1,889,568 * 2 = 3,779,136

Then multiply by 4: 3,779,136 * 2 = 7,558,272

Wait, no. Wait, 1,889,568 * 8 is equal to 1,889,568 * 2 * 2 * 2.

First, 1,889,568 * 2 = 3,779,136

Then, 3,779,136 * 2 = 7,558,272

Then, 7,558,272 * 2 = 15,116,544. Therefore, 1,889,568 * 8 = 15,116,544.

Now add that to the previous result of 18,895,680 + 15,116,544.

Let's compute:

18,895,680 + 15,116,544:

18,895,680 + 15,000,000 = 33,895,680

Then add 116,544: 33,895,680 + 116,544 = 34,012,224.

So 18^6 is 34,012,224. Let me check the units digit here. The number is 34,012,224. The last digit is 4. Yes! So that's the units digit. So, this direct computation also confirms it.

So, even though computing 18^6 directly is a bit of a pain, it's still manageable, and the units digit is indeed 4. Therefore, I can be absolutely sure that the units digit of 18^6 is 4.

I think I've verified it in multiple ways: by recognizing the cycle of units digits for powers of 8, by using modular arithmetic, by breaking down the exponentiation step by step and only keeping track of the units digit, and even by computing the entire number and checking the last digit. All methods lead to the same answer. I don't see any flaws in my reasoning, so I feel confident that 4 is the correct units digit.

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

To find the **units digit of** $ 18^6 $, we can utilize the fact that **only the units digit of the base affects the units digit of the result** when raising to a power. This is because multiplying any number by another only affects the units digit in a predictable way based on the units digits of the operands.

---

### Step 1: Reduce the Base

We observe that:
$$
18 \equiv 8 \pmod{10}
$$

Therefore:
$$
18^6 \equiv 8^6 \pmod{10}
$$

This simplifies the problem to finding the **units digit of $ 8^6 $**.

---

### Step 2: Identify the Pattern in the Units Digits of Powers of 8

Let's compute the units digit of successive powers of 8:

- $ 8^1 = 8 $ → units digit: **8**
- $ 8^2 = 64 $ → units digit: **4**
- $ 8^3 = 512 $ → units digit: **2**
- $ 8^4 = 4096 $ → units digit: **6**
- $ 8^5 = 32768 $ → units digit: **8**
- $ 8^6 = 262144 $ → units digit: **4**

The units digits of the powers of 8 follow a **repeating cycle** of length **4**:
$$
\text{Cycle: } 8, 4, 2, 6
$$

---

### Step 3: Use the Cycle to Find the Units Digit of $ 8^6 $

To find the position of $ 8^6 $ in the cycle, compute the exponent modulo 4:

$$
6 \mod 4 = 2
$$

This corresponds to the **second** digit in the cycle. Referring back to the pattern:
- Position 1: 8
- Position 2: 4

Thus, the **units digit of $ 8^6 $** is **4**.

---

### Step 4: Final Answer

Since:
$$
18^6 \equiv 8^6 \pmod{10}
$$

And the units digit of $ 8^6 $ is **4**, the units digit of $ 18^6 $ is also:

$$
\boxed{4}
$$

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 the general form of a circle is $(x - h)^2 + (y - k)^2 = r^2$, where $(h, k)$ is the center and $r$ is the radius. Then there's the ellipse, which is similar but with different coefficients for the x and y terms: $\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1$. For a hyperbola, it's either $\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1$ or the other way around with y first. A parabola is different; it's something like $(y - k)^2 = 4p(x - h)$ or similar, involving only one squared term.

Looking at the given equation, let me write it again to see more clearly: $\left(\frac{x}{2} - 3\right)^2 + y^2 = 10$. Hmm, so there are two squared terms, x and y, each raised to the second power. That makes me think it might be a circle or an ellipse.

But wait, in the standard circle equation, both terms are squared and have the same coefficient. Here, though, the x term is $\left(\frac{x}{2} - 3\right)^2$, which can be rewritten as $\left(\frac{1}{2}(x - 6)\right)^2$. Let me check that: $\frac{x}{2} - 3$ is the same as $\frac{1}{2}(x - 6)$, right? Because if I factor out 1/2, it's $\frac{1}{2}x - \frac{6}{2}$ which is $\frac{x}{2} - 3$. Yes, so squaring that gives $\left(\frac{1}{2}(x - 6)\right)^2 = \frac{1}{4}(x - 6)^2$. Therefore, the equation can be rewritten as $\frac{1}{4}(x - 6)^2 + y^2 = 10$.

So if I want to compare this to the standard forms, let me manipulate it to look more like the standard equations. Let me multiply both sides by 4 to eliminate the fraction. That would give me $(x - 6)^2 + 4y^2 = 40$. Now, this looks similar to the standard form of an ellipse, which is $\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1$.

To get this into that form, I can divide both sides by 40. Let me do that: $\frac{(x - 6)^2}{40} + \frac{4y^2}{40} = 1$, which simplifies to $\frac{(x - 6)^2}{40} + \frac{y^2}{10} = 1$.

Now, this is definitely the standard form of an ellipse. The denominators under each squared term are 40 and 10, so they are both positive, which is consistent with an ellipse. If they were the same, it would be a circle, but here they are different, so it's an ellipse.

Wait a second, but let me check if there's another possibility. For example, if the equation had a minus sign between the terms, it would be a hyperbola. But here it's a plus sign, so that rules out hyperbola. Also, since both terms are squared and positive, adding up to 10, which is a positive number, there's no way this could be a line or two lines or a point or empty set.

Let me think again. If I had, say, $\left(\frac{x}{2} - 3\right)^2 + y^2 = 0$, then it would be a single point because both squared terms can't be negative. But here, the right-hand side is 10, which is positive, so there are solutions. So the graph isn't empty.

Is there any chance it's a circle? Well, if the coefficients of the squared terms were the same, then yes. But in this case, after rearranging, the coefficients under $(x - 6)^2$ and $y^2$ are different (40 and 10). Therefore, it's an ellipse.

Just to confirm, maybe I can think about the definitions. An ellipse is the set of points where the sum of the distances from two foci is constant. A circle is a special case of an ellipse where the two foci coincide. Since here the equation isn't a circle, but an ellipse with different axes lengths.

Alternatively, if I want to check the axes, the major and minor axes. In the standard ellipse equation, the major axis is along the x-axis if the denominator under the x-term is larger. Here, 40 is bigger than 10, so the major axis is along the x-direction, and the minor axis along y.

So, given that, the equation represents an ellipse.

Wait, let me check again the original equation. The original equation is $\left(\frac{x}{2} - 3\right)^2 + y^2 = 10$. If I consider this, the term $\left(\frac{x}{2} - 3\right)^2$ is equivalent to $\left(\frac{x - 6}{2}\right)^2$, which is $\frac{(x - 6)^2}{4}$. So the equation is $\frac{(x - 6)^2}{4} + y^2 = 10$. Then, if I write it as $\frac{(x - 6)^2}{40} + \frac{y^2}{10} = 1$, that's correct because dividing numerator and denominator by 10? Wait, hold on:

Wait, original equation after multiplying by 4 was $(x - 6)^2 + 4y^2 = 40$. Then, dividing both sides by 40 gives $\frac{(x - 6)^2}{40} + \frac{4y^2}{40} = 1$, which is $\frac{(x - 6)^2}{40} + \frac{y^2}{10} = 1$. Yes, that's correct. Therefore, the denominators are 40 and 10, so the major axis is along the x-axis, with semi-major axis length sqrt(40), and semi-minor axis sqrt(10).

Therefore, this is definitely an ellipse.

But let me also check if there's a possibility of it being a circle. For instance, if the equation was $(x - 6)^2 + y^2 = 10$, that would be a circle with radius sqrt(10). However, in our case, the coefficient in front of the x term is 1/4, so when we adjust it, the denominators are different. Therefore, it's not a circle.

Alternatively, if I had, for example, $\left(\frac{x}{2} - 3\right)^2 + y^2 = 10$, and if I had both terms scaled by the same factor, but here they are scaled differently.

Therefore, I can confirm that this equation represents an ellipse.

Just to be thorough, let me think about other possibilities. Could this be a parabola? A parabola has only one squared term, so no. Could it be two lines? For that, the equation would need to factor into two linear equations. For example, something like (ax + by + c)(dx + ey + f) = 0. But in our case, the equation is a sum of squares equals a constant. If we tried to rearrange it, we might get something like $\left(\frac{x}{2} - 3\right)^2 = 10 - y^2$, which would imply that both sides must be non-negative. However, unless 10 - y^2 is non-negative, which would restrict y to be between -sqrt(10) and sqrt(10). But even so, this is still an ellipse, not two lines.

Alternatively, if the equation had a squared term and a linear term, maybe it could be a parabola, but here there are two squared terms. Therefore, no.

Similarly, if there were a negative sign, it would be a hyperbola, but here it's positive. So, hyperbola is out.

Therefore, the only possible conic section here is an ellipse.

Wait, just to make sure, let me think about the definition of an ellipse. The general second-degree equation is $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$. For it to be an ellipse, the discriminant $B^2 - 4AC$ should be less than 0.

Let me try to convert the given equation into general form. Starting with $\left(\frac{x}{2} - 3\right)^2 + y^2 = 10$. Expanding the x term: $\left(\frac{x}{2} - 3\right)^2 = \frac{x^2}{4} - 3x + 9$. So the entire equation becomes $\frac{x^2}{4} - 3x + 9 + y^2 = 10$. Subtract 10 from both sides: $\frac{x^2}{4} - 3x + y^2 - 1 = 0$.

Multiplying all terms by 4 to eliminate fractions: $x^2 - 12x + 4y^2 - 4 = 0$. So the general form is $x^2 + 4y^2 - 12x - 4 = 0$.

Comparing this to $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$, we have:

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

So, the discriminant is $B^2 - 4AC = 0 - 4(1)(4) = -16$, which is less than 0. Therefore, according to the discriminant test for conic sections, this is an ellipse.

So that confirms it. Since the discriminant is negative, it's an ellipse.

Therefore, the answer is an ellipse.

**Final Answer**
\boxed{ellipse}

To determine the nature of the graph represented by the equation:

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

we begin by interpreting and simplifying this expression to compare it with known conic section forms.

---

### Step 1: Analyze the Structure

The equation contains **two squared terms**, $\left(\frac{x}{2} - 3\right)^2$ and $y^2$, with a constant on the right-hand side. This hints at a **conic section**.

We rewrite 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:

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

---

### Step 2: Eliminate Fractions for Comparison

To align with the standard form of a conic, we eliminate the denominator:

$$
(x - 6)^2 + 4y^2 = 40
$$

Now divide through by 40 to normalize it to the standard ellipse form:

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

---

### Step 3: Classify the Conic

This equation fits the standard form of an **ellipse**:

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

where $a^2 = 40$, $b^2 = 10$, and the center is at $(h, k) = (6, 0)$. Since $a^2 > b^2$, the major axis is along the x-direction.

Also, the equation has positive coefficients for both squared terms and is not equal to zero or negative, meaning the graph exists in the plane and is not empty.

---

### Step 4: Discriminant Verification (Optional but Useful)

To double-check, we translate the equation into the general second-degree form:

$$
\left(\frac{x}{2} - 3\right)^2 + y^2 = 10 \Rightarrow \frac{x^2}{4} - 3x + 9 + y^2 = 10
$$

Subtract 10 from both sides:

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

Multiply by 4 to eliminate the fraction:

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

Compare with the general conic form:

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

We identify:
- $A = 1$, $C = 4$, $B = 0$

Then the discriminant is:

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

Since the discriminant is negative, this is confirmed to be an **ellipse**.

---

### Final Conclusion

Based on both algebraic transformation and discriminant verification, the graph of the given equation is an **ellipse**, centered at $(6, 0)$, with the x-axis as the major axis.

$$
\boxed{ellipse}
$$

[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 the general form of a circle is $(x - h)^2 + (y - k)^2 = r^2$, where $(h, k)$ is the center and $r$ is the radius. Then there's the ellipse, which is similar but with different coefficients for the x and y terms: $\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1$. For a hyperbola, it's either $\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1$ or the other way around with y first. A parabola is different; it's something like $(y - k)^2 = 4p(x - h)$ or similar, involving only one squared term.

Looking at the given equation, let me write it again to see more clearly: $\left(\frac{x}{2} - 3\right)^2 + y^2 = 10$. Hmm, so there are two squared terms, x and y, each raised to the second power. That makes me think it might be a circle or an ellipse.

But wait, in the standard circle equation, both terms are squared and have the same coefficient. Here, though, the x term is $\left(\frac{x}{2} - 3\right)^2$, which can be rewritten as $\left(\frac{1}{2}(x - 6)\right)^2$. Let me check that: $\frac{x}{2} - 3$ is the same as $\frac{1}{2}(x - 6)$, right? Because if I factor out 1/2, it's $\frac{1}{2}x - \frac{6}{2}$ which is $\frac{x}{2} - 3$. Yes, so squaring that gives $\left(\frac{1}{2}(x - 6)\right)^2 = \frac{1}{4}(x - 6)^2$. Therefore, the equation can be rewritten as $\frac{1}{4}(x - 6)^2 + y^2 = 10$.

So if I want to compare this to the standard forms, let me manipulate it to look more like the standard equations. Let me multiply both sides by 4 to eliminate the fraction. That would give me $(x - 6)^2 + 4y^2 = 40$. Now, this looks similar to the standard form of an ellipse, which is $\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1$.

To get this into that form, I can divide both sides by 40. Let me do that: $\frac{(x - 6)^2}{40} + \frac{4y^2}{40} = 1$, which simplifies to $\frac{(x - 6)^2}{40} + \frac{y^2}{10} = 1$.

Now, this is definitely the standard form of an ellipse. The denominators under each squared term are 40 and 10, so they are both positive, which is consistent with an ellipse. If they were the same, it would be a circle, but here they are different, so it's an ellipse.

Wait a second, but let me check if there's another possibility. For example, if the equation had a minus sign between the terms, it would be a hyperbola. But here it's a plus sign, so that rules out hyperbola. Also, since both terms are squared and positive, adding up to 10, which is a positive number, there's no way this could be a line or two lines or a point or empty set.

Let me think again. If I had, say, $\left(\frac{x}{2} - 3\right)^2 + y^2 = 0$, then it would be a single point because both squared terms can't be negative. But here, the right-hand side is 10, which is positive, so there are solutions. So the graph isn't empty.

Is there any chance it's a circle? Well, if the coefficients of the squared terms were the same, then yes. But in this case, after rearranging, the coefficients under $(x - 6)^2$ and $y^2$ are different (40 and 10). Therefore, it's an ellipse.

Just to confirm, maybe I can think about the definitions. An ellipse is the set of points where the sum of the distances from two foci is constant. A circle is a special case of an ellipse where the two foci coincide. Since here the equation isn't a circle, but an ellipse with different axes lengths.

Alternatively, if I want to check the axes, the major and minor axes. In the standard ellipse equation, the major axis is along the x-axis if the denominator under the x-term is larger. Here, 40 is bigger than 10, so the major axis is along the x-direction, and the minor axis along y.

So, given that, the equation represents an ellipse.

Wait, let me check again the original equation. The original equation is $\left(\frac{x}{2} - 3\right)^2 + y^2 = 10$. If I consider this, the term $\left(\frac{x}{2} - 3\right)^2$ is equivalent to $\left(\frac{x - 6}{2}\right)^2$, which is $\frac{(x - 6)^2}{4}$. So the equation is $\frac{(x - 6)^2}{4} + y^2 = 10$. Then, if I write it as $\frac{(x - 6)^2}{40} + \frac{y^2}{10} = 1$, that's correct because dividing numerator and denominator by 10? Wait, hold on:

Wait, original equation after multiplying by 4 was $(x - 6)^2 + 4y^2 = 40$. Then, dividing both sides by 40 gives $\frac{(x - 6)^2}{40} + \frac{4y^2}{40} = 1$, which is $\frac{(x - 6)^2}{40} + \frac{y^2}{10} = 1$. Yes, that's correct. Therefore, the denominators are 40 and 10, so the major axis is along the x-axis, with semi-major axis length sqrt(40), and semi-minor axis sqrt(10).

Therefore, this is definitely an ellipse.

But let me also check if there's a possibility of it being a circle. For instance, if the equation was $(x - 6)^2 + y^2 = 10$, that would be a circle with radius sqrt(10). However, in our case, the coefficient in front of the x term is 1/4, so when we adjust it, the denominators are different. Therefore, it's not a circle.

Alternatively, if I had, for example, $\left(\frac{x}{2} - 3\right)^2 + y^2 = 10$, and if I had both terms scaled by the same factor, but here they are scaled differently.

Therefore, I can confirm that this equation represents an ellipse.

Just to be thorough, let me think about other possibilities. Could this be a parabola? A parabola has only one squared term, so no. Could it be two lines? For that, the equation would need to factor into two linear equations. For example, something like (ax + by + c)(dx + ey + f) = 0. But in our case, the equation is a sum of squares equals a constant. If we tried to rearrange it, we might get something like $\left(\frac{x}{2} - 3\right)^2 = 10 - y^2$, which would imply that both sides must be non-negative. However, unless 10 - y^2 is non-negative, which would restrict y to be between -sqrt(10) and sqrt(10). But even so, this is still an ellipse, not two lines.

Alternatively, if the equation had a squared term and a linear term, maybe it could be a parabola, but here there are two squared terms. Therefore, no.

Similarly, if there were a negative sign, it would be a hyperbola, but here it's positive. So, hyperbola is out.

Therefore, the only possible conic section here is an ellipse.

Wait, just to make sure, let me think about the definition of an ellipse. The general second-degree equation is $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$. For it to be an ellipse, the discriminant $B^2 - 4AC$ should be less than 0.

Let me try to convert the given equation into general form. Starting with $\left(\frac{x}{2} - 3\right)^2 + y^2 = 10$. Expanding the x term: $\left(\frac{x}{2} - 3\right)^2 = \frac{x^2}{4} - 3x + 9$. So the entire equation becomes $\frac{x^2}{4} - 3x + 9 + y^2 = 10$. Subtract 10 from both sides: $\frac{x^2}{4} - 3x + y^2 - 1 = 0$.

Multiplying all terms by 4 to eliminate fractions: $x^2 - 12x + 4y^2 - 4 = 0$. So the general form is $x^2 + 4y^2 - 12x - 4 = 0$.

Comparing this to $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$, we have:

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

So, the discriminant is $B^2 - 4AC = 0 - 4(1)(4) = -16$, which is less than 0. Therefore, according to the discriminant test for conic sections, this is an ellipse.

So that confirms it. Since the discriminant is negative, it's an ellipse.

Therefore, the answer is an ellipse.

**Final Answer**
\boxed{ellipse}

To determine the nature of the graph represented by the equation:

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

we begin by interpreting and simplifying this expression to compare it with known conic section forms.

---

### Step 1: Analyze the Structure

The equation contains **two squared terms**, $\left(\frac{x}{2} - 3\right)^2$ and $y^2$, with a constant on the right-hand side. This hints at a **conic section**.

We rewrite 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:

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

---

### Step 2: Eliminate Fractions for Comparison

To align with the standard form of a conic, we eliminate the denominator:

$$
(x - 6)^2 + 4y^2 = 40
$$

Now divide through by 40 to normalize it to the standard ellipse form:

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

---

### Step 3: Classify the Conic

This equation fits the standard form of an **ellipse**:

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

where $a^2 = 40$, $b^2 = 10$, and the center is at $(h, k) = (6, 0)$. Since $a^2 > b^2$, the major axis is along the x-direction.

Also, the equation has positive coefficients for both squared terms and is not equal to zero or negative, meaning the graph exists in the plane and is not empty.

---

### Step 4: Discriminant Verification (Optional but Useful)

To double-check, we translate the equation into the general second-degree form:

$$
\left(\frac{x}{2} - 3\right)^2 + y^2 = 10 \Rightarrow \frac{x^2}{4} - 3x + 9 + y^2 = 10
$$

Subtract 10 from both sides:

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

Multiply by 4 to eliminate the fraction:

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

Compare with the general conic form:

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

We identify:
- $A = 1$, $C = 4$, $B = 0$

Then the discriminant is:

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

Since the discriminant is negative, this is confirmed to be an **ellipse**.

---

### Final Conclusion

Based on both algebraic transformation and discriminant verification, the graph of the given equation is an **ellipse**, centered at $(6, 0)$, with the x-axis as the major axis.

$$
\boxed{ellipse}
$$

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, let's see. The problem is: Three plus the reciprocal of a number equals 7 divided by that number. I need to find the number. Hmm, let me try to translate this into an equation.

First, let me denote the number as something. Let me call it x. So, the number is x. Then, the reciprocal of the number would be 1/x. The problem says three plus the reciprocal of the number equals 7 divided by that number.

So, let me write that out. Three is 3, so 3 plus 1/x equals 7 divided by x. So, the equation is:

3 + (1/x) = 7/x

Alright, so now I need to solve for x. Let me think about how to do this. Since there are fractions here, maybe I can get rid of the denominators by multiplying both sides of the equation by x. That should eliminate the denominators. Let me try that.

Multiplying both sides by x:

x * [3 + (1/x)] = x * (7/x)

Let me simplify both sides. On the left side, distributing the x:

x * 3 + x * (1/x) = 3x + 1

On the right side, x times 7 over x is 7, because the x cancels out. So:

3x + 1 = 7

Okay, that seems simpler. Now, solving for x. Let me subtract 1 from both sides:

3x + 1 - 1 = 7 - 1

Which simplifies to:

3x = 6

Then, divide both sides by 3:

x = 6 / 3

So, x = 2

Wait, let me check if that makes sense. Let me substitute x = 2 back into the original equation to verify.

Original equation: 3 + (1/x) = 7/x

Left side: 3 + (1/2) = 3.5 or 7/2

Right side: 7 / 2 = 3.5

Yes, both sides are equal. So, x = 2 is indeed the solution.

Wait, but let me think again. What if there was another solution? Because sometimes when you multiply both sides by x, you might assume x is not zero, but since reciprocal is involved, x can't be zero anyway. So, x ≠ 0, so that's okay.

But let me check if there's another possible solution. Let me try solving the equation again step by step. Maybe I made a mistake?

Original equation:

3 + 1/x = 7/x

Subtract 1/x from both sides:

3 = 7/x - 1/x

Which is 3 = (7 - 1)/x => 3 = 6/x

Then, solving for x: 3 = 6/x => x = 6/3 => x = 2

Same result. So, that's consistent.

Alternatively, if I didn't multiply both sides by x, but instead brought all terms to one side:

3 + 1/x - 7/x = 0

Which simplifies to 3 - 6/x = 0

Then, 3 = 6/x => same as before.

So, again, x = 2.

Therefore, the solution is 2.

But wait, sometimes when you have equations with reciprocals, there might be extraneous solutions or something? But in this case, since we only had one solution, and it checks out, so that's the answer.

Alternatively, let me think if there is a possibility of another solution. Let me suppose that maybe I made an error in assuming that multiplying both sides by x is allowed. But as long as x is not zero, which it can't be because of reciprocal, so multiplying both sides by x is valid.

So, no, I don't think there's another solution. So, x must be 2.

Let me also think if there's any alternative way to approach this problem. Maybe by letting the number be something else? But no, I think the way I did it is straightforward.

Alternatively, maybe graphing? If I consider the left side as a function f(x) = 3 + 1/x and the right side as g(x) = 7/x. The intersection point would be the solution. But since I already solved algebraically, that's probably unnecessary.

But just for thoroughness, let's think. For x positive, as x approaches zero from the right, f(x) approaches infinity, and g(x) also approaches infinity. But which one is higher? Let me check for very small x, say x approaching 0+. Then 3 + 1/x is dominated by 1/x, which is positive and large, while 7/x is also positive and large. Let me compare their rates. Let me take x approaching 0+, then f(x) - g(x) = 3 + 1/x - 7/x = 3 - 6/x. As x approaches 0+, 3 - 6/x approaches negative infinity. So, f(x) - g(x) approaches negative infinity. So, for very small x, f(x) < g(x). At x = 2, they are equal. What about as x approaches infinity? Both f(x) and g(x) approach 3 and 0, respectively. So, f(x) approaches 3, and g(x) approaches 0, so f(x) > g(x) for large x. Therefore, since f(x) - g(x) goes from negative infinity at x approaching 0+ to positive infinity as x approaches infinity, and since it's continuous for x > 0 (except at x=0), there must be exactly one solution where they cross. Which is at x=2. So that confirms there's only one solution.

What about negative numbers? The problem didn't specify if the number is positive or negative. Let me check if there's a negative solution. Suppose x is negative. Let me try solving the equation again, but considering x negative.

Original equation: 3 + 1/x = 7/x

Subtract 1/x from both sides: 3 = 6/x => x = 6/3 = 2. Wait, so even if I consider x negative, solving gives me x=2, which is positive. So, there's no negative solution? Because if I rearrange the equation, 3 = 6/x implies x = 2, no matter what. Therefore, x must be 2. Therefore, there is only one solution.

But just to be thorough, let me plug in a negative number. Suppose x = -1. Then, left side: 3 + 1/(-1) = 3 -1 = 2. Right side: 7 / (-1) = -7. 2 ≠ -7. How about x = -2? Left side: 3 + 1/(-2) = 3 - 0.5 = 2.5. Right side: 7 / (-2) = -3.5. Not equal. What about x = -something else? Let me try solving for x negative. Let me go back to equation:

Original equation: 3 + 1/x = 7/x

Multiply both sides by x, assuming x ≠ 0:

3x + 1 = 7

Which gives 3x = 6 => x = 2. So, regardless of sign, the algebra leads to x=2. Therefore, x must be 2. So, even if I consider negative numbers, the algebra shows that x must be positive 2. Therefore, there are no negative solutions.

Therefore, the only solution is x=2.

Another check: Let me think if there was a miscalculation when I multiplied both sides by x. Let me go through that step again.

Original equation:

3 + (1/x) = 7/x

Multiply both sides by x:

Left side: 3 * x + (1/x) * x = 3x + 1

Right side: (7/x) * x = 7

So, 3x + 1 = 7. So that is correct.

Therefore, solving 3x +1 = 7 => 3x = 6 => x = 2. Correct.

Therefore, all steps check out.

So, the answer is 2.

Wait, but just to make sure there are no alternative interpretations of the original problem. Let me check the wording again: "Three plus the reciprocal of a number equals 7 divided by that number." So, three plus (reciprocal of a number) equals (7 divided by that number). Which is 3 + (1/x) = 7/x. So, that is correct.

Alternatively, could someone misinterpret it as "three plus the reciprocal of a number equals 7 divided by that number," but maybe "three plus the reciprocal of a number" could be misread as (3 + reciprocal) versus something else? But I don't think so. The wording is straightforward.

Alternatively, maybe someone might think that "three plus the reciprocal of a number" is 3 + reciprocal(3 + number)? But that would be a different interpretation. But the original sentence is: "Three plus the reciprocal of a number equals 7 divided by that number." So, "the reciprocal of a number" is 1/x, so three plus that. So, no, the correct interpretation is 3 + 1/x.

Therefore, I think my solution is correct.

So, the number is 2.

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

We are given the equation:

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

Our goal is to find the value of $ x $.

---

### Step 1: Eliminate the fractions

To simplify the equation, we can multiply both sides by $ x $, assuming $ x \ne 0 $ (since a reciprocal is undefined when the number is zero):

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

Now, simplify both sides:

- Left side: $ 3x + 1 $
- Right side: $ 7 $

This gives the simplified equation:

$$
3x + 1 = 7
$$

---

### Step 2: Solve for $ x $

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.

---

### Final Answer

$$
\boxed{2}
$$

[Assistant]
Explanation:
Okay, let's see. The problem is: Three plus the reciprocal of a number equals 7 divided by that number. I need to find the number. Hmm, let me try to translate this into an equation.

First, let me denote the number as something. Let me call it x. So, the number is x. Then, the reciprocal of the number would be 1/x. The problem says three plus the reciprocal of the number equals 7 divided by that number.

So, let me write that out. Three is 3, so 3 plus 1/x equals 7 divided by x. So, the equation is:

3 + (1/x) = 7/x

Alright, so now I need to solve for x. Let me think about how to do this. Since there are fractions here, maybe I can get rid of the denominators by multiplying both sides of the equation by x. That should eliminate the denominators. Let me try that.

Multiplying both sides by x:

x * [3 + (1/x)] = x * (7/x)

Let me simplify both sides. On the left side, distributing the x:

x * 3 + x * (1/x) = 3x + 1

On the right side, x times 7 over x is 7, because the x cancels out. So:

3x + 1 = 7

Okay, that seems simpler. Now, solving for x. Let me subtract 1 from both sides:

3x + 1 - 1 = 7 - 1

Which simplifies to:

3x = 6

Then, divide both sides by 3:

x = 6 / 3

So, x = 2

Wait, let me check if that makes sense. Let me substitute x = 2 back into the original equation to verify.

Original equation: 3 + (1/x) = 7/x

Left side: 3 + (1/2) = 3.5 or 7/2

Right side: 7 / 2 = 3.5

Yes, both sides are equal. So, x = 2 is indeed the solution.

Wait, but let me think again. What if there was another solution? Because sometimes when you multiply both sides by x, you might assume x is not zero, but since reciprocal is involved, x can't be zero anyway. So, x ≠ 0, so that's okay.

But let me check if there's another possible solution. Let me try solving the equation again step by step. Maybe I made a mistake?

Original equation:

3 + 1/x = 7/x

Subtract 1/x from both sides:

3 = 7/x - 1/x

Which is 3 = (7 - 1)/x => 3 = 6/x

Then, solving for x: 3 = 6/x => x = 6/3 => x = 2

Same result. So, that's consistent.

Alternatively, if I didn't multiply both sides by x, but instead brought all terms to one side:

3 + 1/x - 7/x = 0

Which simplifies to 3 - 6/x = 0

Then, 3 = 6/x => same as before.

So, again, x = 2.

Therefore, the solution is 2.

But wait, sometimes when you have equations with reciprocals, there might be extraneous solutions or something? But in this case, since we only had one solution, and it checks out, so that's the answer.

Alternatively, let me think if there is a possibility of another solution. Let me suppose that maybe I made an error in assuming that multiplying both sides by x is allowed. But as long as x is not zero, which it can't be because of reciprocal, so multiplying both sides by x is valid.

So, no, I don't think there's another solution. So, x must be 2.

Let me also think if there's any alternative way to approach this problem. Maybe by letting the number be something else? But no, I think the way I did it is straightforward.

Alternatively, maybe graphing? If I consider the left side as a function f(x) = 3 + 1/x and the right side as g(x) = 7/x. The intersection point would be the solution. But since I already solved algebraically, that's probably unnecessary.

But just for thoroughness, let's think. For x positive, as x approaches zero from the right, f(x) approaches infinity, and g(x) also approaches infinity. But which one is higher? Let me check for very small x, say x approaching 0+. Then 3 + 1/x is dominated by 1/x, which is positive and large, while 7/x is also positive and large. Let me compare their rates. Let me take x approaching 0+, then f(x) - g(x) = 3 + 1/x - 7/x = 3 - 6/x. As x approaches 0+, 3 - 6/x approaches negative infinity. So, f(x) - g(x) approaches negative infinity. So, for very small x, f(x) < g(x). At x = 2, they are equal. What about as x approaches infinity? Both f(x) and g(x) approach 3 and 0, respectively. So, f(x) approaches 3, and g(x) approaches 0, so f(x) > g(x) for large x. Therefore, since f(x) - g(x) goes from negative infinity at x approaching 0+ to positive infinity as x approaches infinity, and since it's continuous for x > 0 (except at x=0), there must be exactly one solution where they cross. Which is at x=2. So that confirms there's only one solution.

What about negative numbers? The problem didn't specify if the number is positive or negative. Let me check if there's a negative solution. Suppose x is negative. Let me try solving the equation again, but considering x negative.

Original equation: 3 + 1/x = 7/x

Subtract 1/x from both sides: 3 = 6/x => x = 6/3 = 2. Wait, so even if I consider x negative, solving gives me x=2, which is positive. So, there's no negative solution? Because if I rearrange the equation, 3 = 6/x implies x = 2, no matter what. Therefore, x must be 2. Therefore, there is only one solution.

But just to be thorough, let me plug in a negative number. Suppose x = -1. Then, left side: 3 + 1/(-1) = 3 -1 = 2. Right side: 7 / (-1) = -7. 2 ≠ -7. How about x = -2? Left side: 3 + 1/(-2) = 3 - 0.5 = 2.5. Right side: 7 / (-2) = -3.5. Not equal. What about x = -something else? Let me try solving for x negative. Let me go back to equation:

Original equation: 3 + 1/x = 7/x

Multiply both sides by x, assuming x ≠ 0:

3x + 1 = 7

Which gives 3x = 6 => x = 2. So, regardless of sign, the algebra leads to x=2. Therefore, x must be 2. So, even if I consider negative numbers, the algebra shows that x must be positive 2. Therefore, there are no negative solutions.

Therefore, the only solution is x=2.

Another check: Let me think if there was a miscalculation when I multiplied both sides by x. Let me go through that step again.

Original equation:

3 + (1/x) = 7/x

Multiply both sides by x:

Left side: 3 * x + (1/x) * x = 3x + 1

Right side: (7/x) * x = 7

So, 3x + 1 = 7. So that is correct.

Therefore, solving 3x +1 = 7 => 3x = 6 => x = 2. Correct.

Therefore, all steps check out.

So, the answer is 2.

Wait, but just to make sure there are no alternative interpretations of the original problem. Let me check the wording again: "Three plus the reciprocal of a number equals 7 divided by that number." So, three plus (reciprocal of a number) equals (7 divided by that number). Which is 3 + (1/x) = 7/x. So, that is correct.

Alternatively, could someone misinterpret it as "three plus the reciprocal of a number equals 7 divided by that number," but maybe "three plus the reciprocal of a number" could be misread as (3 + reciprocal) versus something else? But I don't think so. The wording is straightforward.

Alternatively, maybe someone might think that "three plus the reciprocal of a number" is 3 + reciprocal(3 + number)? But that would be a different interpretation. But the original sentence is: "Three plus the reciprocal of a number equals 7 divided by that number." So, "the reciprocal of a number" is 1/x, so three plus that. So, no, the correct interpretation is 3 + 1/x.

Therefore, I think my solution is correct.

So, the number is 2.

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

We are given the equation:

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

Our goal is to find the value of $ x $.

---

### Step 1: Eliminate the fractions

To simplify the equation, we can multiply both sides by $ x $, assuming $ x \ne 0 $ (since a reciprocal is undefined when the number is zero):

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

Now, simplify both sides:

- Left side: $ 3x + 1 $
- Right side: $ 7 $

This gives the simplified equation:

$$
3x + 1 = 7
$$

---

### Step 2: Solve for $ x $

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.

---

### 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 $x = \boxed{2}$.


Explanation:
Okay, so I need to find the greatest possible perimeter of a triangle where two sides are each 8 units long, and the third side has a whole number length. Let me think about how to approach this.

First, I remember that in any triangle, the sum of the lengths of any two sides must be greater than the length of the remaining side. This is called the triangle inequality theorem. So, if two sides are 8 units each, let me denote the third side as 'x'.

So, the sides of the triangle are 8, 8, and x. Now, applying the triangle inequality theorem, I need to make sure that the sum of any two sides is greater than the third side. Let me write down the inequalities:

1. 8 + 8 > x
2. 8 + x > 8
3. 8 + x > 8

Wait, actually, the second and third inequalities are the same because the sides are symmetric. Let me check that again.

For the first inequality: 8 + 8 > x => 16 > x, so x must be less than 16.

For the second inequality: 8 + x > 8 => x > 0. But since x is a length, it's already positive, so this doesn't really restrict it. Similarly, the third inequality is the same as the second one.

Therefore, the main restriction here is that x has to be less than 16. But also, since the third side must be a positive length, x has to be greater than 0. However, since we are looking for a whole number length, x can be any integer from 1 up to 15, right?

But wait, hold on. Let me make sure. The triangle inequality theorem requires that the sum of the two smaller sides must be greater than the largest side. If the third side is longer than 8, then the two sides of 8 would have to add up to more than that third side. If the third side is shorter than 8, then the two sides of 8 would already be the larger sides. Let me think.

If x is the third side, then depending on whether x is longer or shorter than 8, the inequalities might be different. But actually, since we have two sides of 8, let me think about the possible ranges for x.

If x is the third side, then:

Case 1: x is the longest side. Then, the sum of the other two sides must be greater than x. So, 8 + 8 > x => 16 > x => x < 16.

Case 2: x is not the longest side. In this case, since the other two sides are both 8, if x is less than 8, then the longest side is 8, so the sum of the other two sides (which would be 8 and x) must be greater than the third side, which is 8. So, 8 + x > 8 => x > 0. Which is already true because x is a positive integer. So, actually, there's no additional constraint here except x has to be positive.

So, combining both cases, the only constraint is that x must be less than 16. Also, x must be a positive integer. So x can be from 1 to 15. However, there is another thing to consider. If x is less than 8, is there any other restriction? Let me check with the triangle inequality for another pair.

Wait, if x is the third side, then the other inequalities would be:

8 + x > 8 (already satisfied), and 8 + x > 8 (same as above). So, no new constraints. Therefore, the only constraint is x < 16. But x also has to be a positive integer. Therefore, x can be any integer from 1 to 15.

But hold on, the problem says the third side has a whole number length. So, the perimeter would be 8 + 8 + x = 16 + x. So, to maximize the perimeter, we need to maximize x. Since x can be at most 15, then the maximum perimeter would be 16 + 15 = 31.

Wait, but let me check if x can be 15. Let me verify the triangle inequality for x = 15. So, sides are 8, 8, 15. Then, 8 + 8 > 15? 16 > 15? Yes, that's true. 8 + 15 > 8? 23 > 8? Yes. 8 + 15 > 8? Same as above. So, yes, it's a valid triangle. Therefore, perimeter is 8 + 8 + 15 = 31.

But wait, let me check if there's any other constraints. For example, in some cases, even if x is less than 16, maybe there's another condition? Wait, maybe not. Because the triangle inequality only requires that the sum of the two smaller sides is greater than the largest side. So, if x is the largest side, then 8 + 8 > x. If x is not the largest side, then the largest side is 8, so the sum of the other two sides (which would be 8 and x) must be greater than 8. Which is automatically true because 8 + x > 8 implies x > 0. So, since x is a positive integer, that's satisfied. Therefore, x can be from 1 to 15. Therefore, the maximum value of x is 15, leading to the maximum perimeter of 31.

But wait, let me think again. Sometimes, there's a confusion with the triangle inequality. Let me check with x = 15. If you have sides 8, 8, 15, then when you try to form a triangle, the sum of the two smaller sides is 8 + 8 = 16, which is greater than 15. So, that works. If x was 16, the sum would be 8 + 8 = 16, which is not greater than 16, so that's degenerate triangle. So, x can't be 16. Therefore, 15 is the maximum.

But then why do I have a feeling that there might be something wrong here? Wait, maybe not. Maybe the answer is indeed 31. However, let me check with another example. Suppose x is 1. Then sides are 8, 8, 1. Then, 8 + 1 > 8? 9 > 8, yes. So that's okay. So, that's a valid triangle. Similarly, for x = 15, it's valid. Therefore, the maximum perimeter is 8 + 8 + 15 = 31. Therefore, the answer is 31.

But let me check again if there's any mistake here. Maybe I need to consider that the third side can't be too short? For example, if x is 1, the triangle is very "flat", but it is still a valid triangle. So, according to the triangle inequality, as long as the sum of any two sides is greater than the third, it's okay. So, even if x is 1, 8 + 1 > 8, which is true. So, yes, that works.

Wait, but maybe there's another way to think about it. If you have two sides of 8, the third side must be less than 16 and greater than... well, greater than 0. But since it's a triangle, the third side also has to be such that the triangle is non-degenerate. But in our case, as long as x < 16, the triangle is non-degenerate.

Alternatively, if we think of the third side, the minimum length would be greater than 0, but practically, the third side must be at least 1 unit. Because if it's 0, it's not a triangle. So, since the problem says the third side has a whole number length, so x is a positive integer, so minimum 1. Therefore, maximum perimeter would be when x is maximum, which is 15. Therefore, 8 + 8 + 15 = 31.

Wait, but here's another thought. If the problem is from an Olympiad or a standard test, maybe they consider something else? Let me think. Maybe there's some other constraint? For example, sometimes problems specify that the triangle is non-degenerate, but we already considered that by having x < 16.

Alternatively, maybe I need to check if the third side can be 15. If you have two sides of 8 and one of 15, then the triangle is valid. Let me imagine it. If you have two sides of 8, and the third side is 15. So, the two sides of 8 would have to be almost overlapping with the third side. But as long as the sum of the two sides is greater than the third, it's okay. So, yes, 8 + 8 > 15 (16 > 15) is true. So, that's okay.

Therefore, seems like 31 is the correct answer. However, maybe there is a mistake here? Let me check with x = 15. Let me calculate the perimeter. 8 + 8 + 15 = 31. If I take x = 16, it's 8 + 8 + 16 = 32, but that's invalid because 8 + 8 = 16, which is not greater than 16, so it's a degenerate triangle. So, not allowed.

Therefore, the answer is 31. Wait, but I recall that sometimes problems might have other constraints. Let me check the problem again.

"Two sides of a triangle are each 8 units long. If the third side has a whole number length, what is the greatest possible perimeter, in units, for the triangle?"

No, there's no other constraints. So, third side is a whole number, meaning integer. So, possible values from 1 to 15 inclusive. Therefore, maximum perimeter is 31.

But let me think again. Suppose someone argues that the third side must be greater than 0, but also, in some contexts, the third side must be such that the triangle is not just a line. But as long as it's greater than 0, it's a valid triangle. So, if x is 1, then it's a valid triangle. So, 31 is the maximum.

Alternatively, maybe the problem is expecting a different answer. Wait, let me check if I made a mistake in the triangle inequality.

The triangle inequality says that the sum of any two sides must be greater than the third side. So, for sides 8, 8, and x:

1. 8 + 8 > x => 16 > x
2. 8 + x > 8 => x > 0
3. 8 + x > 8 => same as above.

So, the only restriction is x < 16 and x > 0, so x can be 1 to 15. Therefore, maximum x is 15, perimeter 31. So, answer is 31.

Wait, but maybe the problem is in the interpretation. If the third side has a whole number length, does that mean that it has to be an integer? So, yes, that's how I interpreted it. So, whole number length means integer, so 1 to 15. Therefore, 31 is correct.

Alternatively, maybe there is a different approach. For example, if we want to maximize the perimeter, we need to maximize x. So, maximum x is less than 16, so maximum integer is 15. Therefore, perimeter is 8 + 8 + 15 = 31.

So, I think that's solid. Therefore, I think the answer is 31.

**Final Answer**
The greatest possible perimeter is \boxed{31}.

To solve this problem, we are given that two sides of a triangle are each 8 units long, and the third side, denoted as $ x $, must have a **whole number length**. Our goal is to find the **greatest possible perimeter** of such a triangle.

---

### Step 1: Apply the Triangle Inequality Theorem

The triangle inequality theorem states that for any triangle with side lengths $ a $, $ b $, and $ c $, the following must be true for all three:

$$
a + b > c \\
a + c > b \\
b + c > a
$$

Given that the two sides are 8 units each, and the third side is $ x $, this becomes:

- $ 8 + 8 > x \Rightarrow 16 > x $ (i.e., $ x < 16 $)
- $ 8 + x > 8 \Rightarrow x > 0 $ (always true since $ x $ is a positive length)
- The third inequality is identical to the second one

So the key constraint is:

$$
0 < x < 16
$$

Since $ x $ must be a **whole number**, the possible values are:

$$
x = 1, 2, 3, \ldots, 15
$$

---

### Step 2: Maximize the Perimeter

The perimeter $ P $ of the triangle is:

$$
P = 8 + 8 + x = 16 + x
$$

To **maximize the perimeter**, we need to choose the largest possible value for $ x $. The maximum integer value of $ x $ that satisfies $ x < 16 $ is:

$$
x = 15
$$

Substitute that into the perimeter formula:

$$
P = 16 + 15 = \boxed{31}
$$

---

### Step 3: Verify Triangle Validity

We should ensure that the triangle with side lengths 8, 8, and 15 forms a valid triangle.

- $ 8 + 8 > 15 \Rightarrow 16 > 15 $ ✅ (True)
- $ 8 + 15 > 8 \Rightarrow 23 > 8 $ ✅ (True)
- $ 8 + 15 > 8 $ (same as above) ✅

So, the triangle inequality is satisfied.

---

### Final Answer

The **greatest possible perimeter** for a triangle with two sides of 8 units and the third side being a whole number is:

$$
\boxed{31}
$$

[Assistant]
Explanation:
Okay, so I need to find the greatest possible perimeter of a triangle where two sides are each 8 units long, and the third side has a whole number length. Let me think about how to approach this.

First, I remember that in any triangle, the sum of the lengths of any two sides must be greater than the length of the remaining side. This is called the triangle inequality theorem. So, if two sides are 8 units each, let me denote the third side as 'x'.

So, the sides of the triangle are 8, 8, and x. Now, applying the triangle inequality theorem, I need to make sure that the sum of any two sides is greater than the third side. Let me write down the inequalities:

1. 8 + 8 > x
2. 8 + x > 8
3. 8 + x > 8

Wait, actually, the second and third inequalities are the same because the sides are symmetric. Let me check that again.

For the first inequality: 8 + 8 > x => 16 > x, so x must be less than 16.

For the second inequality: 8 + x > 8 => x > 0. But since x is a length, it's already positive, so this doesn't really restrict it. Similarly, the third inequality is the same as the second one.

Therefore, the main restriction here is that x has to be less than 16. But also, since the third side must be a positive length, x has to be greater than 0. However, since we are looking for a whole number length, x can be any integer from 1 up to 15, right?

But wait, hold on. Let me make sure. The triangle inequality theorem requires that the sum of the two smaller sides must be greater than the largest side. If the third side is longer than 8, then the two sides of 8 would have to add up to more than that third side. If the third side is shorter than 8, then the two sides of 8 would already be the larger sides. Let me think.

If x is the third side, then depending on whether x is longer or shorter than 8, the inequalities might be different. But actually, since we have two sides of 8, let me think about the possible ranges for x.

If x is the third side, then:

Case 1: x is the longest side. Then, the sum of the other two sides must be greater than x. So, 8 + 8 > x => 16 > x => x < 16.

Case 2: x is not the longest side. In this case, since the other two sides are both 8, if x is less than 8, then the longest side is 8, so the sum of the other two sides (which would be 8 and x) must be greater than the third side, which is 8. So, 8 + x > 8 => x > 0. Which is already true because x is a positive integer. So, actually, there's no additional constraint here except x has to be positive.

So, combining both cases, the only constraint is that x must be less than 16. Also, x must be a positive integer. So x can be from 1 to 15. However, there is another thing to consider. If x is less than 8, is there any other restriction? Let me check with the triangle inequality for another pair.

Wait, if x is the third side, then the other inequalities would be:

8 + x > 8 (already satisfied), and 8 + x > 8 (same as above). So, no new constraints. Therefore, the only constraint is x < 16. But x also has to be a positive integer. Therefore, x can be any integer from 1 to 15.

But hold on, the problem says the third side has a whole number length. So, the perimeter would be 8 + 8 + x = 16 + x. So, to maximize the perimeter, we need to maximize x. Since x can be at most 15, then the maximum perimeter would be 16 + 15 = 31.

Wait, but let me check if x can be 15. Let me verify the triangle inequality for x = 15. So, sides are 8, 8, 15. Then, 8 + 8 > 15? 16 > 15? Yes, that's true. 8 + 15 > 8? 23 > 8? Yes. 8 + 15 > 8? Same as above. So, yes, it's a valid triangle. Therefore, perimeter is 8 + 8 + 15 = 31.

But wait, let me check if there's any other constraints. For example, in some cases, even if x is less than 16, maybe there's another condition? Wait, maybe not. Because the triangle inequality only requires that the sum of the two smaller sides is greater than the largest side. So, if x is the largest side, then 8 + 8 > x. If x is not the largest side, then the largest side is 8, so the sum of the other two sides (which would be 8 and x) must be greater than 8. Which is automatically true because 8 + x > 8 implies x > 0. So, since x is a positive integer, that's satisfied. Therefore, x can be from 1 to 15. Therefore, the maximum value of x is 15, leading to the maximum perimeter of 31.

But wait, let me think again. Sometimes, there's a confusion with the triangle inequality. Let me check with x = 15. If you have sides 8, 8, 15, then when you try to form a triangle, the sum of the two smaller sides is 8 + 8 = 16, which is greater than 15. So, that works. If x was 16, the sum would be 8 + 8 = 16, which is not greater than 16, so that's degenerate triangle. So, x can't be 16. Therefore, 15 is the maximum.

But then why do I have a feeling that there might be something wrong here? Wait, maybe not. Maybe the answer is indeed 31. However, let me check with another example. Suppose x is 1. Then sides are 8, 8, 1. Then, 8 + 1 > 8? 9 > 8, yes. So that's okay. So, that's a valid triangle. Similarly, for x = 15, it's valid. Therefore, the maximum perimeter is 8 + 8 + 15 = 31. Therefore, the answer is 31.

But let me check again if there's any mistake here. Maybe I need to consider that the third side can't be too short? For example, if x is 1, the triangle is very "flat", but it is still a valid triangle. So, according to the triangle inequality, as long as the sum of any two sides is greater than the third, it's okay. So, even if x is 1, 8 + 1 > 8, which is true. So, yes, that works.

Wait, but maybe there's another way to think about it. If you have two sides of 8, the third side must be less than 16 and greater than... well, greater than 0. But since it's a triangle, the third side also has to be such that the triangle is non-degenerate. But in our case, as long as x < 16, the triangle is non-degenerate.

Alternatively, if we think of the third side, the minimum length would be greater than 0, but practically, the third side must be at least 1 unit. Because if it's 0, it's not a triangle. So, since the problem says the third side has a whole number length, so x is a positive integer, so minimum 1. Therefore, maximum perimeter would be when x is maximum, which is 15. Therefore, 8 + 8 + 15 = 31.

Wait, but here's another thought. If the problem is from an Olympiad or a standard test, maybe they consider something else? Let me think. Maybe there's some other constraint? For example, sometimes problems specify that the triangle is non-degenerate, but we already considered that by having x < 16.

Alternatively, maybe I need to check if the third side can be 15. If you have two sides of 8 and one of 15, then the triangle is valid. Let me imagine it. If you have two sides of 8, and the third side is 15. So, the two sides of 8 would have to be almost overlapping with the third side. But as long as the sum of the two sides is greater than the third, it's okay. So, yes, 8 + 8 > 15 (16 > 15) is true. So, that's okay.

Therefore, seems like 31 is the correct answer. However, maybe there is a mistake here? Let me check with x = 15. Let me calculate the perimeter. 8 + 8 + 15 = 31. If I take x = 16, it's 8 + 8 + 16 = 32, but that's invalid because 8 + 8 = 16, which is not greater than 16, so it's a degenerate triangle. So, not allowed.

Therefore, the answer is 31. Wait, but I recall that sometimes problems might have other constraints. Let me check the problem again.

"Two sides of a triangle are each 8 units long. If the third side has a whole number length, what is the greatest possible perimeter, in units, for the triangle?"

No, there's no other constraints. So, third side is a whole number, meaning integer. So, possible values from 1 to 15 inclusive. Therefore, maximum perimeter is 31.

But let me think again. Suppose someone argues that the third side must be greater than 0, but also, in some contexts, the third side must be such that the triangle is not just a line. But as long as it's greater than 0, it's a valid triangle. So, if x is 1, then it's a valid triangle. So, 31 is the maximum.

Alternatively, maybe the problem is expecting a different answer. Wait, let me check if I made a mistake in the triangle inequality.

The triangle inequality says that the sum of any two sides must be greater than the third side. So, for sides 8, 8, and x:

1. 8 + 8 > x => 16 > x
2. 8 + x > 8 => x > 0
3. 8 + x > 8 => same as above.

So, the only restriction is x < 16 and x > 0, so x can be 1 to 15. Therefore, maximum x is 15, perimeter 31. So, answer is 31.

Wait, but maybe the problem is in the interpretation. If the third side has a whole number length, does that mean that it has to be an integer? So, yes, that's how I interpreted it. So, whole number length means integer, so 1 to 15. Therefore, 31 is correct.

Alternatively, maybe there is a different approach. For example, if we want to maximize the perimeter, we need to maximize x. So, maximum x is less than 16, so maximum integer is 15. Therefore, perimeter is 8 + 8 + 15 = 31.

So, I think that's solid. Therefore, I think the answer is 31.

**Final Answer**
The greatest possible perimeter is \boxed{31}.

To solve this problem, we are given that two sides of a triangle are each 8 units long, and the third side, denoted as $ x $, must have a **whole number length**. Our goal is to find the **greatest possible perimeter** of such a triangle.

---

### Step 1: Apply the Triangle Inequality Theorem

The triangle inequality theorem states that for any triangle with side lengths $ a $, $ b $, and $ c $, the following must be true for all three:

$$
a + b > c \\
a + c > b \\
b + c > a
$$

Given that the two sides are 8 units each, and the third side is $ x $, this becomes:

- $ 8 + 8 > x \Rightarrow 16 > x $ (i.e., $ x < 16 $)
- $ 8 + x > 8 \Rightarrow x > 0 $ (always true since $ x $ is a positive length)
- The third inequality is identical to the second one

So the key constraint is:

$$
0 < x < 16
$$

Since $ x $ must be a **whole number**, the possible values are:

$$
x = 1, 2, 3, \ldots, 15
$$

---

### Step 2: Maximize the Perimeter

The perimeter $ P $ of the triangle is:

$$
P = 8 + 8 + x = 16 + x
$$

To **maximize the perimeter**, we need to choose the largest possible value for $ x $. The maximum integer value of $ x $ that satisfies $ x < 16 $ is:

$$
x = 15
$$

Substitute that into the perimeter formula:

$$
P = 16 + 15 = \boxed{31}
$$

---

### Step 3: Verify Triangle Validity

We should ensure that the triangle with side lengths 8, 8, and 15 forms a valid triangle.

- $ 8 + 8 > 15 \Rightarrow 16 > 15 $ ✅ (True)
- $ 8 + 15 > 8 \Rightarrow 23 > 8 $ ✅ (True)
- $ 8 + 15 > 8 $ (same as above) ✅

So, the triangle inequality is satisfied.

---

### Final Answer

The **greatest possible perimeter** for a triangle with two sides of 8 units and the third side being a whole number is:

$$
\boxed{31}
$$

Solution: The Triangle Inequality says that the sum of the lengths of any two sides must be greater than the length of the third side. That means $8+8=16$ must be greater than the length of the third side. The third side has a whole number length, so the greatest possible length is 15 units. That makes the perimeter $8+8+15=\boxed{31}$ units.