Explanation:
Okay, so I need to figure out how many distinct arrangements there are for putting 2 indistinguishable IMO gold medals and 2 indistinguishable IPhO gold medals in a row. Let me think about this step by step.

First, let me make sure I understand the problem correctly. There are four medals in total: two of one type (IMO gold) and two of another type (IPhO gold). The medals of the same type are indistinguishable, meaning that swapping two IMO medals doesn't create a new arrangement. Similarly for the IPhO medals. The question is asking for the number of distinct ways to arrange these four medals in a row.

Hmm, so this seems like a permutations problem with identical items. I remember that when you have items that are not all distinct, the number of distinct arrangements is given by the factorial of the total number of items divided by the product of the factorials of the numbers of each identical item. The formula is something like:

Number of arrangements = n! / (n1! * n2! * ... * nk!)

Where n is the total number of items, and n1, n2, ..., nk are the numbers of each identical item.

In this case, we have 4 medals total. There are two types: IMO and IPhO. Each has 2 medals. So applying the formula, it should be 4! divided by (2! * 2!). Let me check that.

Calculating 4! is 24. Then 2! is 2, so 2! * 2! is 2 * 2 = 4. Then 24 divided by 4 is 6. So does that mean there are 6 distinct arrangements?

Wait, let me verify this by actually listing all possible arrangements to make sure I didn't make a mistake. Sometimes with small numbers, it's easy to list them out and count.

Let me denote the IMO medals as I and the IPhO medals as P. Since they are indistinct, we can't tell the difference between the two I's or the two P's.

So the possible arrangements are all the different ways to arrange two I's and two P's. Let me list them:

1. I, I, P, P
2. I, P, I, P
3. I, P, P, I
4. P, I, I, P
5. P, I, P, I
6. P, P, I, I

Is that all? Let me check if there are more. Let's think of it as permutations of the letters I and P with two of each. The number of unique permutations is indeed 4! / (2!2!) = 6, which matches the list above. So that seems correct.

But wait, let me think again. Is there a possibility that the problem is different? For example, maybe the medals are placed in a row, but there's some other constraint? The problem says "put 2 of his indistinguishable IMO gold medals and 2 of his indistinguishable IPhO gold medals in one row." So it's just arranging four medals, two of each type, in a row. So yeah, the answer should be 6.

But let me think again if there's another way to approach this. Maybe using combinations? For example, choosing positions for the IMO medals and the rest will be IPhO.

So, there are 4 positions. We need to choose 2 positions to place the IMO medals, and the remaining 2 will be IPhO. The number of ways to choose 2 positions out of 4 is C(4,2), which is 6. That's the same answer as before. So that confirms it again.

Alternatively, if I think of it as permutations with identical elements, the formula is the same. So both methods give me 6. Therefore, I think the answer is 6.

But just to be thorough, let me think if there's any chance that the medals are considered different in some way. The problem says they are indistinguishable. So, for example, if the IMO medals are different from each other, but the problem says they are indistinguishable. Similarly for IPhO. So no, there's no difference between the medals of the same type. Therefore, the answer is indeed 6.

Wait, but maybe the problem is in another language? The original problem mentions IMO and IPhO. But the user wrote the problem in English, so I think that's fine. The key is that the medals are indistinct within their types.

Alternatively, maybe there's a trick here? For example, if the medals are placed in a row, but maybe the row has some symmetry? But no, the problem doesn't mention anything about rotations or reflections being considered the same. It just says "in one row," so I think linear arrangements where the order matters, and the only thing that matters is the sequence of medals.

Therefore, I think my reasoning is correct. The answer is 6. Therefore, the number of distinct arrangements is 6.

But just to make sure, let me think of another approach. Suppose I try to count the number of permutations step by step.

First, the first position can be either I or P. Let's consider cases.

Case 1: First position is I.

Then, we have one I and two P's left. The remaining three positions need to be filled with one I and two P's. The number of arrangements for this case is the number of ways to arrange one I and two P's in three positions, which is C(3,1) = 3.

Case 2: First position is P.

Similarly, then we have two I's and one P left. The number of arrangements for this case is the number of ways to arrange two I's and one P in three positions, which is C(3,2) = 3.

Therefore, total number of arrangements is 3 + 3 = 6. Yep, same answer again. So that's consistent.

Alternatively, if I think recursively, but that might be overcomplicating. But just to check, suppose I think of the first position: 2 choices (I or P). Then, depending on that, the next positions have different numbers. But as above, it still leads to 6.

Alternatively, maybe using generating functions? Although that might be more complex. The exponential generating function for each type is (1 + x)^2, but since we have two types, maybe not. Wait, actually, generating functions for permutations with identical items... Maybe not necessary here.

Alternatively, think of the problem as arranging letters. If we have the word "IIP", but with two I's and two P's. The number of distinct permutations is 4! / (2!2!) = 6. Yep, same thing.

Therefore, all methods lead to 6. Therefore, I think the answer is 6. Therefore, the number of distinct arrangements is 6.

But just to make sure, let me list all the possibilities again:

1. II PP
2. I P I P
3. I PP I
4. P II P
5. P I P I
6. PP II

Yes, that's six. Each of these is unique. For example, the first one is two I's followed by two P's. The second is alternating I and P starting with I. The third is I, then two P's, then I. The fourth is P, then two I's, then P. The fifth is alternating starting with P. The sixth is two P's followed by two I's. So that's all six. There's no other arrangement possible. For example, if I try to make something like I P P I, that's already listed as number 3. Similarly, P I I P is number 4. So yeah, all possibilities are covered.

Therefore, I think I can confidently say that the answer is 6. Therefore, the number of distinct arrangements is 6.

**Final Answer**
The number of distinct arrangements is \boxed{6}.

To determine the number of distinct arrangements of 2 indistinguishable IMO gold medals and 2 indistinguishable IPhO gold medals in a row, we approach the problem using the concept of **permutations with identical items**.

---

### Step 1: Understanding the Problem

We are arranging **4 medals** in a row:
- 2 of them are **indistinguishable IMO gold medals** (denoted as `I`)
- 2 of them are **indistinguishable IPhO gold medals** (denoted as `P`)

Since the medals of the same type are **indistinguishable**, swapping two `I`s or two `P`s does **not** result in a new arrangement.

---

### Step 2: Applying the Permutation Formula

The general formula for the number of distinct permutations of a multiset is:

$$
\text{Number of arrangements} = \frac{n!}{n_1! \cdot n_2! \cdot \ldots \cdot n_k!}
$$

Where:
- $ n $ is the total number of items (here, $ n = 4 $)
- $ n_1, n_2, \ldots, n_k $ are the counts of each type of indistinguishable item

In this case:
- $ n = 4 $
- $ n_1 = 2 $ (number of `I`s)
- $ n_2 = 2 $ (number of `P`s)

So the number of distinct arrangements is:

$$
\frac{4!}{2! \cdot 2!} = \frac{24}{2 \cdot 2} = \frac{24}{4} = 6
$$

---

### Step 3: Verifying by Enumeration

To confirm, we can list all the distinct arrangements of the letters `I` and `P` with two of each:

1. `I I P P`
2. `I P I P`
3. `I P P I`
4. `P I I P`
5. `P I P I`
6. `P P I I`

These are all the unique permutations, and there are exactly **6** of them.

---

### Final Answer

$$
\boxed{6}
$$

[Assistant]
Explanation:
Okay, so I need to figure out how many distinct arrangements there are for putting 2 indistinguishable IMO gold medals and 2 indistinguishable IPhO gold medals in a row. Let me think about this step by step.

First, let me make sure I understand the problem correctly. There are four medals in total: two of one type (IMO gold) and two of another type (IPhO gold). The medals of the same type are indistinguishable, meaning that swapping two IMO medals doesn't create a new arrangement. Similarly for the IPhO medals. The question is asking for the number of distinct ways to arrange these four medals in a row.

Hmm, so this seems like a permutations problem with identical items. I remember that when you have items that are not all distinct, the number of distinct arrangements is given by the factorial of the total number of items divided by the product of the factorials of the numbers of each identical item. The formula is something like:

Number of arrangements = n! / (n1! * n2! * ... * nk!)

Where n is the total number of items, and n1, n2, ..., nk are the numbers of each identical item.

In this case, we have 4 medals total. There are two types: IMO and IPhO. Each has 2 medals. So applying the formula, it should be 4! divided by (2! * 2!). Let me check that.

Calculating 4! is 24. Then 2! is 2, so 2! * 2! is 2 * 2 = 4. Then 24 divided by 4 is 6. So does that mean there are 6 distinct arrangements?

Wait, let me verify this by actually listing all possible arrangements to make sure I didn't make a mistake. Sometimes with small numbers, it's easy to list them out and count.

Let me denote the IMO medals as I and the IPhO medals as P. Since they are indistinct, we can't tell the difference between the two I's or the two P's.

So the possible arrangements are all the different ways to arrange two I's and two P's. Let me list them:

1. I, I, P, P
2. I, P, I, P
3. I, P, P, I
4. P, I, I, P
5. P, I, P, I
6. P, P, I, I

Is that all? Let me check if there are more. Let's think of it as permutations of the letters I and P with two of each. The number of unique permutations is indeed 4! / (2!2!) = 6, which matches the list above. So that seems correct.

But wait, let me think again. Is there a possibility that the problem is different? For example, maybe the medals are placed in a row, but there's some other constraint? The problem says "put 2 of his indistinguishable IMO gold medals and 2 of his indistinguishable IPhO gold medals in one row." So it's just arranging four medals, two of each type, in a row. So yeah, the answer should be 6.

But let me think again if there's another way to approach this. Maybe using combinations? For example, choosing positions for the IMO medals and the rest will be IPhO.

So, there are 4 positions. We need to choose 2 positions to place the IMO medals, and the remaining 2 will be IPhO. The number of ways to choose 2 positions out of 4 is C(4,2), which is 6. That's the same answer as before. So that confirms it again.

Alternatively, if I think of it as permutations with identical elements, the formula is the same. So both methods give me 6. Therefore, I think the answer is 6.

But just to be thorough, let me think if there's any chance that the medals are considered different in some way. The problem says they are indistinguishable. So, for example, if the IMO medals are different from each other, but the problem says they are indistinguishable. Similarly for IPhO. So no, there's no difference between the medals of the same type. Therefore, the answer is indeed 6.

Wait, but maybe the problem is in another language? The original problem mentions IMO and IPhO. But the user wrote the problem in English, so I think that's fine. The key is that the medals are indistinct within their types.

Alternatively, maybe there's a trick here? For example, if the medals are placed in a row, but maybe the row has some symmetry? But no, the problem doesn't mention anything about rotations or reflections being considered the same. It just says "in one row," so I think linear arrangements where the order matters, and the only thing that matters is the sequence of medals.

Therefore, I think my reasoning is correct. The answer is 6. Therefore, the number of distinct arrangements is 6.

But just to make sure, let me think of another approach. Suppose I try to count the number of permutations step by step.

First, the first position can be either I or P. Let's consider cases.

Case 1: First position is I.

Then, we have one I and two P's left. The remaining three positions need to be filled with one I and two P's. The number of arrangements for this case is the number of ways to arrange one I and two P's in three positions, which is C(3,1) = 3.

Case 2: First position is P.

Similarly, then we have two I's and one P left. The number of arrangements for this case is the number of ways to arrange two I's and one P in three positions, which is C(3,2) = 3.

Therefore, total number of arrangements is 3 + 3 = 6. Yep, same answer again. So that's consistent.

Alternatively, if I think recursively, but that might be overcomplicating. But just to check, suppose I think of the first position: 2 choices (I or P). Then, depending on that, the next positions have different numbers. But as above, it still leads to 6.

Alternatively, maybe using generating functions? Although that might be more complex. The exponential generating function for each type is (1 + x)^2, but since we have two types, maybe not. Wait, actually, generating functions for permutations with identical items... Maybe not necessary here.

Alternatively, think of the problem as arranging letters. If we have the word "IIP", but with two I's and two P's. The number of distinct permutations is 4! / (2!2!) = 6. Yep, same thing.

Therefore, all methods lead to 6. Therefore, I think the answer is 6. Therefore, the number of distinct arrangements is 6.

But just to make sure, let me list all the possibilities again:

1. II PP
2. I P I P
3. I PP I
4. P II P
5. P I P I
6. PP II

Yes, that's six. Each of these is unique. For example, the first one is two I's followed by two P's. The second is alternating I and P starting with I. The third is I, then two P's, then I. The fourth is P, then two I's, then P. The fifth is alternating starting with P. The sixth is two P's followed by two I's. So that's all six. There's no other arrangement possible. For example, if I try to make something like I P P I, that's already listed as number 3. Similarly, P I I P is number 4. So yeah, all possibilities are covered.

Therefore, I think I can confidently say that the answer is 6. Therefore, the number of distinct arrangements is 6.

**Final Answer**
The number of distinct arrangements is \boxed{6}.

To determine the number of distinct arrangements of 2 indistinguishable IMO gold medals and 2 indistinguishable IPhO gold medals in a row, we approach the problem using the concept of **permutations with identical items**.

---

### Step 1: Understanding the Problem

We are arranging **4 medals** in a row:
- 2 of them are **indistinguishable IMO gold medals** (denoted as `I`)
- 2 of them are **indistinguishable IPhO gold medals** (denoted as `P`)

Since the medals of the same type are **indistinguishable**, swapping two `I`s or two `P`s does **not** result in a new arrangement.

---

### Step 2: Applying the Permutation Formula

The general formula for the number of distinct permutations of a multiset is:

$$
\text{Number of arrangements} = \frac{n!}{n_1! \cdot n_2! \cdot \ldots \cdot n_k!}
$$

Where:
- $ n $ is the total number of items (here, $ n = 4 $)
- $ n_1, n_2, \ldots, n_k $ are the counts of each type of indistinguishable item

In this case:
- $ n = 4 $
- $ n_1 = 2 $ (number of `I`s)
- $ n_2 = 2 $ (number of `P`s)

So the number of distinct arrangements is:

$$
\frac{4!}{2! \cdot 2!} = \frac{24}{2 \cdot 2} = \frac{24}{4} = 6
$$

---

### Step 3: Verifying by Enumeration

To confirm, we can list all the distinct arrangements of the letters `I` and `P` with two of each:

1. `I I P P`
2. `I P I P`
3. `I P P I`
4. `P I I P`
5. `P I P I`
6. `P P I I`

These are all the unique permutations, and there are exactly **6** of them.

---

### Final Answer

$$
\boxed{6}
$$

Solution: There are 4 slots in the row.  He can choose two of them for his IMO medals in $\binom{4}{2} = \boxed{6}$ ways.


Explanation:
Okay, so I need to figure out what the largest eight-digit number is in base 2, and then convert that to base 10. Let me start by recalling what an eight-digit number in base 2 would look like.

In any base system, the number of digits determines the range of values. For example, in base 10, an eight-digit number ranges from 10,000,000 to 99,999,999. Similarly, in base 2, an eight-digit number would start with 1 followed by seven more digits. But since it's base 2, each digit can only be 0 or 1.

So, the largest eight-digit base 2 number would be the one where all the digits are 1s. That makes sense because each digit represents a power of 2, and having a 1 in the highest place value gives the maximum value. Let me confirm that.

An eight-digit binary number has positions from the 7th power of 2 down to the 0th power. Wait, let me think. The rightmost digit is 2^0, then moving left, it's 2^1, 2^2, up to 2^7 for the leftmost digit. So, an eight-digit binary number would be from 2^7 down to 2^0.

Therefore, the largest eight-digit binary number would be 11111111 in binary. Let me check that. If I have eight 1s, that's the maximum possible value for an eight-digit binary number.

Now, I need to convert this binary number to decimal (base 10). How do I do that? Well, each digit represents a power of 2. Starting from the right, the first digit is 2^0, then 2^1, up to 2^7 for the leftmost digit.

So, let's write out the binary number 11111111 and convert each digit to its decimal equivalent.

Breaking it down:

From left to right, the digits are:

1 (2^7), 1 (2^6), 1 (2^5), 1 (2^4), 1 (2^3), 1 (2^2), 1 (2^1), 1 (2^0).

So, the decimal value would be the sum of all these powers of 2.

Calculating each term:

2^7 = 128

2^6 = 64

2^5 = 32

2^4 = 16

2^3 = 8

2^2 = 4

2^1 = 2

2^0 = 1

Adding them all up: 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1.

Let me add them step by step:

Start with 128 + 64 = 192

192 + 32 = 224

224 + 16 = 240

240 + 8 = 248

248 + 4 = 252

252 + 2 = 254

254 + 1 = 255

So, the total is 255.

Wait, so the largest eight-digit binary number is 255 in decimal? Let me check if that's correct.

Alternatively, I remember that 2^n - 1 is the value of a binary number with n ones. For example, 2^8 - 1 = 256 - 1 = 255. But wait, here we have eight digits, so n is 8? Wait, but 2^8 is 256, so 2^8 - 1 is 255. But does that correspond to an eight-digit binary number?

Wait, let me think again. The number of digits in binary is related to the exponent. For example, the number 2^7 is 10000000 in binary, which is an eight-digit number. So, the maximum eight-digit binary number is 2^8 - 1, which is 255. That seems correct.

But let me verify with another approach. If I have an eight-digit binary number, the smallest eight-digit number is 10000000, which is 128 in decimal. The largest is 11111111, which is 255. So, the range of eight-digit binary numbers is from 128 to 255 in decimal. Therefore, 255 is indeed the largest.

Alternatively, if I think about how binary numbers work, each digit doubles the previous. So, the first digit (leftmost) is 2^7, then each subsequent digit adds half of the previous. But since all digits are 1s, it's the sum of a geometric series.

The formula for the sum of a geometric series is S = a(r^n - 1)/(r - 1), where a is the first term, r is the common ratio, and n is the number of terms.

In this case, the first term a is 1 (for 2^7?), wait, maybe I need to adjust. Let me think.

Actually, the sum of 2^7 + 2^6 + ... + 2^0 is a geometric series with first term 2^0 = 1, ratio 2, and number of terms 8.

Wait, but if I use the formula S = (r^n - 1)/(r - 1). So, substituting r = 2, n = 8, we get (2^8 - 1)/(2 - 1) = (256 - 1)/1 = 255.

Yes, that's the same result. So that confirms it.

Alternatively, if I think of the binary number 11111111, it's equivalent to 255 in decimal.

Therefore, the answer should be 255.

But let me just make sure I didn't make a mistake in the number of digits. For example, sometimes people get confused about how many digits correspond to a certain number of bits.

In binary, the number of digits (bits) required to represent a number is determined by the highest power of 2. For example, 2^7 is 128, which is an 8-digit binary number (10000000). So, 2^7 is the smallest 8-digit binary number, and 2^8 - 1 is the largest 8-digit binary number.

Therefore, 2^8 - 1 = 256 - 1 = 255.

Yes, that seems consistent.

Alternatively, if I think of the number of bits: an n-bit number can represent numbers from 0 to 2^n - 1. So, an 8-bit number can represent numbers from 0 to 255. Therefore, the maximum value is 255.

Therefore, the largest eight-digit base 2 integer is 255 in base 10.

I think that's solid. Let me check with a smaller example to ensure my reasoning is correct.

For example, the largest 1-digit binary number is 1, which is 1 in decimal. Using the formula, 2^1 - 1 = 1. Correct.

The largest 2-digit binary number is 11, which is 3 in decimal. 2^2 - 1 = 3. Correct.

Similarly, 3-digit binary: 111 is 7, which is 2^3 - 1 = 7. Correct.

So, the pattern holds. Therefore, for 8 digits, it's 2^8 - 1 = 255.

Therefore, the answer is 255.

**Final Answer**
The largest eight-digit base 2 integer expressed in base 10 is \boxed{255}.

To determine the largest eight-digit base 2 (binary) integer and express it in base 10, we begin by understanding the structure of binary numbers.

---

### Understanding Binary Digits

In base 2, each digit represents a power of 2, starting from the rightmost digit as $2^0$, then $2^1$, $2^2$, and so on. An eight-digit binary number has digits from $2^7$ (leftmost) down to $2^0$ (rightmost).

The **largest** eight-digit binary number is the one where **all digits are 1**, i.e., `11111111` in binary.

---

### Converting Binary to Decimal

To convert `11111111` from binary to decimal, we sum the values of each bit:

$$
1 \cdot 2^7 + 1 \cdot 2^6 + 1 \cdot 2^5 + 1 \cdot 2^4 + 1 \cdot 2^3 + 1 \cdot 2^2 + 1 \cdot 2^1 + 1 \cdot 2^0
$$

$$
= 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = 255
$$

---

### General Formula

This is a well-known result in binary arithmetic: the largest $n$-digit binary number is:

$$
2^n - 1
$$

For $n = 8$:

$$
2^8 - 1 = 256 - 1 = 255
$$

This confirms our earlier calculation.

---

### Final Answer

$$
\boxed{255}
$$

[Assistant]
Explanation:
Okay, so I need to figure out what the largest eight-digit number is in base 2, and then convert that to base 10. Let me start by recalling what an eight-digit number in base 2 would look like.

In any base system, the number of digits determines the range of values. For example, in base 10, an eight-digit number ranges from 10,000,000 to 99,999,999. Similarly, in base 2, an eight-digit number would start with 1 followed by seven more digits. But since it's base 2, each digit can only be 0 or 1.

So, the largest eight-digit base 2 number would be the one where all the digits are 1s. That makes sense because each digit represents a power of 2, and having a 1 in the highest place value gives the maximum value. Let me confirm that.

An eight-digit binary number has positions from the 7th power of 2 down to the 0th power. Wait, let me think. The rightmost digit is 2^0, then moving left, it's 2^1, 2^2, up to 2^7 for the leftmost digit. So, an eight-digit binary number would be from 2^7 down to 2^0.

Therefore, the largest eight-digit binary number would be 11111111 in binary. Let me check that. If I have eight 1s, that's the maximum possible value for an eight-digit binary number.

Now, I need to convert this binary number to decimal (base 10). How do I do that? Well, each digit represents a power of 2. Starting from the right, the first digit is 2^0, then 2^1, up to 2^7 for the leftmost digit.

So, let's write out the binary number 11111111 and convert each digit to its decimal equivalent.

Breaking it down:

From left to right, the digits are:

1 (2^7), 1 (2^6), 1 (2^5), 1 (2^4), 1 (2^3), 1 (2^2), 1 (2^1), 1 (2^0).

So, the decimal value would be the sum of all these powers of 2.

Calculating each term:

2^7 = 128

2^6 = 64

2^5 = 32

2^4 = 16

2^3 = 8

2^2 = 4

2^1 = 2

2^0 = 1

Adding them all up: 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1.

Let me add them step by step:

Start with 128 + 64 = 192

192 + 32 = 224

224 + 16 = 240

240 + 8 = 248

248 + 4 = 252

252 + 2 = 254

254 + 1 = 255

So, the total is 255.

Wait, so the largest eight-digit binary number is 255 in decimal? Let me check if that's correct.

Alternatively, I remember that 2^n - 1 is the value of a binary number with n ones. For example, 2^8 - 1 = 256 - 1 = 255. But wait, here we have eight digits, so n is 8? Wait, but 2^8 is 256, so 2^8 - 1 is 255. But does that correspond to an eight-digit binary number?

Wait, let me think again. The number of digits in binary is related to the exponent. For example, the number 2^7 is 10000000 in binary, which is an eight-digit number. So, the maximum eight-digit binary number is 2^8 - 1, which is 255. That seems correct.

But let me verify with another approach. If I have an eight-digit binary number, the smallest eight-digit number is 10000000, which is 128 in decimal. The largest is 11111111, which is 255. So, the range of eight-digit binary numbers is from 128 to 255 in decimal. Therefore, 255 is indeed the largest.

Alternatively, if I think about how binary numbers work, each digit doubles the previous. So, the first digit (leftmost) is 2^7, then each subsequent digit adds half of the previous. But since all digits are 1s, it's the sum of a geometric series.

The formula for the sum of a geometric series is S = a(r^n - 1)/(r - 1), where a is the first term, r is the common ratio, and n is the number of terms.

In this case, the first term a is 1 (for 2^7?), wait, maybe I need to adjust. Let me think.

Actually, the sum of 2^7 + 2^6 + ... + 2^0 is a geometric series with first term 2^0 = 1, ratio 2, and number of terms 8.

Wait, but if I use the formula S = (r^n - 1)/(r - 1). So, substituting r = 2, n = 8, we get (2^8 - 1)/(2 - 1) = (256 - 1)/1 = 255.

Yes, that's the same result. So that confirms it.

Alternatively, if I think of the binary number 11111111, it's equivalent to 255 in decimal.

Therefore, the answer should be 255.

But let me just make sure I didn't make a mistake in the number of digits. For example, sometimes people get confused about how many digits correspond to a certain number of bits.

In binary, the number of digits (bits) required to represent a number is determined by the highest power of 2. For example, 2^7 is 128, which is an 8-digit binary number (10000000). So, 2^7 is the smallest 8-digit binary number, and 2^8 - 1 is the largest 8-digit binary number.

Therefore, 2^8 - 1 = 256 - 1 = 255.

Yes, that seems consistent.

Alternatively, if I think of the number of bits: an n-bit number can represent numbers from 0 to 2^n - 1. So, an 8-bit number can represent numbers from 0 to 255. Therefore, the maximum value is 255.

Therefore, the largest eight-digit base 2 integer is 255 in base 10.

I think that's solid. Let me check with a smaller example to ensure my reasoning is correct.

For example, the largest 1-digit binary number is 1, which is 1 in decimal. Using the formula, 2^1 - 1 = 1. Correct.

The largest 2-digit binary number is 11, which is 3 in decimal. 2^2 - 1 = 3. Correct.

Similarly, 3-digit binary: 111 is 7, which is 2^3 - 1 = 7. Correct.

So, the pattern holds. Therefore, for 8 digits, it's 2^8 - 1 = 255.

Therefore, the answer is 255.

**Final Answer**
The largest eight-digit base 2 integer expressed in base 10 is \boxed{255}.

To determine the largest eight-digit base 2 (binary) integer and express it in base 10, we begin by understanding the structure of binary numbers.

---

### Understanding Binary Digits

In base 2, each digit represents a power of 2, starting from the rightmost digit as $2^0$, then $2^1$, $2^2$, and so on. An eight-digit binary number has digits from $2^7$ (leftmost) down to $2^0$ (rightmost).

The **largest** eight-digit binary number is the one where **all digits are 1**, i.e., `11111111` in binary.

---

### Converting Binary to Decimal

To convert `11111111` from binary to decimal, we sum the values of each bit:

$$
1 \cdot 2^7 + 1 \cdot 2^6 + 1 \cdot 2^5 + 1 \cdot 2^4 + 1 \cdot 2^3 + 1 \cdot 2^2 + 1 \cdot 2^1 + 1 \cdot 2^0
$$

$$
= 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = 255
$$

---

### General Formula

This is a well-known result in binary arithmetic: the largest $n$-digit binary number is:

$$
2^n - 1
$$

For $n = 8$:

$$
2^8 - 1 = 256 - 1 = 255
$$

This confirms our earlier calculation.

---

### Final Answer

$$
\boxed{255}
$$

Solution: The largest eight-digit base 2 integer is 1 less than the smallest nine-digit base 2 integer, which is $$ 100000000_{2} = 1 \cdot 2^8 = 256. $$Thus, the largest eight-digit base 2 integer is $256 - 1 = \boxed{255}$.


Explanation:
Okay, so I need to figure out what (4 5/8)^55 multiplied by (8/37)^55 is. Hmm, let me start by recalling some exponent rules. When you have two numbers raised to the same exponent, you can combine them by multiplying the bases first and then raising to that exponent. So, (a^b) * (c^b) = (a*c)^b. Is that right? Let me check with smaller numbers. Like, 2^3 * 3^3 should be (2*3)^3 = 6^3 = 216. And 8*27 is 216. Yeah, that works. So, applying that here, I can combine the two bases first.

First, let me convert the mixed number 4 5/8 into an improper fraction. To do that, I multiply the whole number part by the denominator and add the numerator. So, 4 * 8 = 32, then add 5, which gives 37. So, 4 5/8 is equal to 37/8. Let me confirm that: 37 divided by 8 is 4.625, and 4 and 5/8 is indeed 4.625. Yep, that's correct.

So now the expression becomes (37/8)^55 multiplied by (8/37)^55. Using the exponent rule I mentioned earlier, this should be equal to ( (37/8) * (8/37) )^55. Let me compute the product inside the parentheses.

Multiplying 37/8 by 8/37. The 37 in the numerator and the 37 in the denominator will cancel out, and the 8 in the numerator and the 8 in the denominator will also cancel out. So, that leaves me with 1. Therefore, the entire expression simplifies to 1^55.

Wait, 1 raised to any power is still 1, right? So, regardless of the exponent, 1^55 is 1. Therefore, the answer should be 1.

But let me double-check my steps to make sure I didn't make a mistake. First, converting 4 5/8 to an improper fraction: 4*8=32, 32+5=37, so 37/8. Correct. Then, multiplying (37/8) * (8/37). The numerators are 37*8 and denominators are 8*37. So, yes, they are equal, so the fraction is 1. Then raising 1 to the 55th power is 1. Seems straightforward.

Is there another way to look at this? Maybe by simplifying before applying the exponent? Let me try that.

Original expression: (4 5/8)^55 * (8/37)^55. If I write 4 5/8 as 37/8, then it's (37/8)^55 * (8/37)^55. Alternatively, I can think of (37/8) and (8/37) as reciprocals of each other. Because 8/37 is 1 divided by (37/8). So, multiplying a number by its reciprocal gives 1. Therefore, (37/8) * (8/37) = 1. Therefore, raising 1 to any power is still 1.

Another way: Let me think about exponents. If I have (a/b)^n * (b/a)^n, that's equal to (a/b * b/a)^n = 1^n = 1. So, regardless of a and b, as long as they are non-zero, this will hold. So, in this case, a is 37 and b is 8. So, same result.

Therefore, regardless of the exponent, as long as the two bases are reciprocals, their product raised to the same exponent is 1.

Wait, but just to make sure that there isn't a trick here. The original problem is written as (4 5/8)^55 multiplied by (8/37)^55. Is there any chance that the exponents are different? No, both are raised to the 55th power. So, the key is that the two bases are reciprocals.

Alternatively, maybe I can write both terms with the same base? Let me try.

First, 4 5/8 is 37/8. Then, 8/37 is already a fraction. So, if I write (37/8)^55 * (8/37)^55. Let me write (37/8)^55 as (37^55)/(8^55) and (8/37)^55 as (8^55)/(37^55). Then multiplying them together: (37^55 / 8^55) * (8^55 / 37^55) = (37^55 * 8^55) / (8^55 * 37^55) = 1.

Yes, that's another way to see it. The numerator and denominator are the same, so they cancel out. Therefore, the result is 1.

So, all methods lead to the conclusion that the value is 1. Therefore, I think the answer is 1.

But just to make sure, let me check with smaller exponents. Let's say instead of 55, use exponent 1. Then, (4 5/8) * (8/37) = (37/8) * (8/37) = 1. If exponent is 2, then [(37/8)*(8/37)]^2 = 1^2 = 1. So, same result. Therefore, regardless of the exponent, it's 1.

Therefore, I think confidently the answer is 1.

**Final Answer**
\boxed{1}

To evaluate the expression:

$$
\left(4\dfrac{5}{8}\right)^{55} \cdot \left(\dfrac{8}{37}\right)^{55}
$$

we begin by simplifying the mixed number $4\dfrac{5}{8}$ into an improper fraction. This is done by multiplying the whole number part by the denominator and adding the numerator:

$$
4\dfrac{5}{8} = \frac{4 \cdot 8 + 5}{8} = \frac{32 + 5}{8} = \frac{37}{8}
$$

So the expression becomes:

$$
\left(\frac{37}{8}\right)^{55} \cdot \left(\frac{8}{37}\right)^{55}
$$

---

### Step 1: Apply the exponent rule

We use the property of exponents:

$$
(a^b) \cdot (c^b) = (a \cdot c)^b
$$

Applying this to our expression:

$$
\left(\frac{37}{8} \cdot \frac{8}{37}\right)^{55}
$$

---

### Step 2: Simplify the base

Now, simplify the product inside the parentheses:

$$
\frac{37}{8} \cdot \frac{8}{37} = \frac{37 \cdot 8}{8 \cdot 37} = \frac{296}{296} = 1
$$

So the expression becomes:

$$
1^{55}
$$

---

### Step 3: Evaluate the final expression

Any number raised to the power of 55 (or any power) is still itself if the base is 1:

$$
1^{55} = 1
$$

---

### Final Answer

$$
\boxed{1}
$$

[Assistant]
Explanation:
Okay, so I need to figure out what (4 5/8)^55 multiplied by (8/37)^55 is. Hmm, let me start by recalling some exponent rules. When you have two numbers raised to the same exponent, you can combine them by multiplying the bases first and then raising to that exponent. So, (a^b) * (c^b) = (a*c)^b. Is that right? Let me check with smaller numbers. Like, 2^3 * 3^3 should be (2*3)^3 = 6^3 = 216. And 8*27 is 216. Yeah, that works. So, applying that here, I can combine the two bases first.

First, let me convert the mixed number 4 5/8 into an improper fraction. To do that, I multiply the whole number part by the denominator and add the numerator. So, 4 * 8 = 32, then add 5, which gives 37. So, 4 5/8 is equal to 37/8. Let me confirm that: 37 divided by 8 is 4.625, and 4 and 5/8 is indeed 4.625. Yep, that's correct.

So now the expression becomes (37/8)^55 multiplied by (8/37)^55. Using the exponent rule I mentioned earlier, this should be equal to ( (37/8) * (8/37) )^55. Let me compute the product inside the parentheses.

Multiplying 37/8 by 8/37. The 37 in the numerator and the 37 in the denominator will cancel out, and the 8 in the numerator and the 8 in the denominator will also cancel out. So, that leaves me with 1. Therefore, the entire expression simplifies to 1^55.

Wait, 1 raised to any power is still 1, right? So, regardless of the exponent, 1^55 is 1. Therefore, the answer should be 1.

But let me double-check my steps to make sure I didn't make a mistake. First, converting 4 5/8 to an improper fraction: 4*8=32, 32+5=37, so 37/8. Correct. Then, multiplying (37/8) * (8/37). The numerators are 37*8 and denominators are 8*37. So, yes, they are equal, so the fraction is 1. Then raising 1 to the 55th power is 1. Seems straightforward.

Is there another way to look at this? Maybe by simplifying before applying the exponent? Let me try that.

Original expression: (4 5/8)^55 * (8/37)^55. If I write 4 5/8 as 37/8, then it's (37/8)^55 * (8/37)^55. Alternatively, I can think of (37/8) and (8/37) as reciprocals of each other. Because 8/37 is 1 divided by (37/8). So, multiplying a number by its reciprocal gives 1. Therefore, (37/8) * (8/37) = 1. Therefore, raising 1 to any power is still 1.

Another way: Let me think about exponents. If I have (a/b)^n * (b/a)^n, that's equal to (a/b * b/a)^n = 1^n = 1. So, regardless of a and b, as long as they are non-zero, this will hold. So, in this case, a is 37 and b is 8. So, same result.

Therefore, regardless of the exponent, as long as the two bases are reciprocals, their product raised to the same exponent is 1.

Wait, but just to make sure that there isn't a trick here. The original problem is written as (4 5/8)^55 multiplied by (8/37)^55. Is there any chance that the exponents are different? No, both are raised to the 55th power. So, the key is that the two bases are reciprocals.

Alternatively, maybe I can write both terms with the same base? Let me try.

First, 4 5/8 is 37/8. Then, 8/37 is already a fraction. So, if I write (37/8)^55 * (8/37)^55. Let me write (37/8)^55 as (37^55)/(8^55) and (8/37)^55 as (8^55)/(37^55). Then multiplying them together: (37^55 / 8^55) * (8^55 / 37^55) = (37^55 * 8^55) / (8^55 * 37^55) = 1.

Yes, that's another way to see it. The numerator and denominator are the same, so they cancel out. Therefore, the result is 1.

So, all methods lead to the conclusion that the value is 1. Therefore, I think the answer is 1.

But just to make sure, let me check with smaller exponents. Let's say instead of 55, use exponent 1. Then, (4 5/8) * (8/37) = (37/8) * (8/37) = 1. If exponent is 2, then [(37/8)*(8/37)]^2 = 1^2 = 1. So, same result. Therefore, regardless of the exponent, it's 1.

Therefore, I think confidently the answer is 1.

**Final Answer**
\boxed{1}

To evaluate the expression:

$$
\left(4\dfrac{5}{8}\right)^{55} \cdot \left(\dfrac{8}{37}\right)^{55}
$$

we begin by simplifying the mixed number $4\dfrac{5}{8}$ into an improper fraction. This is done by multiplying the whole number part by the denominator and adding the numerator:

$$
4\dfrac{5}{8} = \frac{4 \cdot 8 + 5}{8} = \frac{32 + 5}{8} = \frac{37}{8}
$$

So the expression becomes:

$$
\left(\frac{37}{8}\right)^{55} \cdot \left(\frac{8}{37}\right)^{55}
$$

---

### Step 1: Apply the exponent rule

We use the property of exponents:

$$
(a^b) \cdot (c^b) = (a \cdot c)^b
$$

Applying this to our expression:

$$
\left(\frac{37}{8} \cdot \frac{8}{37}\right)^{55}
$$

---

### Step 2: Simplify the base

Now, simplify the product inside the parentheses:

$$
\frac{37}{8} \cdot \frac{8}{37} = \frac{37 \cdot 8}{8 \cdot 37} = \frac{296}{296} = 1
$$

So the expression becomes:

$$
1^{55}
$$

---

### Step 3: Evaluate the final expression

Any number raised to the power of 55 (or any power) is still itself if the base is 1:

$$
1^{55} = 1
$$

---

### Final Answer

$$
\boxed{1}
$$

Solution:
First we convert $4\dfrac{5}{8}$ into an improper fraction: \[4\dfrac{5}{8} = 4 + \dfrac{5}{8} = \dfrac{32}{8} + \dfrac{5}{8} = \dfrac{37}{8}.\]We discover that $4\dfrac{5}{8}$ and $\dfrac{8}{37}$ are in fact reciprocals of each other. Using the fact that $(ab)^n = a^nb^n$, we get our answer: \[
\left(4\dfrac{5}{8}\right)^{55} \cdot \left(\dfrac{8}{37}\right)^{55} = \left(4\dfrac{5}{8} \cdot \dfrac{8}{37}\right)^{55} = 1^{55} = \boxed{1}.\]


Explanation:
Okay, so I need to find the minimum value of the expression $(a + b + c + d)\left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} \right)$ where $a, b, c, d$ are positive real numbers. Hmm, let me think about how to approach this.

First, I remember that inequalities like Cauchy-Schwarz or AM-GM might be useful here. Let me recall what those are. The Cauchy-Schwarz inequality states that for any real numbers, $(\sum u_i v_i)^2 \leq (\sum u_i^2)(\sum v_i^2)$. But I'm not sure if that's directly applicable here. Maybe the AM-GM inequality? The AM-GM inequality says that the arithmetic mean is greater than or equal to the geometric mean. For positive numbers, that is.

Alternatively, I remember that for two positive numbers, the product $(x + y)(\frac{1}{x} + \frac{1}{y})$ has a minimum value of 4, achieved when $x = y$. Maybe this problem is a generalization of that? If I consider the case where all variables are equal, maybe that gives the minimum?

Let me test that idea. Suppose $a = b = c = d = k$, where $k$ is a positive real number. Then the expression becomes:

$$
(4k)\left( \frac{4}{k} \right) = 4k \times \frac{4}{k} = 16
$$

So, if all variables are equal, the value is 16. Is this the minimum? Maybe, but I need to check if there's a possibility of getting a lower value if the variables aren't equal.

Let me try with different values. Suppose I take $a = b = c = 1$ and $d$ approaching 0. Then the first part of the product, $a + b + c + d$, approaches 3 + 0 = 3. The second part, $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d}$, approaches 1 + 1 + 1 + infinity = infinity. So the product approaches infinity. Similarly, if one variable becomes very large, then the first part becomes large and the second part becomes small, but I think the product might still go to infinity. Wait, let me check.

Suppose $a$ approaches infinity, and $b = c = d = 1$. Then the first part is $a + 1 + 1 + 1 = a + 3$, which approaches infinity. The second part is $\frac{1}{a} + 1 + 1 + 1 = \frac{1}{a} + 3$, which approaches 3. So the product approaches infinity * 3 = infinity. So making variables very large or very small seems to make the product larger. So maybe the minimum occurs when all variables are equal?

But wait, maybe there's a case where some variables are different but the product is less than 16? Let me try with two variables equal and the other two different. For example, let me take $a = b = 1$ and $c = d = t$, where $t$ is some positive number. Then compute the expression:

First part: $1 + 1 + t + t = 2 + 2t$

Second part: $1 + 1 + \frac{1}{t} + \frac{1}{t} = 2 + \frac{2}{t}$

So the product is $(2 + 2t)\left(2 + \frac{2}{t}\right)$. Let me compute that:

Multiply out:

$2*2 + 2*\frac{2}{t} + 2t*2 + 2t*\frac{2}{t}$

Which is $4 + \frac{4}{t} + 4t + 4$

Simplify: $8 + 4t + \frac{4}{t}$

Now, to find the minimum of $8 + 4t + \frac{4}{t}$. Let me set $f(t) = 4t + \frac{4}{t}$. Then the minimum of $f(t)$ occurs where derivative is zero. But since this is a simple expression, maybe I can use AM-GM here. For positive t, $4t + \frac{4}{t} \geq 2\sqrt{4t * \frac{4}{t}} = 2\sqrt{16} = 8$. So the minimum of $f(t)$ is 8, achieved when $4t = \frac{4}{t}$, so $t^2 = 1$, so $t = 1$. Therefore, the total expression is 8 + 8 = 16. So even in this case, the minimum is 16. So even when I set two variables to 1 and the other two to t, the minimum is still 16. So maybe 16 is indeed the minimum?

But let me try another case. Suppose three variables are equal and one is different. Let me take $a = b = c = 1$ and $d = t$. Then compute the expression:

First part: $1 + 1 + 1 + t = 3 + t$

Second part: $1 + 1 + 1 + \frac{1}{t} = 3 + \frac{1}{t}$

Product: $(3 + t)\left(3 + \frac{1}{t}\right)$. Let me expand this:

$3*3 + 3*\frac{1}{t} + t*3 + t*\frac{1}{t} = 9 + \frac{3}{t} + 3t + 1 = 10 + 3t + \frac{3}{t}$

Now, find the minimum of $10 + 3t + \frac{3}{t}$. Let me set $f(t) = 3t + \frac{3}{t}$. Using AM-GM, $3t + \frac{3}{t} \geq 2\sqrt{3t * \frac{3}{t}} = 2\sqrt{9} = 6$. Therefore, the minimum of $f(t)$ is 6, achieved when $3t = \frac{3}{t}$, so $t^2 = 1$, so $t = 1$. Therefore, the total expression is 10 + 6 = 16. Again, 16. So even in this case, the minimum is 16.

Hmm, so maybe regardless of how I set variables, the minimum is 16? But let me try with different numbers. Suppose I take $a = 2$, $b = 2$, $c = 1$, $d = 1$. Then compute the product:

First part: 2 + 2 + 1 + 1 = 6

Second part: 1/2 + 1/2 + 1 + 1 = 1 + 2 = 3

Product: 6 * 3 = 18, which is larger than 16. So that's worse.

What if I take $a = 1$, $b = 2$, $c = 3$, $d = 6$. Let me compute:

First part: 1 + 2 + 3 + 6 = 12

Second part: 1 + 1/2 + 1/3 + 1/6 = 1 + 0.5 + 0.333... + 0.166... ≈ 2. So 12 * 2 = 24, which is even larger.

Alternatively, maybe take some variables unequal but not too much. Let me try $a = 1$, $b = 1$, $c = 2$, $d = 2$. Then first part: 1 + 1 + 2 + 2 = 6. Second part: 1 + 1 + 1/2 + 1/2 = 3. Product is 6 * 3 = 18 again. Still higher than 16.

Alternatively, maybe take three variables equal and one different but not 1? Let me try $a = b = c = 2$, $d = t$. Then first part: 2 + 2 + 2 + t = 6 + t. Second part: 1/2 + 1/2 + 1/2 + 1/t = 3/2 + 1/t. Product: (6 + t)(3/2 + 1/t). Let me compute this:

Multiply out: 6*(3/2) + 6*(1/t) + t*(3/2) + t*(1/t) = 9 + 6/t + (3t)/2 + 1 = 10 + 6/t + (3t)/2.

To find the minimum of this expression. Let me set derivative with respect to t to zero. But maybe AM-GM? Let me see. Let me write it as 10 + 6/t + (3t)/2. Let me consider the terms 6/t and (3t)/2. Let me set x = t. Then the expression is 10 + 6/x + (3x)/2. Let me find the minimum of 6/x + (3x)/2. Let me take derivative: -6/x² + 3/2. Setting to zero: -6/x² + 3/2 = 0 => 3/2 = 6/x² => x² = 6 / (3/2) = 4 => x = 2. Therefore, minimum occurs at x = 2. Then the value is 6/2 + (3*2)/2 = 3 + 3 = 6. Therefore, total expression is 10 + 6 = 16. Again, 16. So even when variables are different, but adjusted properly, the minimum is still 16. So seems like 16 is the minimum.

But wait, how can I be sure that there isn't some configuration where variables are not equal, but the product is less than 16? Maybe I need to use some inequality to prove that the minimum is indeed 16.

Let me recall that for positive real numbers, the Cauchy-Schwarz inequality states that $(\sum u_i v_i)^2 \leq (\sum u_i^2)(\sum v_i^2)$. But how can I apply that here?

Alternatively, maybe use the Cauchy-Schwarz in the form $(\sum a_i)(\sum \frac{1}{a_i}) \geq n^2$, where n is the number of terms. Wait, if that's the case, then for n=4, the minimum would be 16. But is that true?

Wait, let me check. For example, if I have two variables, then $(a + b)(1/a + 1/b) \geq 4$. Which is true because expanding gives 2 + a/b + b/a ≥ 4, since a/b + b/a ≥ 2. Similarly, for n variables, $(a_1 + a_2 + ... + a_n)(1/a_1 + 1/a_2 + ... + 1/a_n) \geq n^2$. Is this a known inequality?

Yes, this is a direct application of the Cauchy-Schwarz inequality. Let me recall that Cauchy-Schwarz can be written as $(\sum u_i v_i)^2 \leq (\sum u_i^2)(\sum v_i^2)$. If I set $u_i = \sqrt{a_i}$ and $v_i = 1/\sqrt{a_i}$, then $\sum u_i v_i = \sum 1 = n$. Then, by Cauchy-Schwarz:

$n^2 \leq (\sum a_i)(\sum 1/a_i)$. Therefore, $(\sum a_i)(\sum 1/a_i) \geq n^2$. Therefore, for n=4, the minimum is 16. Therefore, the minimum value is 16, achieved when all $a_i$ are equal. Therefore, the answer is 16.

But let me check if this is correct. Let me verify with the case when n=2. If n=2, then $(a + b)(1/a + 1/b) = 2 + a/b + b/a ≥ 4$, which is correct. So yes, the inequality holds. Therefore, for n=4, the minimum is 16. Therefore, regardless of the values of a, b, c, d (as long as they are positive), the product is at least 16, and equality holds when all variables are equal.

Therefore, the minimum value is 16.

But let me think again. Is there a case where variables are not equal but the product is still 16? For example, if some variables are equal and others are different? Wait, according to the Cauchy-Schwarz inequality, equality holds when $u_i / v_i$ is constant. In our case, $u_i = \sqrt{a_i}$, $v_i = 1/\sqrt{a_i}$. Therefore, $u_i / v_i = a_i$. Therefore, equality holds when all $a_i$ are equal. Therefore, only when all variables are equal. Therefore, the minimum is achieved only when all variables are equal. Therefore, the minimum value is 16.

Therefore, the answer is 16. So I think that's the answer.

But just to make sure, let me try with different numbers. Suppose I take $a = 2$, $b = 2$, $c = 2$, $d = 2$. Then the product is (8)(4/2) = 8 * 2 = 16. Wait, no. Wait, (a + b + c + d) is 8, and (1/a + 1/b + 1/c + 1/d) is 4*(1/2) = 2. So 8 * 2 = 16. Correct. If I take $a = 1$, $b = 1$, $c = 1$, $d = 1$, then (4)(4) = 16. So same value. If I take $a = 1$, $b = 1$, $c = 1$, $d = 1$, same thing. If I take $a = 1$, $b = 1$, $c = 1$, $d = 1$, same. So equality holds when all variables are equal. Therefore, the minimum is indeed 16.

Therefore, I think the answer is 16. So I can write that as the final answer.

**Final Answer**
\boxed{16}

To find the minimum value of the expression

$$
(a + b + c + d)\left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} \right),
$$

where $ a, b, c, d $ are positive real numbers, we can apply a powerful inequality known as the **Cauchy-Schwarz Inequality**.

---

### Step 1: Apply the Cauchy-Schwarz Inequality

The Cauchy-Schwarz Inequality in the form:

$$
\left( \sum_{i=1}^n u_i v_i \right)^2 \leq \left( \sum_{i=1}^n u_i^2 \right)\left( \sum_{i=1}^n v_i^2 \right)
$$

can be used with the choice:

- $ u_i = \sqrt{a_i} $
- $ v_i = \frac{1}{\sqrt{a_i}} $

Then:

$$
\sum_{i=1}^n u_i v_i = \sum_{i=1}^n \sqrt{a_i} \cdot \frac{1}{\sqrt{a_i}} = \sum_{i=1}^n 1 = n
$$

So:

$$
n^2 \leq \left( \sum_{i=1}^n a_i \right)\left( \sum_{i=1}^n \frac{1}{a_i} \right)
$$

This gives us the inequality:

$$
(a + b + c + d)\left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} \right) \geq 4^2 = 16
$$

---

### Step 2: Equality Condition

Equality in the Cauchy-Schwarz inequality holds when the vectors $ (u_1, u_2, \dots, u_n) $ and $ (v_1, v_2, \dots, v_n) $ are proportional. In our case, this means:

$$
\frac{u_i}{v_i} = \frac{\sqrt{a_i}}{1/\sqrt{a_i}} = a_i \text{ is constant for all } i
$$

So, all $ a_i $ must be equal. That is, the minimum is achieved when $ a = b = c = d $.

---

### Step 3: Compute the Minimum Value

Let $ a = b = c = d = k $, where $ k > 0 $. Then:

$$
(a + b + c + d) = 4k, \quad \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} \right) = \frac{4}{k}
$$

So the product becomes:

$$
(4k)\left( \frac{4}{k} \right) = 16
$$

---

### Final Conclusion

The minimum value of the expression is:

$$
\boxed{16}
$$

This minimum is achieved when all variables are equal, i.e., $ a = b = c = d $.

[Assistant]
Explanation:
Okay, so I need to find the minimum value of the expression $(a + b + c + d)\left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} \right)$ where $a, b, c, d$ are positive real numbers. Hmm, let me think about how to approach this.

First, I remember that inequalities like Cauchy-Schwarz or AM-GM might be useful here. Let me recall what those are. The Cauchy-Schwarz inequality states that for any real numbers, $(\sum u_i v_i)^2 \leq (\sum u_i^2)(\sum v_i^2)$. But I'm not sure if that's directly applicable here. Maybe the AM-GM inequality? The AM-GM inequality says that the arithmetic mean is greater than or equal to the geometric mean. For positive numbers, that is.

Alternatively, I remember that for two positive numbers, the product $(x + y)(\frac{1}{x} + \frac{1}{y})$ has a minimum value of 4, achieved when $x = y$. Maybe this problem is a generalization of that? If I consider the case where all variables are equal, maybe that gives the minimum?

Let me test that idea. Suppose $a = b = c = d = k$, where $k$ is a positive real number. Then the expression becomes:

$$
(4k)\left( \frac{4}{k} \right) = 4k \times \frac{4}{k} = 16
$$

So, if all variables are equal, the value is 16. Is this the minimum? Maybe, but I need to check if there's a possibility of getting a lower value if the variables aren't equal.

Let me try with different values. Suppose I take $a = b = c = 1$ and $d$ approaching 0. Then the first part of the product, $a + b + c + d$, approaches 3 + 0 = 3. The second part, $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d}$, approaches 1 + 1 + 1 + infinity = infinity. So the product approaches infinity. Similarly, if one variable becomes very large, then the first part becomes large and the second part becomes small, but I think the product might still go to infinity. Wait, let me check.

Suppose $a$ approaches infinity, and $b = c = d = 1$. Then the first part is $a + 1 + 1 + 1 = a + 3$, which approaches infinity. The second part is $\frac{1}{a} + 1 + 1 + 1 = \frac{1}{a} + 3$, which approaches 3. So the product approaches infinity * 3 = infinity. So making variables very large or very small seems to make the product larger. So maybe the minimum occurs when all variables are equal?

But wait, maybe there's a case where some variables are different but the product is less than 16? Let me try with two variables equal and the other two different. For example, let me take $a = b = 1$ and $c = d = t$, where $t$ is some positive number. Then compute the expression:

First part: $1 + 1 + t + t = 2 + 2t$

Second part: $1 + 1 + \frac{1}{t} + \frac{1}{t} = 2 + \frac{2}{t}$

So the product is $(2 + 2t)\left(2 + \frac{2}{t}\right)$. Let me compute that:

Multiply out:

$2*2 + 2*\frac{2}{t} + 2t*2 + 2t*\frac{2}{t}$

Which is $4 + \frac{4}{t} + 4t + 4$

Simplify: $8 + 4t + \frac{4}{t}$

Now, to find the minimum of $8 + 4t + \frac{4}{t}$. Let me set $f(t) = 4t + \frac{4}{t}$. Then the minimum of $f(t)$ occurs where derivative is zero. But since this is a simple expression, maybe I can use AM-GM here. For positive t, $4t + \frac{4}{t} \geq 2\sqrt{4t * \frac{4}{t}} = 2\sqrt{16} = 8$. So the minimum of $f(t)$ is 8, achieved when $4t = \frac{4}{t}$, so $t^2 = 1$, so $t = 1$. Therefore, the total expression is 8 + 8 = 16. So even in this case, the minimum is 16. So even when I set two variables to 1 and the other two to t, the minimum is still 16. So maybe 16 is indeed the minimum?

But let me try another case. Suppose three variables are equal and one is different. Let me take $a = b = c = 1$ and $d = t$. Then compute the expression:

First part: $1 + 1 + 1 + t = 3 + t$

Second part: $1 + 1 + 1 + \frac{1}{t} = 3 + \frac{1}{t}$

Product: $(3 + t)\left(3 + \frac{1}{t}\right)$. Let me expand this:

$3*3 + 3*\frac{1}{t} + t*3 + t*\frac{1}{t} = 9 + \frac{3}{t} + 3t + 1 = 10 + 3t + \frac{3}{t}$

Now, find the minimum of $10 + 3t + \frac{3}{t}$. Let me set $f(t) = 3t + \frac{3}{t}$. Using AM-GM, $3t + \frac{3}{t} \geq 2\sqrt{3t * \frac{3}{t}} = 2\sqrt{9} = 6$. Therefore, the minimum of $f(t)$ is 6, achieved when $3t = \frac{3}{t}$, so $t^2 = 1$, so $t = 1$. Therefore, the total expression is 10 + 6 = 16. Again, 16. So even in this case, the minimum is 16.

Hmm, so maybe regardless of how I set variables, the minimum is 16? But let me try with different numbers. Suppose I take $a = 2$, $b = 2$, $c = 1$, $d = 1$. Then compute the product:

First part: 2 + 2 + 1 + 1 = 6

Second part: 1/2 + 1/2 + 1 + 1 = 1 + 2 = 3

Product: 6 * 3 = 18, which is larger than 16. So that's worse.

What if I take $a = 1$, $b = 2$, $c = 3$, $d = 6$. Let me compute:

First part: 1 + 2 + 3 + 6 = 12

Second part: 1 + 1/2 + 1/3 + 1/6 = 1 + 0.5 + 0.333... + 0.166... ≈ 2. So 12 * 2 = 24, which is even larger.

Alternatively, maybe take some variables unequal but not too much. Let me try $a = 1$, $b = 1$, $c = 2$, $d = 2$. Then first part: 1 + 1 + 2 + 2 = 6. Second part: 1 + 1 + 1/2 + 1/2 = 3. Product is 6 * 3 = 18 again. Still higher than 16.

Alternatively, maybe take three variables equal and one different but not 1? Let me try $a = b = c = 2$, $d = t$. Then first part: 2 + 2 + 2 + t = 6 + t. Second part: 1/2 + 1/2 + 1/2 + 1/t = 3/2 + 1/t. Product: (6 + t)(3/2 + 1/t). Let me compute this:

Multiply out: 6*(3/2) + 6*(1/t) + t*(3/2) + t*(1/t) = 9 + 6/t + (3t)/2 + 1 = 10 + 6/t + (3t)/2.

To find the minimum of this expression. Let me set derivative with respect to t to zero. But maybe AM-GM? Let me see. Let me write it as 10 + 6/t + (3t)/2. Let me consider the terms 6/t and (3t)/2. Let me set x = t. Then the expression is 10 + 6/x + (3x)/2. Let me find the minimum of 6/x + (3x)/2. Let me take derivative: -6/x² + 3/2. Setting to zero: -6/x² + 3/2 = 0 => 3/2 = 6/x² => x² = 6 / (3/2) = 4 => x = 2. Therefore, minimum occurs at x = 2. Then the value is 6/2 + (3*2)/2 = 3 + 3 = 6. Therefore, total expression is 10 + 6 = 16. Again, 16. So even when variables are different, but adjusted properly, the minimum is still 16. So seems like 16 is the minimum.

But wait, how can I be sure that there isn't some configuration where variables are not equal, but the product is less than 16? Maybe I need to use some inequality to prove that the minimum is indeed 16.

Let me recall that for positive real numbers, the Cauchy-Schwarz inequality states that $(\sum u_i v_i)^2 \leq (\sum u_i^2)(\sum v_i^2)$. But how can I apply that here?

Alternatively, maybe use the Cauchy-Schwarz in the form $(\sum a_i)(\sum \frac{1}{a_i}) \geq n^2$, where n is the number of terms. Wait, if that's the case, then for n=4, the minimum would be 16. But is that true?

Wait, let me check. For example, if I have two variables, then $(a + b)(1/a + 1/b) \geq 4$. Which is true because expanding gives 2 + a/b + b/a ≥ 4, since a/b + b/a ≥ 2. Similarly, for n variables, $(a_1 + a_2 + ... + a_n)(1/a_1 + 1/a_2 + ... + 1/a_n) \geq n^2$. Is this a known inequality?

Yes, this is a direct application of the Cauchy-Schwarz inequality. Let me recall that Cauchy-Schwarz can be written as $(\sum u_i v_i)^2 \leq (\sum u_i^2)(\sum v_i^2)$. If I set $u_i = \sqrt{a_i}$ and $v_i = 1/\sqrt{a_i}$, then $\sum u_i v_i = \sum 1 = n$. Then, by Cauchy-Schwarz:

$n^2 \leq (\sum a_i)(\sum 1/a_i)$. Therefore, $(\sum a_i)(\sum 1/a_i) \geq n^2$. Therefore, for n=4, the minimum is 16. Therefore, the minimum value is 16, achieved when all $a_i$ are equal. Therefore, the answer is 16.

But let me check if this is correct. Let me verify with the case when n=2. If n=2, then $(a + b)(1/a + 1/b) = 2 + a/b + b/a ≥ 4$, which is correct. So yes, the inequality holds. Therefore, for n=4, the minimum is 16. Therefore, regardless of the values of a, b, c, d (as long as they are positive), the product is at least 16, and equality holds when all variables are equal.

Therefore, the minimum value is 16.

But let me think again. Is there a case where variables are not equal but the product is still 16? For example, if some variables are equal and others are different? Wait, according to the Cauchy-Schwarz inequality, equality holds when $u_i / v_i$ is constant. In our case, $u_i = \sqrt{a_i}$, $v_i = 1/\sqrt{a_i}$. Therefore, $u_i / v_i = a_i$. Therefore, equality holds when all $a_i$ are equal. Therefore, only when all variables are equal. Therefore, the minimum is achieved only when all variables are equal. Therefore, the minimum value is 16.

Therefore, the answer is 16. So I think that's the answer.

But just to make sure, let me try with different numbers. Suppose I take $a = 2$, $b = 2$, $c = 2$, $d = 2$. Then the product is (8)(4/2) = 8 * 2 = 16. Wait, no. Wait, (a + b + c + d) is 8, and (1/a + 1/b + 1/c + 1/d) is 4*(1/2) = 2. So 8 * 2 = 16. Correct. If I take $a = 1$, $b = 1$, $c = 1$, $d = 1$, then (4)(4) = 16. So same value. If I take $a = 1$, $b = 1$, $c = 1$, $d = 1$, same thing. If I take $a = 1$, $b = 1$, $c = 1$, $d = 1$, same. So equality holds when all variables are equal. Therefore, the minimum is indeed 16.

Therefore, I think the answer is 16. So I can write that as the final answer.

**Final Answer**
\boxed{16}

To find the minimum value of the expression

$$
(a + b + c + d)\left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} \right),
$$

where $ a, b, c, d $ are positive real numbers, we can apply a powerful inequality known as the **Cauchy-Schwarz Inequality**.

---

### Step 1: Apply the Cauchy-Schwarz Inequality

The Cauchy-Schwarz Inequality in the form:

$$
\left( \sum_{i=1}^n u_i v_i \right)^2 \leq \left( \sum_{i=1}^n u_i^2 \right)\left( \sum_{i=1}^n v_i^2 \right)
$$

can be used with the choice:

- $ u_i = \sqrt{a_i} $
- $ v_i = \frac{1}{\sqrt{a_i}} $

Then:

$$
\sum_{i=1}^n u_i v_i = \sum_{i=1}^n \sqrt{a_i} \cdot \frac{1}{\sqrt{a_i}} = \sum_{i=1}^n 1 = n
$$

So:

$$
n^2 \leq \left( \sum_{i=1}^n a_i \right)\left( \sum_{i=1}^n \frac{1}{a_i} \right)
$$

This gives us the inequality:

$$
(a + b + c + d)\left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} \right) \geq 4^2 = 16
$$

---

### Step 2: Equality Condition

Equality in the Cauchy-Schwarz inequality holds when the vectors $ (u_1, u_2, \dots, u_n) $ and $ (v_1, v_2, \dots, v_n) $ are proportional. In our case, this means:

$$
\frac{u_i}{v_i} = \frac{\sqrt{a_i}}{1/\sqrt{a_i}} = a_i \text{ is constant for all } i
$$

So, all $ a_i $ must be equal. That is, the minimum is achieved when $ a = b = c = d $.

---

### Step 3: Compute the Minimum Value

Let $ a = b = c = d = k $, where $ k > 0 $. Then:

$$
(a + b + c + d) = 4k, \quad \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} \right) = \frac{4}{k}
$$

So the product becomes:

$$
(4k)\left( \frac{4}{k} \right) = 16
$$

---

### Final Conclusion

The minimum value of the expression is:

$$
\boxed{16}
$$

This minimum is achieved when all variables are equal, i.e., $ a = b = c = d $.

Solution:
By AM-GM,
\[a + b + c + d \ge 4 \sqrt[4]{abcd},\]and
\[\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} \ge 4 \sqrt[4]{\frac{1}{abcd}},\]so
\[(a + b + c + d) \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} \right) \ge 4 \sqrt[4]{abcd} \cdot 4 \sqrt[4]{\frac{1}{abcd}} = 16.\]Equality occurs when $a = b = c = d,$ so the minimum value is $\boxed{16}.$


Explanation:
Okay, so I need to solve this equation: (6r² - 19r - 7)/(2r - 7) = 4r - 3. Hmm, let me think about how to approach this.

First, I remember that when there's a fraction equal to something, maybe I can cross-multiply or eliminate the denominator. Let me try that. If I multiply both sides by (2r - 7), that should get rid of the denominator. Let me write that down:

Multiply both sides by (2r - 7):

6r² - 19r - 7 = (4r - 3)(2r - 7)

Okay, now I need to expand the right side. Let me do that step by step.

First, multiply 4r by (2r - 7):

4r * 2r = 8r²

4r * (-7) = -28r

Then, multiply -3 by (2r - 7):

-3 * 2r = -6r

-3 * (-7) = 21

So combining all those terms:

8r² - 28r - 6r + 21

Combine like terms: -28r -6r is -34r

So the right side becomes 8r² - 34r + 21

So now the equation is:

6r² - 19r - 7 = 8r² - 34r + 21

Hmm, let me bring all terms to one side so that I can solve for r. Let me subtract the right side from both sides:

6r² - 19r - 7 - 8r² + 34r - 21 = 0

Wait, let me check that again. If I subtract 8r² -34r +21 from both sides, it's:

Left side minus right side equals zero:

(6r² - 19r -7) - (8r² -34r +21) = 0

Which is 6r² -19r -7 -8r² +34r -21 = 0

Now combine like terms:

6r² -8r² = -2r²

-19r +34r = 15r

-7 -21 = -28

So the equation simplifies to:

-2r² +15r -28 = 0

Hmm, maybe I should multiply both sides by -1 to make the coefficient of r² positive:

2r² -15r +28 = 0

Now, I need to solve this quadratic equation. Let me check if it factors. Let me think of two numbers that multiply to 2*28=56 and add up to -15. Wait, since the middle term is -15r and the constant term is +28, the numbers should multiply to +56 and add to -15. Let me think... factors of 56: 1 & 56, 2 & 28, 4 & 14, 7 & 8.

Looking for two numbers that add up to 15 (since both are negative, their sum would be -15). Let me check 7 and 8: 7 + 8 = 15. So if I use -7 and -8, they multiply to +56 and add to -15.

So, let me rewrite the middle term:

2r² -7r -8r +28 = 0

Group the first two and last two terms:

(2r² -7r) - (8r -28) = 0

Wait, actually, let me check grouping:

Wait, 2r² -7r -8r +28. So group as (2r² -7r) + (-8r +28).

Factor out common terms:

From the first group: r(2r -7)

From the second group: -4(2r -7). Wait, let me check:

-8r +28 = -4*2r + (-4)*(-7) = -4(2r -7). Yes, that's right.

So now we have:

r(2r -7) -4(2r -7) = 0

Factor out (2r -7):

(2r -7)(r -4) = 0

So, the solutions are when 2r -7 = 0 or r -4 = 0.

Therefore, r = 7/2 or r = 4.

But wait, I need to check if these solutions make the original denominator zero. The original equation has a denominator of 2r -7. If 2r -7 = 0, then r = 7/2. So, r = 7/2 would make the denominator zero, which is undefined. Therefore, r = 7/2 is an extraneous solution and must be rejected.

So the only valid solution is r = 4.

Wait, let me check that again. Let me verify both solutions in the original equation to make sure.

First, check r = 4:

Left side: (6*(4)^2 -19*(4) -7)/(2*(4) -7) = (6*16 -76 -7)/(8 -7) = (96 -76 -7)/1 = (13)/1 = 13

Right side: 4*(4) -3 = 16 -3 = 13. Okay, so that works.

Now check r = 7/2:

Left side denominator is 2*(7/2) -7 = 7 -7 = 0. So division by zero, which is undefined. Therefore, r = 7/2 is not a valid solution. So only r = 4 is the solution.

Wait, but let me check if I did the algebra correctly. Let me go back through the steps again to make sure I didn't make a mistake.

Original equation: (6r² -19r -7)/(2r -7) = 4r -3

Multiply both sides by (2r -7):

6r² -19r -7 = (4r -3)(2r -7)

Expanding the right side:

First, 4r*2r = 8r²

4r*(-7) = -28r

-3*2r = -6r

-3*(-7) = +21

Combine terms: 8r² -28r -6r +21 = 8r² -34r +21. Correct.

Left side is 6r² -19r -7. So subtracting right side:

6r² -19r -7 -8r² +34r -21 = -2r² +15r -28 = 0. Multiply by -1: 2r² -15r +28 = 0. Factored into (2r -7)(r -4) = 0. Correct. So solutions r = 7/2 and r = 4. Then check for extraneous solutions. Correct, r = 7/2 makes denominator zero, so only r = 4 is valid.

Therefore, the answer is r = 4.

But just to be thorough, let me try plugging r = 4 into the original equation again:

Numerator: 6*(16) -19*(4) -7 = 96 -76 -7 = 13. Denominator: 2*4 -7 = 8 -7 = 1. So 13/1 = 13. Right side: 4*4 -3 = 16 -3 = 13. Yep, matches.

So I think that's correct. Therefore, the value of r is 4.

**Final Answer**
The value of $ r $ is \boxed{4}.

To solve the equation:

$$
\frac{6r^2 - 19r - 7}{2r - 7} = 4r - 3
$$

we begin by eliminating the denominator. Multiply both sides of the equation by $ 2r - 7 $, assuming $ 2r - 7 \neq 0 $ (we'll check this later):

$$
6r^2 - 19r - 7 = (4r - 3)(2r - 7)
$$

---

### Step 1: Expand the Right-Hand Side

Use the distributive property to expand the product on the right:

$$
(4r - 3)(2r - 7) = 4r \cdot 2r + 4r \cdot (-7) + (-3) \cdot 2r + (-3) \cdot (-7)
$$
$$
= 8r^2 - 28r - 6r + 21 = 8r^2 - 34r + 21
$$

So the equation becomes:

$$
6r^2 - 19r - 7 = 8r^2 - 34r + 21
$$

---

### Step 2: Move All Terms to One Side

Subtract the right-hand side from both sides:

$$
6r^2 - 19r - 7 - (8r^2 - 34r + 21) = 0
$$
$$
6r^2 - 19r - 7 - 8r^2 + 34r - 21 = 0
$$

Combine like terms:

$$
(6r^2 - 8r^2) + (-19r + 34r) + (-7 - 21) = 0
$$
$$
-2r^2 + 15r - 28 = 0
$$

Multiply both sides by $-1$ to simplify:

$$
2r^2 - 15r + 28 = 0
$$

---

### Step 3: Solve the Quadratic Equation

We factor the quadratic:

$$
2r^2 - 15r + 28 = (2r - 7)(r - 4) = 0
$$

Set each factor equal to zero:

$$
2r - 7 = 0 \Rightarrow r = \frac{7}{2}
$$
$$
r - 4 = 0 \Rightarrow r = 4
$$

---

### Step 4: Check for Extraneous Solutions

Recall that the original equation has a denominator $ 2r - 7 $. If $ r = \frac{7}{2} $, then the denominator becomes zero, which is undefined. Therefore, $ r = \frac{7}{2} $ is **not** a valid solution.

Now check $ r = 4 $ in the original equation:

**Left-hand side:**

$$
\frac{6(4)^2 - 19(4) - 7}{2(4) - 7} = \frac{96 - 76 - 7}{8 - 7} = \frac{13}{1} = 13
$$

**Right-hand side:**

$$
4(4) - 3 = 16 - 3 = 13
$$

Both sides are equal, so $ r = 4 $ is a valid solution.

---

### Final Answer

$$
\boxed{4}
$$

[Assistant]
Explanation:
Okay, so I need to solve this equation: (6r² - 19r - 7)/(2r - 7) = 4r - 3. Hmm, let me think about how to approach this.

First, I remember that when there's a fraction equal to something, maybe I can cross-multiply or eliminate the denominator. Let me try that. If I multiply both sides by (2r - 7), that should get rid of the denominator. Let me write that down:

Multiply both sides by (2r - 7):

6r² - 19r - 7 = (4r - 3)(2r - 7)

Okay, now I need to expand the right side. Let me do that step by step.

First, multiply 4r by (2r - 7):

4r * 2r = 8r²

4r * (-7) = -28r

Then, multiply -3 by (2r - 7):

-3 * 2r = -6r

-3 * (-7) = 21

So combining all those terms:

8r² - 28r - 6r + 21

Combine like terms: -28r -6r is -34r

So the right side becomes 8r² - 34r + 21

So now the equation is:

6r² - 19r - 7 = 8r² - 34r + 21

Hmm, let me bring all terms to one side so that I can solve for r. Let me subtract the right side from both sides:

6r² - 19r - 7 - 8r² + 34r - 21 = 0

Wait, let me check that again. If I subtract 8r² -34r +21 from both sides, it's:

Left side minus right side equals zero:

(6r² - 19r -7) - (8r² -34r +21) = 0

Which is 6r² -19r -7 -8r² +34r -21 = 0

Now combine like terms:

6r² -8r² = -2r²

-19r +34r = 15r

-7 -21 = -28

So the equation simplifies to:

-2r² +15r -28 = 0

Hmm, maybe I should multiply both sides by -1 to make the coefficient of r² positive:

2r² -15r +28 = 0

Now, I need to solve this quadratic equation. Let me check if it factors. Let me think of two numbers that multiply to 2*28=56 and add up to -15. Wait, since the middle term is -15r and the constant term is +28, the numbers should multiply to +56 and add to -15. Let me think... factors of 56: 1 & 56, 2 & 28, 4 & 14, 7 & 8.

Looking for two numbers that add up to 15 (since both are negative, their sum would be -15). Let me check 7 and 8: 7 + 8 = 15. So if I use -7 and -8, they multiply to +56 and add to -15.

So, let me rewrite the middle term:

2r² -7r -8r +28 = 0

Group the first two and last two terms:

(2r² -7r) - (8r -28) = 0

Wait, actually, let me check grouping:

Wait, 2r² -7r -8r +28. So group as (2r² -7r) + (-8r +28).

Factor out common terms:

From the first group: r(2r -7)

From the second group: -4(2r -7). Wait, let me check:

-8r +28 = -4*2r + (-4)*(-7) = -4(2r -7). Yes, that's right.

So now we have:

r(2r -7) -4(2r -7) = 0

Factor out (2r -7):

(2r -7)(r -4) = 0

So, the solutions are when 2r -7 = 0 or r -4 = 0.

Therefore, r = 7/2 or r = 4.

But wait, I need to check if these solutions make the original denominator zero. The original equation has a denominator of 2r -7. If 2r -7 = 0, then r = 7/2. So, r = 7/2 would make the denominator zero, which is undefined. Therefore, r = 7/2 is an extraneous solution and must be rejected.

So the only valid solution is r = 4.

Wait, let me check that again. Let me verify both solutions in the original equation to make sure.

First, check r = 4:

Left side: (6*(4)^2 -19*(4) -7)/(2*(4) -7) = (6*16 -76 -7)/(8 -7) = (96 -76 -7)/1 = (13)/1 = 13

Right side: 4*(4) -3 = 16 -3 = 13. Okay, so that works.

Now check r = 7/2:

Left side denominator is 2*(7/2) -7 = 7 -7 = 0. So division by zero, which is undefined. Therefore, r = 7/2 is not a valid solution. So only r = 4 is the solution.

Wait, but let me check if I did the algebra correctly. Let me go back through the steps again to make sure I didn't make a mistake.

Original equation: (6r² -19r -7)/(2r -7) = 4r -3

Multiply both sides by (2r -7):

6r² -19r -7 = (4r -3)(2r -7)

Expanding the right side:

First, 4r*2r = 8r²

4r*(-7) = -28r

-3*2r = -6r

-3*(-7) = +21

Combine terms: 8r² -28r -6r +21 = 8r² -34r +21. Correct.

Left side is 6r² -19r -7. So subtracting right side:

6r² -19r -7 -8r² +34r -21 = -2r² +15r -28 = 0. Multiply by -1: 2r² -15r +28 = 0. Factored into (2r -7)(r -4) = 0. Correct. So solutions r = 7/2 and r = 4. Then check for extraneous solutions. Correct, r = 7/2 makes denominator zero, so only r = 4 is valid.

Therefore, the answer is r = 4.

But just to be thorough, let me try plugging r = 4 into the original equation again:

Numerator: 6*(16) -19*(4) -7 = 96 -76 -7 = 13. Denominator: 2*4 -7 = 8 -7 = 1. So 13/1 = 13. Right side: 4*4 -3 = 16 -3 = 13. Yep, matches.

So I think that's correct. Therefore, the value of r is 4.

**Final Answer**
The value of $ r $ is \boxed{4}.

To solve the equation:

$$
\frac{6r^2 - 19r - 7}{2r - 7} = 4r - 3
$$

we begin by eliminating the denominator. Multiply both sides of the equation by $ 2r - 7 $, assuming $ 2r - 7 \neq 0 $ (we'll check this later):

$$
6r^2 - 19r - 7 = (4r - 3)(2r - 7)
$$

---

### Step 1: Expand the Right-Hand Side

Use the distributive property to expand the product on the right:

$$
(4r - 3)(2r - 7) = 4r \cdot 2r + 4r \cdot (-7) + (-3) \cdot 2r + (-3) \cdot (-7)
$$
$$
= 8r^2 - 28r - 6r + 21 = 8r^2 - 34r + 21
$$

So the equation becomes:

$$
6r^2 - 19r - 7 = 8r^2 - 34r + 21
$$

---

### Step 2: Move All Terms to One Side

Subtract the right-hand side from both sides:

$$
6r^2 - 19r - 7 - (8r^2 - 34r + 21) = 0
$$
$$
6r^2 - 19r - 7 - 8r^2 + 34r - 21 = 0
$$

Combine like terms:

$$
(6r^2 - 8r^2) + (-19r + 34r) + (-7 - 21) = 0
$$
$$
-2r^2 + 15r - 28 = 0
$$

Multiply both sides by $-1$ to simplify:

$$
2r^2 - 15r + 28 = 0
$$

---

### Step 3: Solve the Quadratic Equation

We factor the quadratic:

$$
2r^2 - 15r + 28 = (2r - 7)(r - 4) = 0
$$

Set each factor equal to zero:

$$
2r - 7 = 0 \Rightarrow r = \frac{7}{2}
$$
$$
r - 4 = 0 \Rightarrow r = 4
$$

---

### Step 4: Check for Extraneous Solutions

Recall that the original equation has a denominator $ 2r - 7 $. If $ r = \frac{7}{2} $, then the denominator becomes zero, which is undefined. Therefore, $ r = \frac{7}{2} $ is **not** a valid solution.

Now check $ r = 4 $ in the original equation:

**Left-hand side:**

$$
\frac{6(4)^2 - 19(4) - 7}{2(4) - 7} = \frac{96 - 76 - 7}{8 - 7} = \frac{13}{1} = 13
$$

**Right-hand side:**

$$
4(4) - 3 = 16 - 3 = 13
$$

Both sides are equal, so $ r = 4 $ is a valid solution.

---

### Final Answer

$$
\boxed{4}
$$

Solution: Factoring the quadratic on the left gives $$\frac{(2r-7)(3r+1)}{2r-7} = 4r-3.$$Canceling the common factor on the left gives $3r + 1 = 4r - 3$.  Solving this equation gives $r = \boxed{4}$.