Explanation:
Okay, so I need to find the least possible sum of a set of distinct positive integers whose product is 84. Let me think about how to approach this.

First, I know that the problem is about factoring 84 into distinct integers and then finding the combination that adds up to the smallest possible sum. Since the integers have to be distinct, I can't just use the same number multiple times.

Let me start by factoring 84 into its prime factors. The prime factorization of 84 is 2 × 2 × 3 × 7. So, 2² × 3 × 7. But since we need distinct integers, I can't just use the prime factors directly because there are two 2s. Maybe I need to combine some of these prime factors to make distinct numbers.

Let me list all the factors of 84 first. The factors are 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84. But since we need distinct integers, and their product is 84, we need to pick a subset of these factors (excluding 1? Wait, no, 1 is allowed as long as it's distinct. But including 1 might not be helpful because multiplying by 1 doesn't change the product, but it adds 1 to the sum. So maybe we can avoid using 1? Let me check.

But maybe using 1 is necessary? Let me think. For example, if I have numbers like 2, 3, 7, their product is 42, which is half of 84. If I want to get 84, maybe I need to multiply one of them by 2? But then I have to make sure that the numbers are distinct. Let me try different combinations.

Alternatively, maybe I can think of breaking down the prime factors into different numbers. Let me try to find all possible sets of distinct integers whose product is 84 and then calculate their sums.

First, let me consider the number of elements in the set. The more numbers I have, the smaller the sum might be? Or maybe not? For example, if I have more numbers, but they are smaller, maybe the sum is smaller. But if I have fewer numbers, maybe they are larger, but the sum could be smaller. So it's a balance.

Let me try different possibilities.

Case 1: Using 1. If I include 1, then the remaining product is 84. So I need to find distinct integers greater than 1 whose product is 84. Let me try that.

For example, 1, 2, 3, 7. Their product is 1×2×3×7=42. Not enough. So need to adjust. Maybe 1, 2, 3, 14? 1×2×3×14=84. The sum is 1+2+3+14=20. Alternatively, 1, 2, 6, 7. Product is 1×2×6×7=84. Sum is 1+2+6+7=16. That's better. What about 1, 3, 4, 7? 1×3×4×7=84. Sum is 1+3+4+7=15. Even better. How about 1, 2, 2, 21? But they have to be distinct, so duplicates aren't allowed. So that's invalid. How about 1, 2, 3, 4, 7? Product is 1×2×3×4×7=168, which is too big. So that's not good.

Alternatively, maybe 1, 2, 3, 14 (sum 20), 1, 2, 6, 7 (sum 16), 1, 3, 4, 7 (sum 15). What about 1, 4, 3, 7? Same as above. What about 1, 2, 4, 21? Product is 1×2×4×21=168. No. 1, 2, 7, 6? Same as 1,2,6,7.

Alternatively, maybe 1, 3, 2, 14? Same as 1,2,3,14.

Is there a way to get a smaller sum? Let me think. What if I use more numbers? Let's try with 5 numbers. For example, 1, 2, 3, 4, 7. But product is 1×2×3×4×7=168. Too big. If I replace 7 with something smaller? But 7 is already the largest factor. Wait, maybe 1, 2, 3, 4, something? Let me check. If I have 1, 2, 3, 4, then the product is 24. To get 84, I need to multiply by 84/24=3.5. Not an integer. So that's not possible.

Alternatively, maybe 1, 2, 3, 7, and then... but that's already four numbers. If I want five numbers, maybe 1, 2, 3, 4, and then... but as above, not possible.

Alternatively, maybe 1, 2, 3, 2, 7? But duplicates again. Not allowed.

So maybe with 4 numbers, the best we can do is 1, 3, 4, 7 with sum 15. Let me check if there are other combinations with 4 numbers.

What about 1, 2, 7, 6? That's 1+2+6+7=16.

What about 1, 2, 3, 14? 1+2+3+14=20.

What about 1, 3, 7, 4? Same as 1,3,4,7.

What about 1, 2, 4, 21? Sum is 1+2+4+21=28. Worse.

How about 1, 3, 2, 14? Same as before.

Alternatively, maybe 1, 2, 3, 4, 7? But that's five numbers with product 168. Not helpful.

Alternatively, maybe 1, 2, 3, 7, and then divide one of the numbers? No, we can't.

So maybe 1, 3, 4, 7 is the best with 4 numbers.

But maybe there's a combination without 1? Let me check.

Case 2: Not using 1. So all numbers are at least 2. Let's see.

What's the product? 84. Let me try different combinations.

First, the prime factors are 2, 2, 3, 7. So we need to combine these into distinct integers.

Let me think. Let's try to split the factors into numbers.

Option 1: 2, 3, 7, 2. But duplicates. Not allowed.

Option 2: Combine two 2s into 4. So 4, 3, 7. Product is 4×3×7=84. Sum is 4+3+7=14. That's better than 15.

Wait, that's a better sum. So 3, 4, 7. Sum is 14.

Is that allowed? Yes, because they are distinct.

So that's better. So maybe this is the minimal sum?

But let me check if there are other combinations.

For example, combining 2 and 3 into 6. Then we have 6, 2, 7. But 2 is already there? Wait, original factors are 2,2,3,7. If I combine 2 and 3 into 6, then we have 6, 2, 7. But there's still another 2. So we have 2, 2, 6, 7. But duplicates again. Not allowed.

Alternatively, combine 2 and 2 into 4, and 3 and 7 into 21. Then we have 4, 21. Product is 4×21=84. Sum is 4+21=25. Worse than 14.

Alternatively, combine 2 and 7 into 14. Then we have 14, 2, 3. But again, duplicates? Wait, original factors are 2,2,3,7. If I combine 2 and 7 into 14, then we have 14, 2, 3. But there's another 2. So duplicates again.

Alternatively, combine 2 and 3 into 6, and 2 and 7 into 14. Then we have 6 and 14. Product is 6×14=84. Sum is 6+14=20. Worse than 14.

Alternatively, combine 3 and 2 into 6, and leave the other 2 and 7. So 6, 2, 7. But duplicates? Wait, there are two 2s? No, original factors are two 2s. If we combine one 2 with 3 into 6, then we have one 2 left. So numbers are 6, 2, 7. All distinct. Wait, that's allowed. Wait, 6, 2, 7. Product is 6×2×7=84. Sum is 6+2+7=15. Which is worse than 14.

So the combination of 4, 3, 7 gives sum 14.

Is there another combination? Let me think.

What if we split into more numbers? For example, 2, 3, 4, 7? Wait, product is 2×3×4×7=168. Too big.

Wait, but if we have 2, 3, 7, and then divide one of them? No, we can't.

Alternatively, maybe 2, 3, 14? Product is 2×3×14=84. Sum is 2+3+14=19. Worse than 14.

Alternatively, 2, 6, 7. Product is 2×6×7=84. Sum is 15.

Alternatively, 3, 4, 7. Sum 14.

What about 2, 2, 3, 7? But duplicates. Not allowed.

Alternatively, 1, 2, 3, 14? Sum 20.

So seems like 3, 4, 7 is better.

Wait, but can we have more numbers? Let me check.

For example, 2, 3, 4, 7? Product is 168. Too big.

If I want to have more numbers, maybe split one of the numbers into smaller ones. For example, take 4 and split into 2 and 2. But duplicates again.

Alternatively, take 3 and split into 1 and 3? But then we have 1. Let me check.

If I have 1, 2, 2, 3, 7. But duplicates again.

Alternatively, 1, 2, 3, 4, 7. Product is 1×2×3×4×7=168. Not 84.

Alternatively, 1, 2, 3, 7, and then divide one of the numbers? Not sure.

Alternatively, maybe 1, 2, 3, 4, and then 7 divided by something? Not helpful.

Alternatively, maybe 1, 2, 3, 7, and then 1? But duplicates.

So maybe not.

Alternatively, maybe 2, 3, 7, and then split 4 into 2 and 2? But duplicates.

So seems like the best combination without 1 is 3, 4, 7 with sum 14.

But let me check if there are other combinations.

What about 2, 3, 14? Sum 19.

Or 2, 6, 7? Sum 15.

Or 3, 4, 7? Sum 14.

What about 1, 3, 4, 7? Sum 15.

So 14 is better.

Is there a way to get even lower? Let me think.

What if we have more numbers? Let me try.

Suppose we have four numbers. Let me think.

If I have numbers like 2, 3, 4, 7. Product is 168. Too big.

If I have numbers like 2, 3, 2, 7. Duplicates.

Alternatively, 2, 3, 1, 14. Sum is 2+3+1+14=20.

Alternatively, 2, 3, 4, 7 divided by something? Not sure.

Alternatively, maybe 2, 3, 7, and then split 4 into 2 and 2? But duplicates.

Alternatively, maybe 1, 2, 3, 4, 7 divided by something? Not helpful.

Alternatively, think of other factor combinations.

Wait, 84 can be written as 2×42, 3×28, 4×21, 6×14, 7×12.

So if we take two numbers: 6 and 14. Sum is 20.

Or 7 and 12. Sum is 19.

Or 4 and 21. Sum 25.

Or 3 and 28. Sum 31.

Or 2 and 42. Sum 44.

So the best two numbers is 7 and 12 with sum 19.

But we already have a better sum with three numbers: 3, 4, 7 (sum 14).

What about three numbers? Let me check all possibilities.

Possible triplets:

- 2, 3, 14: product 84, sum 19

- 2, 4, 10.5: Not integer.

- 2, 6, 7: product 84, sum 15

- 3, 4, 7: product 84, sum 14

- 3, 2, 14: same as first

- 4, 3, 7: same as third

- 1, 2, 42: sum 45

- 1, 3, 28: sum 32

- 1, 4, 21: sum 26

- 1, 6, 14: sum 21

- 1, 7, 12: sum 20

So the best triplet is 3, 4, 7 with sum 14.

What about four numbers? Let me check if there's a combination with four numbers that gives a lower sum than 14.

Earlier, I thought of 1, 3, 4, 7 with sum 15.

Is there another combination?

For example, 2, 3, 4, 7: product is 168. Not 84.

If I take 2, 3, 2, 7: duplicates.

If I take 1, 2, 3, 14: sum 20.

If I take 1, 2, 6, 7: sum 16.

If I take 1, 3, 4, 7: sum 15.

If I take 1, 2, 4, 21: sum 28.

So no, the best four-number combination is 15.

What about five numbers? Let me check.

For example, 1, 2, 3, 4, 7: product is 1×2×3×4×7=168. Too big.

If I want product 84, maybe divide one of the numbers? Not sure.

Alternatively, 1, 2, 3, 7, and then... but need to have product 84. Let me think.

If I have 1, 2, 3, 7, then the product is 42. So need another number which is 84 / 42 = 2. But 2 is already there. So duplicate. Not allowed.

Alternatively, replace 7 with something else? Let me think.

If I have 1, 2, 3, 4, then product is 24. Need 84 / 24 = 3.5. Not integer.

Alternatively, 1, 2, 3, 6. Product is 36. Need 84 / 36 = 2.333... Not integer.

Alternatively, 1, 2, 3, 7. Product 42. Need 2.

Alternatively, 1, 2, 4, 7. Product 56. Need 84 / 56 = 1.5. Not integer.

Alternatively, 1, 3, 4, 7. Product 84. Wait, that's four numbers. So that's already considered.

So five numbers seems not possible.

What about six numbers? Let me think.

For example, 1, 1, 2, 3, 7, 2. But duplicates.

Alternatively, 1, 2, 3, 7, and then split into smaller numbers. But need to keep product 84.

For example, 1, 2, 3, 7, and then split 2 into 1 and 2? But duplicates again.

Not helpful.

So seems like the minimal sum is 14 with the set {3, 4, 7}.

But let me check if there are other triplets.

For example, 2, 3, 14 (sum 19), 2, 6, 7 (sum 15), 3, 4, 7 (sum 14).

Is there another triplet?

What about 1, 2, 42? Sum 45.

Or 1, 3, 28? Sum 32.

Or 1, 4, 21? Sum 26.

Or 1, 6, 14? Sum 21.

Or 1, 7, 12? Sum 20.

So no, the triplet with sum 14 is the best.

Wait, but what about using more numbers? For example, if I use four numbers, but not including 1.

Is that possible? Let me check.

For example, 2, 3, 4, 7. Product is 168. Too big.

If I can split one of the numbers into smaller ones. For example, split 4 into 2 and 2. But duplicates.

Alternatively, split 7 into 1 and 7. But then we have 1.

So 1, 2, 2, 3, 7. Duplicates again.

Alternatively, split 3 into 1 and 3. Then 1, 2, 1, 3, 7. Duplicates.

Not helpful.

Alternatively, split 2 into 1 and 2. Then 1, 1, 2, 3, 7. Duplicates.

So seems like not possible.

Alternatively, maybe 2, 3, 7, and then split 2 into 1 and 2? But again duplicates.

So no.

Therefore, the minimal sum is 14.

But let me check if there is another combination with three numbers.

For example, 2, 3, 14 (sum 19), 2, 7, 6 (sum 15), 3, 4, 7 (sum 14).

Is there another triplet?

What about 2, 2, 21? Duplicates.

Or 3, 2, 14? Same as before.

Or 4, 3, 7? Same as before.

So no.

Alternatively, 1, 2, 3, 14? But that's four numbers.

So seems like 14 is the minimal.

But let me check if there is a combination with more numbers but smaller sum.

For example, if I have 1, 2, 3, 4, 7. Product is 168. If I can divide one of the numbers by something? Not sure.

Alternatively, maybe 1, 2, 3, 4, and then replace 7 with something? Not helpful.

Alternatively, think of other factor combinations.

Wait, 84 can be written as 1×2×3×14, sum 20.

Or 1×2×6×7, sum 16.

Or 1×3×4×7, sum 15.

Or 2×3×14, sum 19.

Or 2×6×7, sum 15.

Or 3×4×7, sum 14.

So the minimal is 14.

But let me check if there is a combination with more numbers. For example, 1, 2, 3, 4, 7 divided by something? Not sure.

Alternatively, think of breaking down numbers further.

For example, take 3, 4, 7. If I can split 4 into 2 and 2, but duplicates.

Alternatively, split 3 into 1 and 3. Then we have 1, 3, 2, 2, 7. Duplicates again.

Alternatively, split 7 into 1 and 7. Then 1, 1, 3, 4, 7. Duplicates.

So no.

Alternatively, split 3 and 4 into smaller numbers.

For example, 2, 3, 2, 7. Duplicates.

So no.

Therefore, I think the minimal sum is 14.

But let me check once again.

Is there a way to have four numbers without 1?

For example, 2, 3, 4, 7. Product is 168. If I can divide one of the numbers by something?

Wait, if I take 2, 3, 4, and then instead of 7, take 7 divided by something? Not sure.

Alternatively, maybe replace 7 with 3 and 7/3? Not integer.

Alternatively, replace 4 with 2 and 2. But duplicates.

So no.

Alternatively, replace 3 with 1 and 3. Then 1, 2, 2, 7. Duplicates.

No.

So no, seems like no way to get four numbers without 1.

Therefore, the minimal sum is 14.

But let me check if there is another triplet.

Wait, 84 divided by 2 is 42. So if I have 2 and then split 42 into two distinct numbers.

For example, 2, 3, 14. Sum 19.

Or 2, 6, 7. Sum 15.

Or 2, 2, 21. Duplicates.

Or 2, 1, 42. Sum 45.

So no.

Alternatively, split 84 into three numbers.

For example, 3, 4, 7.

Or 2, 6, 7.

Or 3, 2, 14.

So the minimal is 14.

Therefore, I think the answer is 14.

But let me check once again if there is a combination with more numbers.

For example, 1, 2, 3, 4, 7. Product is 168. If I can divide one of the numbers by 2?

But then I need to adjust.

Suppose I take 1, 2, 3, 4, and then instead of 7, take 7 divided by something.

But 84 divided by (1×2×3×4) is 84 / 24 = 3.5. Not integer.

Alternatively, replace 4 with 3. Then numbers are 1, 2, 3, 3, 7. Duplicates.

No.

Alternatively, replace 2 with 1. Then numbers are 1, 1, 3, 4, 7. Duplicates.

No.

Alternatively, replace 3 with 2. Then numbers are 1, 2, 2, 4, 7. Duplicates.

No.

So no.

Alternatively, think of other numbers.

For example, 1, 2, 3, 7, and then 84 divided by (1×2×3×7) = 84 / 42 = 2. But 2 is already there.

So duplicates.

Alternatively, replace 7 with 2 and 7/2? Not integer.

No.

So no.

Therefore, I think the minimal sum is indeed 14.

But let me check if there is a combination with 3 numbers where one of them is 1.

For example, 1, 4, 21. Sum 26.

Or 1, 3, 28. Sum 32.

Or 1, 2, 42. Sum 45.

Or 1, 6, 14. Sum 21.

Or 1, 7, 12. Sum 20.

So none of these are better than 14.

Therefore, the minimal sum is 14.

Wait, but let me check if there is a combination with 3 numbers where one of them is 1.

But even if I take 1, 2, 42. Sum is 45.

No.

Alternatively, 1, 3, 28. Sum 32.

No.

So no.

Therefore, the answer is 14.

But let me check once again.

Original prime factors: 2, 2, 3, 7.

If I combine the two 2s into 4, then I have 3, 4, 7.

Yes, that's 3×4×7=84.

Sum is 3+4+7=14.

Alternatively, if I combine one 2 with 3 into 6, then I have 2, 6, 7.

Sum is 2+6+7=15.

Which is worse.

Alternatively, combine one 2 with 7 into 14, then have 2, 3, 14. Sum 19.

So 14 is better.

Alternatively, combine 3 and 7 into 21, then have 2, 2, 21. Duplicates.

Not allowed.

Alternatively, combine 2 and 3 into 6, and 2 and 7 into 14. Then have 6 and 14. Product 6×14=84. Sum 20.

So worse.

Therefore, the minimal sum is 14.

Therefore, the answer is 14.

**Final Answer**
The least possible sum of these integers is \boxed{14}.

To solve the problem of finding the **least possible sum** of a set of **distinct positive integers** whose **product is 84**, we begin by analyzing the prime factorization of 84:

$$
84 = 2^2 \times 3 \times 7
$$

Our goal is to **group these prime factors** into **distinct integers** such that their **product is 84**, and their **sum is minimized**.

---

### Step 1: Consider All Possible Sets of Distinct Integers

We want to find all combinations of **distinct positive integers** whose product is 84, and then compute their sums to find the **minimum**.

Let’s explore different groupings of the prime factors into distinct integers:

#### Option 1: Use 3, 4, and 7
- Product: $3 \times 4 \times 7 = 84$
- Sum: $3 + 4 + 7 = 14$

#### Option 2: Use 2, 6, and 7
- Product: $2 \times 6 \times 7 = 84$
- Sum: $2 + 6 + 7 = 15$

#### Option 3: Use 2, 3, and 14
- Product: $2 \times 3 \times 14 = 84$
- Sum: $2 + 3 + 14 = 19$

#### Option 4: Use 1, 2, 6, and 7
- Product: $1 \times 2 \times 6 \times 7 = 84$
- Sum: $1 + 2 + 6 + 7 = 16$

#### Option 5: Use 1, 3, 4, and 7
- Product: $1 \times 3 \times 4 \times 7 = 84$
- Sum: $1 + 3 + 4 + 7 = 15$

#### Option 6: Use 1, 2, 3, and 14
- Product: $1 \times 2 \times 3 \times 14 = 84$
- Sum: $1 + 2 + 3 + 14 = 20$

---

### Step 2: Compare All Valid Combinations

From the above, the **smallest sum** is achieved with the set:

$$
\{3, 4, 7\}
$$

- Product: $3 \times 4 \times 7 = 84$
- Sum: $3 + 4 + 7 = 14$

All other combinations either have a **larger sum** or **duplicate numbers**, which are not allowed.

---

### Final Answer

$$
\boxed{14}
$$

[Assistant]
Explanation:
Okay, so I need to find the least possible sum of a set of distinct positive integers whose product is 84. Let me think about how to approach this.

First, I know that the problem is about factoring 84 into distinct integers and then finding the combination that adds up to the smallest possible sum. Since the integers have to be distinct, I can't just use the same number multiple times.

Let me start by factoring 84 into its prime factors. The prime factorization of 84 is 2 × 2 × 3 × 7. So, 2² × 3 × 7. But since we need distinct integers, I can't just use the prime factors directly because there are two 2s. Maybe I need to combine some of these prime factors to make distinct numbers.

Let me list all the factors of 84 first. The factors are 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84. But since we need distinct integers, and their product is 84, we need to pick a subset of these factors (excluding 1? Wait, no, 1 is allowed as long as it's distinct. But including 1 might not be helpful because multiplying by 1 doesn't change the product, but it adds 1 to the sum. So maybe we can avoid using 1? Let me check.

But maybe using 1 is necessary? Let me think. For example, if I have numbers like 2, 3, 7, their product is 42, which is half of 84. If I want to get 84, maybe I need to multiply one of them by 2? But then I have to make sure that the numbers are distinct. Let me try different combinations.

Alternatively, maybe I can think of breaking down the prime factors into different numbers. Let me try to find all possible sets of distinct integers whose product is 84 and then calculate their sums.

First, let me consider the number of elements in the set. The more numbers I have, the smaller the sum might be? Or maybe not? For example, if I have more numbers, but they are smaller, maybe the sum is smaller. But if I have fewer numbers, maybe they are larger, but the sum could be smaller. So it's a balance.

Let me try different possibilities.

Case 1: Using 1. If I include 1, then the remaining product is 84. So I need to find distinct integers greater than 1 whose product is 84. Let me try that.

For example, 1, 2, 3, 7. Their product is 1×2×3×7=42. Not enough. So need to adjust. Maybe 1, 2, 3, 14? 1×2×3×14=84. The sum is 1+2+3+14=20. Alternatively, 1, 2, 6, 7. Product is 1×2×6×7=84. Sum is 1+2+6+7=16. That's better. What about 1, 3, 4, 7? 1×3×4×7=84. Sum is 1+3+4+7=15. Even better. How about 1, 2, 2, 21? But they have to be distinct, so duplicates aren't allowed. So that's invalid. How about 1, 2, 3, 4, 7? Product is 1×2×3×4×7=168, which is too big. So that's not good.

Alternatively, maybe 1, 2, 3, 14 (sum 20), 1, 2, 6, 7 (sum 16), 1, 3, 4, 7 (sum 15). What about 1, 4, 3, 7? Same as above. What about 1, 2, 4, 21? Product is 1×2×4×21=168. No. 1, 2, 7, 6? Same as 1,2,6,7.

Alternatively, maybe 1, 3, 2, 14? Same as 1,2,3,14.

Is there a way to get a smaller sum? Let me think. What if I use more numbers? Let's try with 5 numbers. For example, 1, 2, 3, 4, 7. But product is 1×2×3×4×7=168. Too big. If I replace 7 with something smaller? But 7 is already the largest factor. Wait, maybe 1, 2, 3, 4, something? Let me check. If I have 1, 2, 3, 4, then the product is 24. To get 84, I need to multiply by 84/24=3.5. Not an integer. So that's not possible.

Alternatively, maybe 1, 2, 3, 7, and then... but that's already four numbers. If I want five numbers, maybe 1, 2, 3, 4, and then... but as above, not possible.

Alternatively, maybe 1, 2, 3, 2, 7? But duplicates again. Not allowed.

So maybe with 4 numbers, the best we can do is 1, 3, 4, 7 with sum 15. Let me check if there are other combinations with 4 numbers.

What about 1, 2, 7, 6? That's 1+2+6+7=16.

What about 1, 2, 3, 14? 1+2+3+14=20.

What about 1, 3, 7, 4? Same as 1,3,4,7.

What about 1, 2, 4, 21? Sum is 1+2+4+21=28. Worse.

How about 1, 3, 2, 14? Same as before.

Alternatively, maybe 1, 2, 3, 4, 7? But that's five numbers with product 168. Not helpful.

Alternatively, maybe 1, 2, 3, 7, and then divide one of the numbers? No, we can't.

So maybe 1, 3, 4, 7 is the best with 4 numbers.

But maybe there's a combination without 1? Let me check.

Case 2: Not using 1. So all numbers are at least 2. Let's see.

What's the product? 84. Let me try different combinations.

First, the prime factors are 2, 2, 3, 7. So we need to combine these into distinct integers.

Let me think. Let's try to split the factors into numbers.

Option 1: 2, 3, 7, 2. But duplicates. Not allowed.

Option 2: Combine two 2s into 4. So 4, 3, 7. Product is 4×3×7=84. Sum is 4+3+7=14. That's better than 15.

Wait, that's a better sum. So 3, 4, 7. Sum is 14.

Is that allowed? Yes, because they are distinct.

So that's better. So maybe this is the minimal sum?

But let me check if there are other combinations.

For example, combining 2 and 3 into 6. Then we have 6, 2, 7. But 2 is already there? Wait, original factors are 2,2,3,7. If I combine 2 and 3 into 6, then we have 6, 2, 7. But there's still another 2. So we have 2, 2, 6, 7. But duplicates again. Not allowed.

Alternatively, combine 2 and 2 into 4, and 3 and 7 into 21. Then we have 4, 21. Product is 4×21=84. Sum is 4+21=25. Worse than 14.

Alternatively, combine 2 and 7 into 14. Then we have 14, 2, 3. But again, duplicates? Wait, original factors are 2,2,3,7. If I combine 2 and 7 into 14, then we have 14, 2, 3. But there's another 2. So duplicates again.

Alternatively, combine 2 and 3 into 6, and 2 and 7 into 14. Then we have 6 and 14. Product is 6×14=84. Sum is 6+14=20. Worse than 14.

Alternatively, combine 3 and 2 into 6, and leave the other 2 and 7. So 6, 2, 7. But duplicates? Wait, there are two 2s? No, original factors are two 2s. If we combine one 2 with 3 into 6, then we have one 2 left. So numbers are 6, 2, 7. All distinct. Wait, that's allowed. Wait, 6, 2, 7. Product is 6×2×7=84. Sum is 6+2+7=15. Which is worse than 14.

So the combination of 4, 3, 7 gives sum 14.

Is there another combination? Let me think.

What if we split into more numbers? For example, 2, 3, 4, 7? Wait, product is 2×3×4×7=168. Too big.

Wait, but if we have 2, 3, 7, and then divide one of them? No, we can't.

Alternatively, maybe 2, 3, 14? Product is 2×3×14=84. Sum is 2+3+14=19. Worse than 14.

Alternatively, 2, 6, 7. Product is 2×6×7=84. Sum is 15.

Alternatively, 3, 4, 7. Sum 14.

What about 2, 2, 3, 7? But duplicates. Not allowed.

Alternatively, 1, 2, 3, 14? Sum 20.

So seems like 3, 4, 7 is better.

Wait, but can we have more numbers? Let me check.

For example, 2, 3, 4, 7? Product is 168. Too big.

If I want to have more numbers, maybe split one of the numbers into smaller ones. For example, take 4 and split into 2 and 2. But duplicates again.

Alternatively, take 3 and split into 1 and 3? But then we have 1. Let me check.

If I have 1, 2, 2, 3, 7. But duplicates again.

Alternatively, 1, 2, 3, 4, 7. Product is 1×2×3×4×7=168. Not 84.

Alternatively, 1, 2, 3, 7, and then divide one of the numbers? Not sure.

Alternatively, maybe 1, 2, 3, 4, and then 7 divided by something? Not helpful.

Alternatively, maybe 1, 2, 3, 7, and then 1? But duplicates.

So maybe not.

Alternatively, maybe 2, 3, 7, and then split 4 into 2 and 2? But duplicates.

So seems like the best combination without 1 is 3, 4, 7 with sum 14.

But let me check if there are other combinations.

What about 2, 3, 14? Sum 19.

Or 2, 6, 7? Sum 15.

Or 3, 4, 7? Sum 14.

What about 1, 3, 4, 7? Sum 15.

So 14 is better.

Is there a way to get even lower? Let me think.

What if we have more numbers? Let me try.

Suppose we have four numbers. Let me think.

If I have numbers like 2, 3, 4, 7. Product is 168. Too big.

If I have numbers like 2, 3, 2, 7. Duplicates.

Alternatively, 2, 3, 1, 14. Sum is 2+3+1+14=20.

Alternatively, 2, 3, 4, 7 divided by something? Not sure.

Alternatively, maybe 2, 3, 7, and then split 4 into 2 and 2? But duplicates.

Alternatively, maybe 1, 2, 3, 4, 7 divided by something? Not helpful.

Alternatively, think of other factor combinations.

Wait, 84 can be written as 2×42, 3×28, 4×21, 6×14, 7×12.

So if we take two numbers: 6 and 14. Sum is 20.

Or 7 and 12. Sum is 19.

Or 4 and 21. Sum 25.

Or 3 and 28. Sum 31.

Or 2 and 42. Sum 44.

So the best two numbers is 7 and 12 with sum 19.

But we already have a better sum with three numbers: 3, 4, 7 (sum 14).

What about three numbers? Let me check all possibilities.

Possible triplets:

- 2, 3, 14: product 84, sum 19

- 2, 4, 10.5: Not integer.

- 2, 6, 7: product 84, sum 15

- 3, 4, 7: product 84, sum 14

- 3, 2, 14: same as first

- 4, 3, 7: same as third

- 1, 2, 42: sum 45

- 1, 3, 28: sum 32

- 1, 4, 21: sum 26

- 1, 6, 14: sum 21

- 1, 7, 12: sum 20

So the best triplet is 3, 4, 7 with sum 14.

What about four numbers? Let me check if there's a combination with four numbers that gives a lower sum than 14.

Earlier, I thought of 1, 3, 4, 7 with sum 15.

Is there another combination?

For example, 2, 3, 4, 7: product is 168. Not 84.

If I take 2, 3, 2, 7: duplicates.

If I take 1, 2, 3, 14: sum 20.

If I take 1, 2, 6, 7: sum 16.

If I take 1, 3, 4, 7: sum 15.

If I take 1, 2, 4, 21: sum 28.

So no, the best four-number combination is 15.

What about five numbers? Let me check.

For example, 1, 2, 3, 4, 7: product is 1×2×3×4×7=168. Too big.

If I want product 84, maybe divide one of the numbers? Not sure.

Alternatively, 1, 2, 3, 7, and then... but need to have product 84. Let me think.

If I have 1, 2, 3, 7, then the product is 42. So need another number which is 84 / 42 = 2. But 2 is already there. So duplicate. Not allowed.

Alternatively, replace 7 with something else? Let me think.

If I have 1, 2, 3, 4, then product is 24. Need 84 / 24 = 3.5. Not integer.

Alternatively, 1, 2, 3, 6. Product is 36. Need 84 / 36 = 2.333... Not integer.

Alternatively, 1, 2, 3, 7. Product 42. Need 2.

Alternatively, 1, 2, 4, 7. Product 56. Need 84 / 56 = 1.5. Not integer.

Alternatively, 1, 3, 4, 7. Product 84. Wait, that's four numbers. So that's already considered.

So five numbers seems not possible.

What about six numbers? Let me think.

For example, 1, 1, 2, 3, 7, 2. But duplicates.

Alternatively, 1, 2, 3, 7, and then split into smaller numbers. But need to keep product 84.

For example, 1, 2, 3, 7, and then split 2 into 1 and 2? But duplicates again.

Not helpful.

So seems like the minimal sum is 14 with the set {3, 4, 7}.

But let me check if there are other triplets.

For example, 2, 3, 14 (sum 19), 2, 6, 7 (sum 15), 3, 4, 7 (sum 14).

Is there another triplet?

What about 1, 2, 42? Sum 45.

Or 1, 3, 28? Sum 32.

Or 1, 4, 21? Sum 26.

Or 1, 6, 14? Sum 21.

Or 1, 7, 12? Sum 20.

So no, the triplet with sum 14 is the best.

Wait, but what about using more numbers? For example, if I use four numbers, but not including 1.

Is that possible? Let me check.

For example, 2, 3, 4, 7. Product is 168. Too big.

If I can split one of the numbers into smaller ones. For example, split 4 into 2 and 2. But duplicates.

Alternatively, split 7 into 1 and 7. But then we have 1.

So 1, 2, 2, 3, 7. Duplicates again.

Alternatively, split 3 into 1 and 3. Then 1, 2, 1, 3, 7. Duplicates.

Not helpful.

Alternatively, split 2 into 1 and 2. Then 1, 1, 2, 3, 7. Duplicates.

So seems like not possible.

Alternatively, maybe 2, 3, 7, and then split 2 into 1 and 2? But again duplicates.

So no.

Therefore, the minimal sum is 14.

But let me check if there is another combination with three numbers.

For example, 2, 3, 14 (sum 19), 2, 7, 6 (sum 15), 3, 4, 7 (sum 14).

Is there another triplet?

What about 2, 2, 21? Duplicates.

Or 3, 2, 14? Same as before.

Or 4, 3, 7? Same as before.

So no.

Alternatively, 1, 2, 3, 14? But that's four numbers.

So seems like 14 is the minimal.

But let me check if there is a combination with more numbers but smaller sum.

For example, if I have 1, 2, 3, 4, 7. Product is 168. If I can divide one of the numbers by something? Not sure.

Alternatively, maybe 1, 2, 3, 4, and then replace 7 with something? Not helpful.

Alternatively, think of other factor combinations.

Wait, 84 can be written as 1×2×3×14, sum 20.

Or 1×2×6×7, sum 16.

Or 1×3×4×7, sum 15.

Or 2×3×14, sum 19.

Or 2×6×7, sum 15.

Or 3×4×7, sum 14.

So the minimal is 14.

But let me check if there is a combination with more numbers. For example, 1, 2, 3, 4, 7 divided by something? Not sure.

Alternatively, think of breaking down numbers further.

For example, take 3, 4, 7. If I can split 4 into 2 and 2, but duplicates.

Alternatively, split 3 into 1 and 3. Then we have 1, 3, 2, 2, 7. Duplicates again.

Alternatively, split 7 into 1 and 7. Then 1, 1, 3, 4, 7. Duplicates.

So no.

Alternatively, split 3 and 4 into smaller numbers.

For example, 2, 3, 2, 7. Duplicates.

So no.

Therefore, I think the minimal sum is 14.

But let me check once again.

Is there a way to have four numbers without 1?

For example, 2, 3, 4, 7. Product is 168. If I can divide one of the numbers by something?

Wait, if I take 2, 3, 4, and then instead of 7, take 7 divided by something? Not sure.

Alternatively, maybe replace 7 with 3 and 7/3? Not integer.

Alternatively, replace 4 with 2 and 2. But duplicates.

So no.

Alternatively, replace 3 with 1 and 3. Then 1, 2, 2, 7. Duplicates.

No.

So no, seems like no way to get four numbers without 1.

Therefore, the minimal sum is 14.

But let me check if there is another triplet.

Wait, 84 divided by 2 is 42. So if I have 2 and then split 42 into two distinct numbers.

For example, 2, 3, 14. Sum 19.

Or 2, 6, 7. Sum 15.

Or 2, 2, 21. Duplicates.

Or 2, 1, 42. Sum 45.

So no.

Alternatively, split 84 into three numbers.

For example, 3, 4, 7.

Or 2, 6, 7.

Or 3, 2, 14.

So the minimal is 14.

Therefore, I think the answer is 14.

But let me check once again if there is a combination with more numbers.

For example, 1, 2, 3, 4, 7. Product is 168. If I can divide one of the numbers by 2?

But then I need to adjust.

Suppose I take 1, 2, 3, 4, and then instead of 7, take 7 divided by something.

But 84 divided by (1×2×3×4) is 84 / 24 = 3.5. Not integer.

Alternatively, replace 4 with 3. Then numbers are 1, 2, 3, 3, 7. Duplicates.

No.

Alternatively, replace 2 with 1. Then numbers are 1, 1, 3, 4, 7. Duplicates.

No.

Alternatively, replace 3 with 2. Then numbers are 1, 2, 2, 4, 7. Duplicates.

No.

So no.

Alternatively, think of other numbers.

For example, 1, 2, 3, 7, and then 84 divided by (1×2×3×7) = 84 / 42 = 2. But 2 is already there.

So duplicates.

Alternatively, replace 7 with 2 and 7/2? Not integer.

No.

So no.

Therefore, I think the minimal sum is indeed 14.

But let me check if there is a combination with 3 numbers where one of them is 1.

For example, 1, 4, 21. Sum 26.

Or 1, 3, 28. Sum 32.

Or 1, 2, 42. Sum 45.

Or 1, 6, 14. Sum 21.

Or 1, 7, 12. Sum 20.

So none of these are better than 14.

Therefore, the minimal sum is 14.

Wait, but let me check if there is a combination with 3 numbers where one of them is 1.

But even if I take 1, 2, 42. Sum is 45.

No.

Alternatively, 1, 3, 28. Sum 32.

No.

So no.

Therefore, the answer is 14.

But let me check once again.

Original prime factors: 2, 2, 3, 7.

If I combine the two 2s into 4, then I have 3, 4, 7.

Yes, that's 3×4×7=84.

Sum is 3+4+7=14.

Alternatively, if I combine one 2 with 3 into 6, then I have 2, 6, 7.

Sum is 2+6+7=15.

Which is worse.

Alternatively, combine one 2 with 7 into 14, then have 2, 3, 14. Sum 19.

So 14 is better.

Alternatively, combine 3 and 7 into 21, then have 2, 2, 21. Duplicates.

Not allowed.

Alternatively, combine 2 and 3 into 6, and 2 and 7 into 14. Then have 6 and 14. Product 6×14=84. Sum 20.

So worse.

Therefore, the minimal sum is 14.

Therefore, the answer is 14.

**Final Answer**
The least possible sum of these integers is \boxed{14}.

To solve the problem of finding the **least possible sum** of a set of **distinct positive integers** whose **product is 84**, we begin by analyzing the prime factorization of 84:

$$
84 = 2^2 \times 3 \times 7
$$

Our goal is to **group these prime factors** into **distinct integers** such that their **product is 84**, and their **sum is minimized**.

---

### Step 1: Consider All Possible Sets of Distinct Integers

We want to find all combinations of **distinct positive integers** whose product is 84, and then compute their sums to find the **minimum**.

Let’s explore different groupings of the prime factors into distinct integers:

#### Option 1: Use 3, 4, and 7
- Product: $3 \times 4 \times 7 = 84$
- Sum: $3 + 4 + 7 = 14$

#### Option 2: Use 2, 6, and 7
- Product: $2 \times 6 \times 7 = 84$
- Sum: $2 + 6 + 7 = 15$

#### Option 3: Use 2, 3, and 14
- Product: $2 \times 3 \times 14 = 84$
- Sum: $2 + 3 + 14 = 19$

#### Option 4: Use 1, 2, 6, and 7
- Product: $1 \times 2 \times 6 \times 7 = 84$
- Sum: $1 + 2 + 6 + 7 = 16$

#### Option 5: Use 1, 3, 4, and 7
- Product: $1 \times 3 \times 4 \times 7 = 84$
- Sum: $1 + 3 + 4 + 7 = 15$

#### Option 6: Use 1, 2, 3, and 14
- Product: $1 \times 2 \times 3 \times 14 = 84$
- Sum: $1 + 2 + 3 + 14 = 20$

---

### Step 2: Compare All Valid Combinations

From the above, the **smallest sum** is achieved with the set:

$$
\{3, 4, 7\}
$$

- Product: $3 \times 4 \times 7 = 84$
- Sum: $3 + 4 + 7 = 14$

All other combinations either have a **larger sum** or **duplicate numbers**, which are not allowed.

---

### Final Answer

$$
\boxed{14}
$$

Solution: We know that the prime factors of the set of numbers must equal the prime factors of 84, which are $2^2\cdot3\cdot7$. The set with the smallest sum would be the factors themselves - 2, 2, 3, and 7. However, the set can't have two 2's since the integers must be distinct, but it can have a 4, 3, and 7 instead.  The sum of those numbers is $\boxed{14}$.  We could also have paired one of the 2's with the 3, to have 2, 6, and 7, but these have sum 15.  Grouping the extra 2 with 7 gives 2, 3, and 14 (which sum to 19), and any other grouping clearly gives a sum higher than 14.


Explanation:
Okay, so I need to solve the equation |x - 4| - 10 = 2 and find the product of all possible values of x. Hmm, let me think. First, I remember that absolute value equations can sometimes have two solutions because the expression inside the absolute value can be positive or negative. Let me start by isolating the absolute value term.

The equation is |x - 4| - 10 = 2. To isolate |x - 4|, I should add 10 to both sides. Let me do that:

|x - 4| - 10 + 10 = 2 + 10

Simplifying both sides:

|x - 4| = 12

Okay, so now the absolute value of (x - 4) is equal to 12. That means that x - 4 can be either 12 or -12 because absolute value represents distance from zero on the number line. So, I can set up two separate equations:

1) x - 4 = 12
2) x - 4 = -12

Now, solving each equation for x.

Starting with the first equation:

x - 4 = 12

Add 4 to both sides:

x = 12 + 4
x = 16

Alright, so one solution is x = 16. Now the second equation:

x - 4 = -12

Again, add 4 to both sides:

x = -12 + 4
x = -8

So the other solution is x = -8. Therefore, the possible values of x are 16 and -8. The question asks for the product of all possible values of x. To find the product, I just multiply these two solutions together:

16 * (-8) = ?

Let me calculate that. 16 times 8 is 128, so 16 times -8 is -128. Therefore, the product is -128.

Wait, let me double-check my steps to make sure I didn't make any mistakes. Starting with the original equation:

|x - 4| - 10 = 2

Adding 10 to both sides gives |x - 4| = 12. That seems right. Then, solving the absolute value equation by splitting into two cases. For |A| = B, A = B or A = -B. So, x - 4 = 12 or x - 4 = -12. Solving both gives x = 16 and x = -8. Multiplying them gives 16 * (-8) = -128. That seems correct.

Alternatively, maybe there's another way to check? Let me plug the solutions back into the original equation to verify.

First, x = 16:

Left side: |16 - 4| - 10 = |12| - 10 = 12 - 10 = 2. Which matches the right side, 2. Good.

Now, x = -8:

Left side: |-8 - 4| - 10 = |-12| - 10 = 12 - 10 = 2. Also matches. Perfect.

So both solutions are valid. Therefore, the product is indeed -128. I think that's the answer. Wait, but just to make sure I didn't mix up any signs. Let me check the multiplication again. 16 times 8 is 128, so 16 times -8 is -128. Yep, that's right.

Alternatively, if I think of the product of the roots of a quadratic equation. Wait, maybe I can think of this equation as a quadratic? Let me try that approach to cross-verify.

Starting from |x - 4| - 10 = 2. We already solved it, but maybe if I remove the absolute value by squaring both sides? Wait, but squaring might complicate things. Let me see.

Alternatively, let me consider that |x - 4| = 12 implies (x - 4)^2 = 12^2 = 144. So expanding that:

(x - 4)^2 = 144

x^2 - 8x + 16 = 144

x^2 - 8x + 16 - 144 = 0

x^2 - 8x - 128 = 0

So, quadratic equation x² - 8x - 128 = 0. If I want the product of the roots, Vieta's formula says that the product is c/a, where the equation is ax² + bx + c = 0. Here, a = 1, b = -8, c = -128. Therefore, product is -128/1 = -128. Which matches the previous result. So that's another way to confirm that the product is -128. Therefore, I can be confident that the answer is correct.

So, the product of all possible values of x is -128.

**Final Answer**
The product of all possible values of $ x $ is \boxed{-128}.

To solve the equation:

$$
|x - 4| - 10 = 2
$$

we begin by isolating the absolute value expression. Add 10 to both sides:

$$
|x - 4| = 12
$$

The absolute value equation $ |x - 4| = 12 $ implies that the expression inside the absolute value, $ x - 4 $, can be either 12 or -12. This gives us two separate linear equations:

1. $ x - 4 = 12 $
2. $ x - 4 = -12 $

---

### Solving Each Equation

**First equation:**

$$
x - 4 = 12 \Rightarrow x = 16
$$

**Second equation:**

$$
x - 4 = -12 \Rightarrow x = -8
$$

So, the two possible values of $ x $ are $ 16 $ and $ -8 $.

---

### Finding the Product

To find the product of all possible values of $ x $, we multiply the two solutions:

$$
16 \times (-8) = -128
$$

---

### Verification via Quadratic Form

We can also verify this by rewriting the absolute value equation as a quadratic. Starting from:

$$
|x - 4| = 12
$$

Square both sides:

$$
(x - 4)^2 = 144
$$

Expanding:

$$
x^2 - 8x + 16 = 144
$$

$$
x^2 - 8x - 128 = 0
$$

Using **Vieta's formulas**, the product of the roots of a quadratic equation $ ax^2 + bx + c = 0 $ is given by $ \frac{c}{a} $. Here, $ a = 1 $, $ c = -128 $, so the product of the roots is:

$$
\frac{-128}{1} = -128
$$

This confirms our earlier result.

---

### Final Answer

$$
\boxed{-128}
$$

[Assistant]
Explanation:
Okay, so I need to solve the equation |x - 4| - 10 = 2 and find the product of all possible values of x. Hmm, let me think. First, I remember that absolute value equations can sometimes have two solutions because the expression inside the absolute value can be positive or negative. Let me start by isolating the absolute value term.

The equation is |x - 4| - 10 = 2. To isolate |x - 4|, I should add 10 to both sides. Let me do that:

|x - 4| - 10 + 10 = 2 + 10

Simplifying both sides:

|x - 4| = 12

Okay, so now the absolute value of (x - 4) is equal to 12. That means that x - 4 can be either 12 or -12 because absolute value represents distance from zero on the number line. So, I can set up two separate equations:

1) x - 4 = 12
2) x - 4 = -12

Now, solving each equation for x.

Starting with the first equation:

x - 4 = 12

Add 4 to both sides:

x = 12 + 4
x = 16

Alright, so one solution is x = 16. Now the second equation:

x - 4 = -12

Again, add 4 to both sides:

x = -12 + 4
x = -8

So the other solution is x = -8. Therefore, the possible values of x are 16 and -8. The question asks for the product of all possible values of x. To find the product, I just multiply these two solutions together:

16 * (-8) = ?

Let me calculate that. 16 times 8 is 128, so 16 times -8 is -128. Therefore, the product is -128.

Wait, let me double-check my steps to make sure I didn't make any mistakes. Starting with the original equation:

|x - 4| - 10 = 2

Adding 10 to both sides gives |x - 4| = 12. That seems right. Then, solving the absolute value equation by splitting into two cases. For |A| = B, A = B or A = -B. So, x - 4 = 12 or x - 4 = -12. Solving both gives x = 16 and x = -8. Multiplying them gives 16 * (-8) = -128. That seems correct.

Alternatively, maybe there's another way to check? Let me plug the solutions back into the original equation to verify.

First, x = 16:

Left side: |16 - 4| - 10 = |12| - 10 = 12 - 10 = 2. Which matches the right side, 2. Good.

Now, x = -8:

Left side: |-8 - 4| - 10 = |-12| - 10 = 12 - 10 = 2. Also matches. Perfect.

So both solutions are valid. Therefore, the product is indeed -128. I think that's the answer. Wait, but just to make sure I didn't mix up any signs. Let me check the multiplication again. 16 times 8 is 128, so 16 times -8 is -128. Yep, that's right.

Alternatively, if I think of the product of the roots of a quadratic equation. Wait, maybe I can think of this equation as a quadratic? Let me try that approach to cross-verify.

Starting from |x - 4| - 10 = 2. We already solved it, but maybe if I remove the absolute value by squaring both sides? Wait, but squaring might complicate things. Let me see.

Alternatively, let me consider that |x - 4| = 12 implies (x - 4)^2 = 12^2 = 144. So expanding that:

(x - 4)^2 = 144

x^2 - 8x + 16 = 144

x^2 - 8x + 16 - 144 = 0

x^2 - 8x - 128 = 0

So, quadratic equation x² - 8x - 128 = 0. If I want the product of the roots, Vieta's formula says that the product is c/a, where the equation is ax² + bx + c = 0. Here, a = 1, b = -8, c = -128. Therefore, product is -128/1 = -128. Which matches the previous result. So that's another way to confirm that the product is -128. Therefore, I can be confident that the answer is correct.

So, the product of all possible values of x is -128.

**Final Answer**
The product of all possible values of $ x $ is \boxed{-128}.

To solve the equation:

$$
|x - 4| - 10 = 2
$$

we begin by isolating the absolute value expression. Add 10 to both sides:

$$
|x - 4| = 12
$$

The absolute value equation $ |x - 4| = 12 $ implies that the expression inside the absolute value, $ x - 4 $, can be either 12 or -12. This gives us two separate linear equations:

1. $ x - 4 = 12 $
2. $ x - 4 = -12 $

---

### Solving Each Equation

**First equation:**

$$
x - 4 = 12 \Rightarrow x = 16
$$

**Second equation:**

$$
x - 4 = -12 \Rightarrow x = -8
$$

So, the two possible values of $ x $ are $ 16 $ and $ -8 $.

---

### Finding the Product

To find the product of all possible values of $ x $, we multiply the two solutions:

$$
16 \times (-8) = -128
$$

---

### Verification via Quadratic Form

We can also verify this by rewriting the absolute value equation as a quadratic. Starting from:

$$
|x - 4| = 12
$$

Square both sides:

$$
(x - 4)^2 = 144
$$

Expanding:

$$
x^2 - 8x + 16 = 144
$$

$$
x^2 - 8x - 128 = 0
$$

Using **Vieta's formulas**, the product of the roots of a quadratic equation $ ax^2 + bx + c = 0 $ is given by $ \frac{c}{a} $. Here, $ a = 1 $, $ c = -128 $, so the product of the roots is:

$$
\frac{-128}{1} = -128
$$

This confirms our earlier result.

---

### Final Answer

$$
\boxed{-128}
$$

Solution: We rearrange the given equation to $|x-4| = 12$.  Thus either $x-4 = 12$, meaning $x = 16$, or $x-4 = -12$, meaning $x=-8$.  Our answer is therefore $16\cdot (-8) = \boxed{-128}$.


Explanation:
Okay, let me try to figure out this problem. So, we have two functions: f(x) = x - 3 and q(x) = bx + 1. The question says that f(q(1)) = -3, and we need to find the value of b. Alright, let me break this down step by step.

First, I remember that when you have a function inside another function, like f(q(1)), you need to evaluate the inner function first and then plug that result into the outer function. So, let me start by finding q(1).

The function q(x) is given as bx + 1. So, if I substitute x with 1, that should be q(1) = b*(1) + 1. Let me write that out:

q(1) = b*1 + 1 = b + 1.

Okay, so q(1) simplifies to b + 1. Now, the next step is to plug this result into the function f(x). The function f(x) is x - 3, so replacing x with q(1) which is b + 1, we get:

f(q(1)) = f(b + 1) = (b + 1) - 3.

Let me compute that: (b + 1) - 3 is the same as b + 1 - 3, which simplifies to b - 2.

So, according to the problem, f(q(1)) equals -3. Therefore, we can set up the equation:

b - 2 = -3.

Now, solving for b, I just need to add 2 to both sides of the equation. Let me do that:

b - 2 + 2 = -3 + 2

Which simplifies to:

b = -1.

Wait, let me check my steps again to make sure I didn't make a mistake. Starting from the beginning:

1. q(1) = b*1 + 1 = b + 1. That seems right.
2. Then f(q(1)) = f(b + 1) = (b + 1) - 3. Yes, because f(x) is x - 3, so substituting x with b + 1 gives that.
3. Simplifying (b + 1) - 3 gives b - 2. Correct.
4. Setting that equal to -3: b - 2 = -3. Solving for b: add 2 to both sides, so b = -3 + 2 = -1. That seems correct.

Hmm, maybe I can verify by plugging b = -1 back into the original functions and checking if f(q(1)) is indeed -3.

Let me do that. If b = -1, then q(x) = -1*x + 1, which is q(x) = -x + 1. Then q(1) would be -1*(1) + 1 = -1 + 1 = 0. Then f(q(1)) is f(0) = 0 - 3 = -3. Which matches the given condition. So that checks out.

Therefore, my answer of b = -1 seems to be correct.

Wait, but let me think again if there's another way this could be approached or if I missed something. For example, could there be a different interpretation of the functions? Let me check the original problem again:

"f(x) = x - 3 and q(x) = bx +1. If f(q(1)) = -3, what is b?"

No, it seems straightforward. There's no ambiguity in the functions. f is a linear function, q is also a linear function with coefficient b. The composition f(q(1)) is evaluated at x=1 for q, then the result is plugged into f.

Another way to think about it: maybe using function composition notation? But I don't think that's necessary here. The problem is just about evaluating the composition at a specific point, which is 1.

Alternatively, could there be a mistake in arithmetic? Let me recheck:

Starting with f(q(1)) = -3.

q(1) = b*1 + 1 = b + 1.

Then f(q(1)) = (b + 1) - 3 = b + 1 - 3 = b - 2.

Set equal to -3: b - 2 = -3 => b = -3 + 2 = -1.

Yes, that's correct.

Alternatively, if I had messed up the signs, maybe? Let me see. Suppose I thought f(q(1)) was something else. But f is x - 3, so if q(1) is some value, say, let's call it y, then f(y) = y - 3. So if y is b + 1, then f(y) is (b + 1) - 3.

No, that's still the same.

Alternatively, maybe I misread the problem? Let me check again. The problem says f(q(1)) = -3. So, f(q(1)) is -3.

Yes. So, unless there's some other hidden step or trick here, which I don't see.

Alternatively, maybe the problem is testing understanding of function composition? For example, maybe someone might confuse q(f(1)) instead of f(q(1))? But the problem clearly states f(q(1)), so that's not the case.

Alternatively, maybe there's a typo in the problem? But assuming the problem is stated correctly, then my answer is correct.

Alternatively, maybe I need to consider that q(x) is a function, so maybe there's more to it? But no, q(x) is given as bx + 1, so it's a linear function.

Alternatively, maybe the problem is expecting a different form? For example, if they wanted the answer in fractions or something? But -1 is an integer, so that's fine.

Alternatively, maybe I need to check if there are multiple solutions? But since it's a linear equation, there should be only one solution.

So, I think my answer is correct. Therefore, the value of b is -1.

**Final Answer**
The value of $ b $ is \boxed{-1}.

To solve the problem, we are given two functions:

- $ f(x) = x - 3 $
- $ q(x) = bx + 1 $

We are also told that:

$$
f(q(1)) = -3
$$

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

---

### Step 1: Evaluate $ q(1) $

We substitute $ x = 1 $ into the function $ q(x) $:

$$
q(1) = b(1) + 1 = b + 1
$$

---

### Step 2: Evaluate $ f(q(1)) $

Now that we know $ q(1) = b + 1 $, we substitute this into the function $ f(x) $:

$$
f(q(1)) = f(b + 1) = (b + 1) - 3
$$

Simplify the expression:

$$
f(q(1)) = b + 1 - 3 = b - 2
$$

---

### Step 3: Set the expression equal to -3

We are told that $ f(q(1)) = -3 $, so:

$$
b - 2 = -3
$$

Solve for $ b $:

$$
b = -3 + 2 = -1
$$

---

### Step 4: Verify the solution

Let’s verify by plugging $ b = -1 $ back into the functions:

- $ q(x) = -x + 1 $
- $ q(1) = -1 + 1 = 0 $
- $ f(q(1)) = f(0) = 0 - 3 = -3 $

This matches the given condition, confirming that our solution is correct.

---

### Final Answer

$$
\boxed{-1}
$$

[Assistant]
Explanation:
Okay, let me try to figure out this problem. So, we have two functions: f(x) = x - 3 and q(x) = bx + 1. The question says that f(q(1)) = -3, and we need to find the value of b. Alright, let me break this down step by step.

First, I remember that when you have a function inside another function, like f(q(1)), you need to evaluate the inner function first and then plug that result into the outer function. So, let me start by finding q(1).

The function q(x) is given as bx + 1. So, if I substitute x with 1, that should be q(1) = b*(1) + 1. Let me write that out:

q(1) = b*1 + 1 = b + 1.

Okay, so q(1) simplifies to b + 1. Now, the next step is to plug this result into the function f(x). The function f(x) is x - 3, so replacing x with q(1) which is b + 1, we get:

f(q(1)) = f(b + 1) = (b + 1) - 3.

Let me compute that: (b + 1) - 3 is the same as b + 1 - 3, which simplifies to b - 2.

So, according to the problem, f(q(1)) equals -3. Therefore, we can set up the equation:

b - 2 = -3.

Now, solving for b, I just need to add 2 to both sides of the equation. Let me do that:

b - 2 + 2 = -3 + 2

Which simplifies to:

b = -1.

Wait, let me check my steps again to make sure I didn't make a mistake. Starting from the beginning:

1. q(1) = b*1 + 1 = b + 1. That seems right.
2. Then f(q(1)) = f(b + 1) = (b + 1) - 3. Yes, because f(x) is x - 3, so substituting x with b + 1 gives that.
3. Simplifying (b + 1) - 3 gives b - 2. Correct.
4. Setting that equal to -3: b - 2 = -3. Solving for b: add 2 to both sides, so b = -3 + 2 = -1. That seems correct.

Hmm, maybe I can verify by plugging b = -1 back into the original functions and checking if f(q(1)) is indeed -3.

Let me do that. If b = -1, then q(x) = -1*x + 1, which is q(x) = -x + 1. Then q(1) would be -1*(1) + 1 = -1 + 1 = 0. Then f(q(1)) is f(0) = 0 - 3 = -3. Which matches the given condition. So that checks out.

Therefore, my answer of b = -1 seems to be correct.

Wait, but let me think again if there's another way this could be approached or if I missed something. For example, could there be a different interpretation of the functions? Let me check the original problem again:

"f(x) = x - 3 and q(x) = bx +1. If f(q(1)) = -3, what is b?"

No, it seems straightforward. There's no ambiguity in the functions. f is a linear function, q is also a linear function with coefficient b. The composition f(q(1)) is evaluated at x=1 for q, then the result is plugged into f.

Another way to think about it: maybe using function composition notation? But I don't think that's necessary here. The problem is just about evaluating the composition at a specific point, which is 1.

Alternatively, could there be a mistake in arithmetic? Let me recheck:

Starting with f(q(1)) = -3.

q(1) = b*1 + 1 = b + 1.

Then f(q(1)) = (b + 1) - 3 = b + 1 - 3 = b - 2.

Set equal to -3: b - 2 = -3 => b = -3 + 2 = -1.

Yes, that's correct.

Alternatively, if I had messed up the signs, maybe? Let me see. Suppose I thought f(q(1)) was something else. But f is x - 3, so if q(1) is some value, say, let's call it y, then f(y) = y - 3. So if y is b + 1, then f(y) is (b + 1) - 3.

No, that's still the same.

Alternatively, maybe I misread the problem? Let me check again. The problem says f(q(1)) = -3. So, f(q(1)) is -3.

Yes. So, unless there's some other hidden step or trick here, which I don't see.

Alternatively, maybe the problem is testing understanding of function composition? For example, maybe someone might confuse q(f(1)) instead of f(q(1))? But the problem clearly states f(q(1)), so that's not the case.

Alternatively, maybe there's a typo in the problem? But assuming the problem is stated correctly, then my answer is correct.

Alternatively, maybe I need to consider that q(x) is a function, so maybe there's more to it? But no, q(x) is given as bx + 1, so it's a linear function.

Alternatively, maybe the problem is expecting a different form? For example, if they wanted the answer in fractions or something? But -1 is an integer, so that's fine.

Alternatively, maybe I need to check if there are multiple solutions? But since it's a linear equation, there should be only one solution.

So, I think my answer is correct. Therefore, the value of b is -1.

**Final Answer**
The value of $ b $ is \boxed{-1}.

To solve the problem, we are given two functions:

- $ f(x) = x - 3 $
- $ q(x) = bx + 1 $

We are also told that:

$$
f(q(1)) = -3
$$

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

---

### Step 1: Evaluate $ q(1) $

We substitute $ x = 1 $ into the function $ q(x) $:

$$
q(1) = b(1) + 1 = b + 1
$$

---

### Step 2: Evaluate $ f(q(1)) $

Now that we know $ q(1) = b + 1 $, we substitute this into the function $ f(x) $:

$$
f(q(1)) = f(b + 1) = (b + 1) - 3
$$

Simplify the expression:

$$
f(q(1)) = b + 1 - 3 = b - 2
$$

---

### Step 3: Set the expression equal to -3

We are told that $ f(q(1)) = -3 $, so:

$$
b - 2 = -3
$$

Solve for $ b $:

$$
b = -3 + 2 = -1
$$

---

### Step 4: Verify the solution

Let’s verify by plugging $ b = -1 $ back into the functions:

- $ q(x) = -x + 1 $
- $ q(1) = -1 + 1 = 0 $
- $ f(q(1)) = f(0) = 0 - 3 = -3 $

This matches the given condition, confirming that our solution is correct.

---

### Final Answer

$$
\boxed{-1}
$$

Solution: We have $q(1) = b\cdot 1 + 1 = b+1$, so $f(q(1)) = f(b+1)$. Applying the definition of $f$, we have $f(q(1)) = f(b+1) = (b+1) - 3 = b-2$.  Therefore, the equation $f(q(1)) = -3$ gives us $b-2 = -3$, so $ b= \boxed{-1}$.


Explanation:
Okay, so I need to find the value of the product abc given these three equations: a + b = 8, b + c = -3, and a + c = -5. Hmm, let me think about how to approach this.

First, I remember that when you have a system of equations with three variables, you can solve for each variable step by step. Since there are three equations, maybe I can use substitution or elimination. Let me try elimination first.

Let me write down the equations again:

1) a + b = 8
2) b + c = -3
3) a + c = -5

So, there are three equations here. Maybe I can solve for one variable in terms of another and substitute. Let me see.

Let me try to solve for one variable. Let's say I want to find the value of a, b, and c. Let me try adding all three equations together. Wait, if I add equations 1, 2, and 3, what would that give me?

Adding left sides: (a + b) + (b + c) + (a + c) = 2a + 2b + 2c
Adding right sides: 8 + (-3) + (-5) = 0

So, 2a + 2b + 2c = 0
Divide both sides by 2: a + b + c = 0

Oh, that's interesting. So the sum of a, b, and c is zero. Let me note that down: a + b + c = 0.

Now, if I know that a + b + c = 0, maybe I can find each variable individually. Let me think.

From equation 1: a + b = 8. If I subtract this from the total sum a + b + c = 0, then I can find c.

So, (a + b + c) - (a + b) = 0 - 8
Which gives c = -8.

Wait, let me check that again. If a + b + c = 0 and a + b = 8, then substituting, 8 + c = 0, so c = -8. Yes, that seems right.

Similarly, if I want to find a, I can use equation 2: b + c = -3. Since I know c is -8, then b + (-8) = -3. So, b = -3 + 8 = 5.

Then, if I know b is 5, I can use equation 1: a + 5 = 8, so a = 8 - 5 = 3.

Alternatively, I can check with equation 3: a + c = -5. If a is 3 and c is -8, then 3 + (-8) = -5, which matches equation 3. Perfect, that checks out.

So, a = 3, b = 5, c = -8.

Therefore, the product abc is 3 * 5 * (-8). Let me compute that.

First, 3 * 5 is 15. Then, 15 * (-8) is -120.

Wait, so is the answer -120? Let me verify again to make sure I didn't make any mistakes.

Let me check all the original equations with these values:

1) a + b = 3 + 5 = 8. Correct.
2) b + c = 5 + (-8) = -3. Correct.
3) a + c = 3 + (-8) = -5. Correct.

All equations are satisfied. So, the product abc is indeed 3 * 5 * (-8) = -120.

But let me think again if there's another way to approach this problem, just to cross-verify. Maybe solving the equations step by step without adding them all together.

Let me try that.

From equation 1: a + b = 8.
From equation 2: b + c = -3.
From equation 3: a + c = -5.

Let me solve equation 1 for a: a = 8 - b.
Then substitute this into equation 3: (8 - b) + c = -5.
Which simplifies to 8 - b + c = -5.
Rearranged: -b + c = -5 - 8 = -13.
So, -b + c = -13. Let me write that as equation 4: -b + c = -13.

Now, equation 2 is b + c = -3.

So, now I have equation 2: b + c = -3
and equation 4: -b + c = -13.

Let me add these two equations together.

Adding left sides: (b + c) + (-b + c) = 2c
Adding right sides: -3 + (-13) = -16
So, 2c = -16 => c = -8.

Which is the same result as before. Then, substitute back into equation 2: b + (-8) = -3 => b = 5. Then, from equation 1, a = 8 - 5 = 3.

Same result. So, abc = 3 * 5 * (-8) = -120.

Alternatively, maybe using substitution with different variables. Let me try solving for another variable.

From equation 3: a + c = -5. Let me solve for a: a = -5 - c.

Then substitute into equation 1: (-5 - c) + b = 8 => b - c = 8 + 5 = 13. So, equation 5: b - c = 13.

From equation 2: b + c = -3.

So, equation 2: b + c = -3
equation 5: b - c = 13.

Adding these two equations: 2b = 10 => b = 5.

Then, substitute back into equation 2: 5 + c = -3 => c = -8. Then, from equation 3: a + (-8) = -5 => a = 3.

Same result again. So, abc is still 3 * 5 * (-8) = -120.

So, all methods lead to the same answer. Therefore, I can be confident that the product abc is -120.

But just to make sure, let me try solving using matrices or another method.

Let me write the system of equations in matrix form.

The equations are:

1) a + b + 0c = 8
2) 0a + b + c = -3
3) a + 0b + c = -5

So, the coefficient matrix is:

[1 1 0]
[0 1 1]
[1 0 1]

And the constants are [8, -3, -5]^T.

To solve this system, we can use Cramer's rule or Gaussian elimination. Let me try Gaussian elimination.

First, write the augmented matrix:

[1 1 0 | 8]
[0 1 1 | -3]
[1 0 1 | -5]

Let me perform row operations.

First, let me eliminate the a term from the third equation. The third equation has a coefficient of 1 for a, and the first equation also has 1. So, subtract the first row from the third row:

Row3 = Row3 - Row1:

[1 0 1 | -5] - [1 1 0 | 8] = [0 -1 1 | -13]

So, the new augmented matrix is:

[1 1 0 | 8]
[0 1 1 | -3]
[0 -1 1 | -13]

Now, let me focus on the second and third rows. Let me eliminate the b term from the third row. The third row has -1 in the b column, and the second row has 1. So, add Row2 to Row3:

Row3 = Row3 + Row2:

[0 -1 1 | -13] + [0 1 1 | -3] = [0 0 2 | -16]

So, the matrix becomes:

[1 1 0 | 8]
[0 1 1 | -3]
[0 0 2 | -16]

Now, this is upper triangular. Let me solve for c first. From the third row: 2c = -16 => c = -8.

Then, substitute c = -8 into the second row: b + (-8) = -3 => b = 5.

Then, substitute b = 5 into the first row: a + 5 = 8 => a = 3.

Same result again. So, abc = 3 * 5 * (-8) = -120.

Therefore, regardless of the method I use, substitution, elimination, matrix operations, I get the same answer.

I think that's thorough enough. I can't see any mistakes in my calculations. All equations check out with the values of a, b, c.

So, the value of the product abc is -120.

**Final Answer**
The value of the product $abc$ is \boxed{-120}.

To find the value of the product $ abc $, we are given the following system of equations:

$$
\begin{align*}
(1) &\quad a + b = 8 \\
(2) &\quad b + c = -3 \\
(3) &\quad a + c = -5
\end{align*}
$$

---

### Step 1: Add all three equations

Adding the left-hand sides and the right-hand sides:

$$
(a + b) + (b + c) + (a + c) = 8 + (-3) + (-5)
$$

$$
2a + 2b + 2c = 0 \Rightarrow a + b + c = 0
$$

This gives us a new equation:

$$
(4) \quad a + b + c = 0
$$

---

### Step 2: Solve for each variable

We can now use this new equation to find the individual values of $ a $, $ b $, and $ c $.

#### Find $ c $:

From equation (1): $ a + b = 8 $

Substitute into equation (4):

$$
8 + c = 0 \Rightarrow c = -8
$$

#### Find $ b $:

From equation (2): $ b + c = -3 $

Substitute $ c = -8 $:

$$
b - 8 = -3 \Rightarrow b = 5
$$

#### Find $ a $:

From equation (1): $ a + b = 8 $

Substitute $ b = 5 $:

$$
a + 5 = 8 \Rightarrow a = 3
$$

---

### Step 3: Compute the product $ abc $

Now that we have:

$$
a = 3, \quad b = 5, \quad c = -8
$$

We compute:

$$
abc = 3 \times 5 \times (-8) = 15 \times (-8) = -120
$$

---

### Final Answer

$$
\boxed{-120}
$$

[Assistant]
Explanation:
Okay, so I need to find the value of the product abc given these three equations: a + b = 8, b + c = -3, and a + c = -5. Hmm, let me think about how to approach this.

First, I remember that when you have a system of equations with three variables, you can solve for each variable step by step. Since there are three equations, maybe I can use substitution or elimination. Let me try elimination first.

Let me write down the equations again:

1) a + b = 8
2) b + c = -3
3) a + c = -5

So, there are three equations here. Maybe I can solve for one variable in terms of another and substitute. Let me see.

Let me try to solve for one variable. Let's say I want to find the value of a, b, and c. Let me try adding all three equations together. Wait, if I add equations 1, 2, and 3, what would that give me?

Adding left sides: (a + b) + (b + c) + (a + c) = 2a + 2b + 2c
Adding right sides: 8 + (-3) + (-5) = 0

So, 2a + 2b + 2c = 0
Divide both sides by 2: a + b + c = 0

Oh, that's interesting. So the sum of a, b, and c is zero. Let me note that down: a + b + c = 0.

Now, if I know that a + b + c = 0, maybe I can find each variable individually. Let me think.

From equation 1: a + b = 8. If I subtract this from the total sum a + b + c = 0, then I can find c.

So, (a + b + c) - (a + b) = 0 - 8
Which gives c = -8.

Wait, let me check that again. If a + b + c = 0 and a + b = 8, then substituting, 8 + c = 0, so c = -8. Yes, that seems right.

Similarly, if I want to find a, I can use equation 2: b + c = -3. Since I know c is -8, then b + (-8) = -3. So, b = -3 + 8 = 5.

Then, if I know b is 5, I can use equation 1: a + 5 = 8, so a = 8 - 5 = 3.

Alternatively, I can check with equation 3: a + c = -5. If a is 3 and c is -8, then 3 + (-8) = -5, which matches equation 3. Perfect, that checks out.

So, a = 3, b = 5, c = -8.

Therefore, the product abc is 3 * 5 * (-8). Let me compute that.

First, 3 * 5 is 15. Then, 15 * (-8) is -120.

Wait, so is the answer -120? Let me verify again to make sure I didn't make any mistakes.

Let me check all the original equations with these values:

1) a + b = 3 + 5 = 8. Correct.
2) b + c = 5 + (-8) = -3. Correct.
3) a + c = 3 + (-8) = -5. Correct.

All equations are satisfied. So, the product abc is indeed 3 * 5 * (-8) = -120.

But let me think again if there's another way to approach this problem, just to cross-verify. Maybe solving the equations step by step without adding them all together.

Let me try that.

From equation 1: a + b = 8.
From equation 2: b + c = -3.
From equation 3: a + c = -5.

Let me solve equation 1 for a: a = 8 - b.
Then substitute this into equation 3: (8 - b) + c = -5.
Which simplifies to 8 - b + c = -5.
Rearranged: -b + c = -5 - 8 = -13.
So, -b + c = -13. Let me write that as equation 4: -b + c = -13.

Now, equation 2 is b + c = -3.

So, now I have equation 2: b + c = -3
and equation 4: -b + c = -13.

Let me add these two equations together.

Adding left sides: (b + c) + (-b + c) = 2c
Adding right sides: -3 + (-13) = -16
So, 2c = -16 => c = -8.

Which is the same result as before. Then, substitute back into equation 2: b + (-8) = -3 => b = 5. Then, from equation 1, a = 8 - 5 = 3.

Same result. So, abc = 3 * 5 * (-8) = -120.

Alternatively, maybe using substitution with different variables. Let me try solving for another variable.

From equation 3: a + c = -5. Let me solve for a: a = -5 - c.

Then substitute into equation 1: (-5 - c) + b = 8 => b - c = 8 + 5 = 13. So, equation 5: b - c = 13.

From equation 2: b + c = -3.

So, equation 2: b + c = -3
equation 5: b - c = 13.

Adding these two equations: 2b = 10 => b = 5.

Then, substitute back into equation 2: 5 + c = -3 => c = -8. Then, from equation 3: a + (-8) = -5 => a = 3.

Same result again. So, abc is still 3 * 5 * (-8) = -120.

So, all methods lead to the same answer. Therefore, I can be confident that the product abc is -120.

But just to make sure, let me try solving using matrices or another method.

Let me write the system of equations in matrix form.

The equations are:

1) a + b + 0c = 8
2) 0a + b + c = -3
3) a + 0b + c = -5

So, the coefficient matrix is:

[1 1 0]
[0 1 1]
[1 0 1]

And the constants are [8, -3, -5]^T.

To solve this system, we can use Cramer's rule or Gaussian elimination. Let me try Gaussian elimination.

First, write the augmented matrix:

[1 1 0 | 8]
[0 1 1 | -3]
[1 0 1 | -5]

Let me perform row operations.

First, let me eliminate the a term from the third equation. The third equation has a coefficient of 1 for a, and the first equation also has 1. So, subtract the first row from the third row:

Row3 = Row3 - Row1:

[1 0 1 | -5] - [1 1 0 | 8] = [0 -1 1 | -13]

So, the new augmented matrix is:

[1 1 0 | 8]
[0 1 1 | -3]
[0 -1 1 | -13]

Now, let me focus on the second and third rows. Let me eliminate the b term from the third row. The third row has -1 in the b column, and the second row has 1. So, add Row2 to Row3:

Row3 = Row3 + Row2:

[0 -1 1 | -13] + [0 1 1 | -3] = [0 0 2 | -16]

So, the matrix becomes:

[1 1 0 | 8]
[0 1 1 | -3]
[0 0 2 | -16]

Now, this is upper triangular. Let me solve for c first. From the third row: 2c = -16 => c = -8.

Then, substitute c = -8 into the second row: b + (-8) = -3 => b = 5.

Then, substitute b = 5 into the first row: a + 5 = 8 => a = 3.

Same result again. So, abc = 3 * 5 * (-8) = -120.

Therefore, regardless of the method I use, substitution, elimination, matrix operations, I get the same answer.

I think that's thorough enough. I can't see any mistakes in my calculations. All equations check out with the values of a, b, c.

So, the value of the product abc is -120.

**Final Answer**
The value of the product $abc$ is \boxed{-120}.

To find the value of the product $ abc $, we are given the following system of equations:

$$
\begin{align*}
(1) &\quad a + b = 8 \\
(2) &\quad b + c = -3 \\
(3) &\quad a + c = -5
\end{align*}
$$

---

### Step 1: Add all three equations

Adding the left-hand sides and the right-hand sides:

$$
(a + b) + (b + c) + (a + c) = 8 + (-3) + (-5)
$$

$$
2a + 2b + 2c = 0 \Rightarrow a + b + c = 0
$$

This gives us a new equation:

$$
(4) \quad a + b + c = 0
$$

---

### Step 2: Solve for each variable

We can now use this new equation to find the individual values of $ a $, $ b $, and $ c $.

#### Find $ c $:

From equation (1): $ a + b = 8 $

Substitute into equation (4):

$$
8 + c = 0 \Rightarrow c = -8
$$

#### Find $ b $:

From equation (2): $ b + c = -3 $

Substitute $ c = -8 $:

$$
b - 8 = -3 \Rightarrow b = 5
$$

#### Find $ a $:

From equation (1): $ a + b = 8 $

Substitute $ b = 5 $:

$$
a + 5 = 8 \Rightarrow a = 3
$$

---

### Step 3: Compute the product $ abc $

Now that we have:

$$
a = 3, \quad b = 5, \quad c = -8
$$

We compute:

$$
abc = 3 \times 5 \times (-8) = 15 \times (-8) = -120
$$

---

### Final Answer

$$
\boxed{-120}
$$

Solution: Adding the first given equation to the second, we have $a+2b+c=5$. Then, subtracting the third given equation from this last equation, we get $2b=10$, so $b=5$. Plugging this value of $b$ into the first given equation, we find that $a=3$. Plugging this value of $a$ into the third given equation, we find that $c=-8$. Thus, the product $abc=3\cdot5\cdot-8=\boxed{-120}$.


Explanation:
Okay, so I need to find three consecutive integers whose product is 120, and when you divide that product by the mean of the three integers, you get 24. Then, I have to find the largest of those three integers. Let me try to break this down step by step.

First, let me recall that consecutive integers are numbers that follow each other in order without gaps. For example, 3, 4, 5 or -2, -1, 0. Since the problem mentions three consecutive integers, I can represent them algebraically. Let me think... If I let the middle integer be 'n', then the three consecutive integers would be n-1, n, and n+1. That seems like a good approach because it centers the numbers around 'n', which might make calculations easier.

So, the three integers are n-1, n, n+1. Their product is given as 120. So, I can write the equation:

(n - 1) * n * (n + 1) = 120

Hmm, that's the product. Let me compute that. First, note that (n - 1)(n + 1) is a difference of squares, which is n² - 1. So, multiplying that by n gives:

n(n² - 1) = 120

Which simplifies to:

n³ - n = 120

So, n³ - n - 120 = 0

Now, I need to solve this cubic equation. Let me see if I can find an integer solution here. Since we're dealing with integers, maybe I can try plugging in some numbers.

Let me try n=5: 5³ - 5 - 120 = 125 - 5 - 120 = 0. Oh! That works. So n=5 is a solution. Let me check if there are other integer solutions. Let me try n=4: 64 - 4 - 120 = -60 ≠ 0. n=6: 216 - 6 - 120 = 90 ≠ 0. n=3: 27 - 3 - 120 = -96 ≠ 0. n=2: 8 - 2 - 120 = -114 ≠ 0. n=1: 1 - 1 - 120 = -120 ≠ 0. n=0: 0 - 0 - 120 = -120 ≠ 0. n=-1: -1 - (-1) - 120 = -1 +1 -120 = -120 ≠ 0. So, n=5 is the only integer solution here. Therefore, the three consecutive integers are 4, 5, 6. Let me check their product: 4*5*6 = 120. Yes, that's correct. So that seems to satisfy the first condition.

But wait, the problem also says that when you divide the product by the mean of the three integers, you get 24. Let me check if that's true for these numbers. First, the mean of the three integers is (4 + 5 + 6)/3 = 15/3 = 5. Then, the product is 120. So, 120 divided by 5 is 24. Yes! That matches the second condition. So, the numbers are indeed 4, 5, 6. Therefore, the largest of the three integers is 6. So, the answer is 6. Wait, but let me make sure there are no other possible sets of consecutive integers that satisfy these conditions. Maybe negative numbers?

Let me think. Suppose the three consecutive integers are negative. Let me try n=-5. Then the numbers would be -6, -5, -4. Their product is (-6)*(-5)*(-4) = (-6)*(-5) is 30, times (-4) is -120. But the product is supposed to be 120, so that's -120, which is not equal to 120. If I take n=-4, then numbers are -5, -4, -3. Their product is (-5)*(-4)*(-3) = -60. Not 120. n=-3: -4, -3, -2. Product is (-4)*(-3)*(-2) = -24. Not 120. n=-2: -3, -2, -1. Product is -6. n=-1: -2, -1, 0. Product is 0. So none of the negative numbers give a positive product of 120. Therefore, the only possible set is 4, 5, 6. Therefore, the largest is 6.

But let me check again if there's another possible solution. Maybe non-integer? Wait, but the problem says three consecutive integers, so they have to be integers. So, n=5 is the only solution. Therefore, the answer is 6. Wait, but let me check again the second condition. The problem says "the product divided by the mean of the three integers is 24". So, product is 120, mean is 5, 120 divided by 5 is 24. Correct. So that's exactly what's given. Therefore, the answer is 6. So, I think that's it.

But just to make sure, let me try to approach the problem another way. Maybe by not assuming the middle number is n. Let me try letting the three consecutive integers be x, x+1, x+2. Then their product is x(x+1)(x+2) = 120. Then, the mean of the three integers is (x + (x+1) + (x+2))/3 = (3x + 3)/3 = x + 1. Therefore, the product divided by the mean is [x(x+1)(x+2)] / (x + 1) = x(x+2) = 24. Wait, that's interesting. So according to this, x(x + 2) = 24. Therefore, x² + 2x - 24 = 0. Let me solve this quadratic equation. The discriminant is 4 + 96 = 100. Therefore, x = [-2 ± 10]/2. So, x = (-2 + 10)/2 = 8/2 = 4, or x = (-2 -10)/2 = -12/2 = -6. Therefore, x is either 4 or -6. Therefore, if x=4, then the numbers are 4, 5, 6. If x=-6, then the numbers are -6, -5, -4. Let me check their product. For -6, -5, -4: product is (-6)*(-5)*(-4) = -120. But the problem says the product is 120. Therefore, this would be -120, which is not equal to 120. Therefore, x=-6 is invalid. Therefore, only x=4 is valid. Therefore, the numbers are 4, 5, 6. Therefore, the largest is 6. So, same answer. Therefore, this method also gives 6. Therefore, that confirms the answer.

Wait, but why did the second approach give me x=-6? Because when I divided the product by the mean, I canceled out the (x+1) term. But if x=-6, then the mean is x + 1 = -5. Then, the product is -120, divided by -5 is 24. Wait, hold on! The problem says "the product divided by the mean of the three integers is 24". So, if the product is -120 and the mean is -5, then -120 divided by -5 is 24. Which is correct. But the problem says "the product of three consecutive integers is 120". But in this case, the product is -120. Therefore, this would not satisfy the first condition. Therefore, even though the second condition is satisfied, the first condition is not. Therefore, x=-6 is invalid. Therefore, only x=4 is valid. Therefore, the answer is 6. Therefore, both methods lead to the same conclusion.

So, the answer is 6. Therefore, the largest of the three consecutive integers is 6. Therefore, I think that's the correct answer.

But let me check once again. Let me check for x=-6. The product is -120, which is not 120. Therefore, invalid. Therefore, only x=4 is valid. Therefore, the numbers are 4, 5, 6. Therefore, the largest is 6. Therefore, the answer is 6. So, I think that's solid.

Alternatively, maybe there's another way to think about it. Let me try. Let me suppose that the three consecutive integers are a, a+1, a+2. Then, their product is a(a+1)(a+2) = 120. Also, the mean is (a + (a+1) + (a+2))/3 = (3a + 3)/3 = a + 1. Therefore, the product divided by the mean is [a(a+1)(a+2)] / (a + 1) = a(a + 2) = 24. Therefore, as before, a(a + 2) = 24. Which gives a² + 2a - 24 = 0. Which gives a = [-2 ± sqrt(4 + 96)] / 2 = [-2 ± sqrt(100)] / 2 = [-2 ± 10]/2. Therefore, a = 4 or a = -6. Then, as before, check the product. For a=4: 4*5*6=120. For a=-6: (-6)*(-5)*(-4)=-120. Therefore, only a=4 is valid. Therefore, the numbers are 4,5,6. Therefore, the largest is 6. So, same answer again.

Therefore, regardless of the method, the answer is 6. Therefore, I think that's correct. Therefore, the answer is 6.

But let me check if there's another possible interpretation of the problem. The problem says "the product of three consecutive integers is 120. That product divided by the mean of the three integers is 24." So, maybe someone could misinterpret "that product" as referring to something else? But no, "that product" refers to the product mentioned before, which is 120. Therefore, 120 divided by the mean is 24. Therefore, the mean is 120 / 24 = 5. Therefore, the mean of the three integers is 5. Therefore, since the mean of three consecutive integers is the middle one, so the middle integer is 5. Therefore, the three integers are 4,5,6. Therefore, the largest is 6. Therefore, that's another way to think about it. So, if the mean is 5, then the numbers are 4,5,6. Therefore, product is 120. Therefore, that's correct. Therefore, answer is 6. Therefore, all methods lead to the same conclusion.

Therefore, I think the answer is 6. Therefore, I can confidently say that the largest of the three consecutive integers is 6.

**Final Answer**
The largest of the three consecutive integers is \boxed{6}.

To solve the problem, we are given two key pieces of information:

1. The **product** of three consecutive integers is **120**.
2. When this product is **divided by the mean** of the three integers, the result is **24**.

---

### Step 1: Represent the Three Consecutive Integers

Let the three consecutive integers be:

$$
x, \quad x+1, \quad x+2
$$

Their **product** is:

$$
x(x+1)(x+2) = 120
$$

Their **mean** is:

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

So, the second condition becomes:

$$
\frac{x(x+1)(x+2)}{x+1} = 24
$$

Simplifying the left-hand side:

$$
\frac{x(x+1)(x+2)}{x+1} = x(x+2)
$$

So we now have the equation:

$$
x(x+2) = 24
$$

---

### Step 2: Solve the Quadratic Equation

$$
x(x+2) = 24 \Rightarrow x^2 + 2x - 24 = 0
$$

Solving this quadratic using the quadratic formula:

$$
x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-24)}}{2(1)} = \frac{-2 \pm \sqrt{4 + 96}}{2} = \frac{-2 \pm \sqrt{100}}{2}
$$

$$
x = \frac{-2 \pm 10}{2}
$$

So the two possible values for $ x $ are:

$$
x = \frac{-2 + 10}{2} = 4 \quad \text{or} \quad x = \frac{-2 - 10}{2} = -6
$$

---

### Step 3: Check Both Solutions

#### Case 1: $ x = 4 $

The three integers are:
$$
4, \quad 5, \quad 6
$$

- Product: $ 4 \times 5 \times 6 = 120 $ ✅
- Mean: $ \frac{4 + 5 + 6}{3} = 5 $
- $ \frac{120}{5} = 24 $ ✅

This satisfies **both** conditions.

#### Case 2: $ x = -6 $

The three integers are:
$$
-6, \quad -5, \quad -4
$$

- Product: $ (-6) \times (-5) \times (-4) = -120 $ ❌ (Does not match 120)
- Mean: $ \frac{-6 + (-5) + (-4)}{3} = -5 $
- $ \frac{-120}{-5} = 24 $ ✅

This satisfies the second condition, but **not** the first. So this solution is **invalid**.

---

### Final Conclusion

The only valid set of three consecutive integers that satisfies both conditions is:

$$
4, \quad 5, \quad 6
$$

Thus, the **largest** of the three integers is:

$$
\boxed{6}
$$

[Assistant]
Explanation:
Okay, so I need to find three consecutive integers whose product is 120, and when you divide that product by the mean of the three integers, you get 24. Then, I have to find the largest of those three integers. Let me try to break this down step by step.

First, let me recall that consecutive integers are numbers that follow each other in order without gaps. For example, 3, 4, 5 or -2, -1, 0. Since the problem mentions three consecutive integers, I can represent them algebraically. Let me think... If I let the middle integer be 'n', then the three consecutive integers would be n-1, n, and n+1. That seems like a good approach because it centers the numbers around 'n', which might make calculations easier.

So, the three integers are n-1, n, n+1. Their product is given as 120. So, I can write the equation:

(n - 1) * n * (n + 1) = 120

Hmm, that's the product. Let me compute that. First, note that (n - 1)(n + 1) is a difference of squares, which is n² - 1. So, multiplying that by n gives:

n(n² - 1) = 120

Which simplifies to:

n³ - n = 120

So, n³ - n - 120 = 0

Now, I need to solve this cubic equation. Let me see if I can find an integer solution here. Since we're dealing with integers, maybe I can try plugging in some numbers.

Let me try n=5: 5³ - 5 - 120 = 125 - 5 - 120 = 0. Oh! That works. So n=5 is a solution. Let me check if there are other integer solutions. Let me try n=4: 64 - 4 - 120 = -60 ≠ 0. n=6: 216 - 6 - 120 = 90 ≠ 0. n=3: 27 - 3 - 120 = -96 ≠ 0. n=2: 8 - 2 - 120 = -114 ≠ 0. n=1: 1 - 1 - 120 = -120 ≠ 0. n=0: 0 - 0 - 120 = -120 ≠ 0. n=-1: -1 - (-1) - 120 = -1 +1 -120 = -120 ≠ 0. So, n=5 is the only integer solution here. Therefore, the three consecutive integers are 4, 5, 6. Let me check their product: 4*5*6 = 120. Yes, that's correct. So that seems to satisfy the first condition.

But wait, the problem also says that when you divide the product by the mean of the three integers, you get 24. Let me check if that's true for these numbers. First, the mean of the three integers is (4 + 5 + 6)/3 = 15/3 = 5. Then, the product is 120. So, 120 divided by 5 is 24. Yes! That matches the second condition. So, the numbers are indeed 4, 5, 6. Therefore, the largest of the three integers is 6. So, the answer is 6. Wait, but let me make sure there are no other possible sets of consecutive integers that satisfy these conditions. Maybe negative numbers?

Let me think. Suppose the three consecutive integers are negative. Let me try n=-5. Then the numbers would be -6, -5, -4. Their product is (-6)*(-5)*(-4) = (-6)*(-5) is 30, times (-4) is -120. But the product is supposed to be 120, so that's -120, which is not equal to 120. If I take n=-4, then numbers are -5, -4, -3. Their product is (-5)*(-4)*(-3) = -60. Not 120. n=-3: -4, -3, -2. Product is (-4)*(-3)*(-2) = -24. Not 120. n=-2: -3, -2, -1. Product is -6. n=-1: -2, -1, 0. Product is 0. So none of the negative numbers give a positive product of 120. Therefore, the only possible set is 4, 5, 6. Therefore, the largest is 6.

But let me check again if there's another possible solution. Maybe non-integer? Wait, but the problem says three consecutive integers, so they have to be integers. So, n=5 is the only solution. Therefore, the answer is 6. Wait, but let me check again the second condition. The problem says "the product divided by the mean of the three integers is 24". So, product is 120, mean is 5, 120 divided by 5 is 24. Correct. So that's exactly what's given. Therefore, the answer is 6. So, I think that's it.

But just to make sure, let me try to approach the problem another way. Maybe by not assuming the middle number is n. Let me try letting the three consecutive integers be x, x+1, x+2. Then their product is x(x+1)(x+2) = 120. Then, the mean of the three integers is (x + (x+1) + (x+2))/3 = (3x + 3)/3 = x + 1. Therefore, the product divided by the mean is [x(x+1)(x+2)] / (x + 1) = x(x+2) = 24. Wait, that's interesting. So according to this, x(x + 2) = 24. Therefore, x² + 2x - 24 = 0. Let me solve this quadratic equation. The discriminant is 4 + 96 = 100. Therefore, x = [-2 ± 10]/2. So, x = (-2 + 10)/2 = 8/2 = 4, or x = (-2 -10)/2 = -12/2 = -6. Therefore, x is either 4 or -6. Therefore, if x=4, then the numbers are 4, 5, 6. If x=-6, then the numbers are -6, -5, -4. Let me check their product. For -6, -5, -4: product is (-6)*(-5)*(-4) = -120. But the problem says the product is 120. Therefore, this would be -120, which is not equal to 120. Therefore, x=-6 is invalid. Therefore, only x=4 is valid. Therefore, the numbers are 4, 5, 6. Therefore, the largest is 6. So, same answer. Therefore, this method also gives 6. Therefore, that confirms the answer.

Wait, but why did the second approach give me x=-6? Because when I divided the product by the mean, I canceled out the (x+1) term. But if x=-6, then the mean is x + 1 = -5. Then, the product is -120, divided by -5 is 24. Wait, hold on! The problem says "the product divided by the mean of the three integers is 24". So, if the product is -120 and the mean is -5, then -120 divided by -5 is 24. Which is correct. But the problem says "the product of three consecutive integers is 120". But in this case, the product is -120. Therefore, this would not satisfy the first condition. Therefore, even though the second condition is satisfied, the first condition is not. Therefore, x=-6 is invalid. Therefore, only x=4 is valid. Therefore, the answer is 6. Therefore, both methods lead to the same conclusion.

So, the answer is 6. Therefore, the largest of the three consecutive integers is 6. Therefore, I think that's the correct answer.

But let me check once again. Let me check for x=-6. The product is -120, which is not 120. Therefore, invalid. Therefore, only x=4 is valid. Therefore, the numbers are 4, 5, 6. Therefore, the largest is 6. Therefore, the answer is 6. So, I think that's solid.

Alternatively, maybe there's another way to think about it. Let me try. Let me suppose that the three consecutive integers are a, a+1, a+2. Then, their product is a(a+1)(a+2) = 120. Also, the mean is (a + (a+1) + (a+2))/3 = (3a + 3)/3 = a + 1. Therefore, the product divided by the mean is [a(a+1)(a+2)] / (a + 1) = a(a + 2) = 24. Therefore, as before, a(a + 2) = 24. Which gives a² + 2a - 24 = 0. Which gives a = [-2 ± sqrt(4 + 96)] / 2 = [-2 ± sqrt(100)] / 2 = [-2 ± 10]/2. Therefore, a = 4 or a = -6. Then, as before, check the product. For a=4: 4*5*6=120. For a=-6: (-6)*(-5)*(-4)=-120. Therefore, only a=4 is valid. Therefore, the numbers are 4,5,6. Therefore, the largest is 6. So, same answer again.

Therefore, regardless of the method, the answer is 6. Therefore, I think that's correct. Therefore, the answer is 6.

But let me check if there's another possible interpretation of the problem. The problem says "the product of three consecutive integers is 120. That product divided by the mean of the three integers is 24." So, maybe someone could misinterpret "that product" as referring to something else? But no, "that product" refers to the product mentioned before, which is 120. Therefore, 120 divided by the mean is 24. Therefore, the mean is 120 / 24 = 5. Therefore, the mean of the three integers is 5. Therefore, since the mean of three consecutive integers is the middle one, so the middle integer is 5. Therefore, the three integers are 4,5,6. Therefore, the largest is 6. Therefore, that's another way to think about it. So, if the mean is 5, then the numbers are 4,5,6. Therefore, product is 120. Therefore, that's correct. Therefore, answer is 6. Therefore, all methods lead to the same conclusion.

Therefore, I think the answer is 6. Therefore, I can confidently say that the largest of the three consecutive integers is 6.

**Final Answer**
The largest of the three consecutive integers is \boxed{6}.

To solve the problem, we are given two key pieces of information:

1. The **product** of three consecutive integers is **120**.
2. When this product is **divided by the mean** of the three integers, the result is **24**.

---

### Step 1: Represent the Three Consecutive Integers

Let the three consecutive integers be:

$$
x, \quad x+1, \quad x+2
$$

Their **product** is:

$$
x(x+1)(x+2) = 120
$$

Their **mean** is:

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

So, the second condition becomes:

$$
\frac{x(x+1)(x+2)}{x+1} = 24
$$

Simplifying the left-hand side:

$$
\frac{x(x+1)(x+2)}{x+1} = x(x+2)
$$

So we now have the equation:

$$
x(x+2) = 24
$$

---

### Step 2: Solve the Quadratic Equation

$$
x(x+2) = 24 \Rightarrow x^2 + 2x - 24 = 0
$$

Solving this quadratic using the quadratic formula:

$$
x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-24)}}{2(1)} = \frac{-2 \pm \sqrt{4 + 96}}{2} = \frac{-2 \pm \sqrt{100}}{2}
$$

$$
x = \frac{-2 \pm 10}{2}
$$

So the two possible values for $ x $ are:

$$
x = \frac{-2 + 10}{2} = 4 \quad \text{or} \quad x = \frac{-2 - 10}{2} = -6
$$

---

### Step 3: Check Both Solutions

#### Case 1: $ x = 4 $

The three integers are:
$$
4, \quad 5, \quad 6
$$

- Product: $ 4 \times 5 \times 6 = 120 $ ✅
- Mean: $ \frac{4 + 5 + 6}{3} = 5 $
- $ \frac{120}{5} = 24 $ ✅

This satisfies **both** conditions.

#### Case 2: $ x = -6 $

The three integers are:
$$
-6, \quad -5, \quad -4
$$

- Product: $ (-6) \times (-5) \times (-4) = -120 $ ❌ (Does not match 120)
- Mean: $ \frac{-6 + (-5) + (-4)}{3} = -5 $
- $ \frac{-120}{-5} = 24 $ ✅

This satisfies the second condition, but **not** the first. So this solution is **invalid**.

---

### Final Conclusion

The only valid set of three consecutive integers that satisfies both conditions is:

$$
4, \quad 5, \quad 6
$$

Thus, the **largest** of the three integers is:

$$
\boxed{6}
$$

Solution: Call the integers $n-1$, $n$, and $n+1$. Their mean is $n$; their product is $(n-1)(n)(n+1)=120$ and their product divided by their mean is $(n-1)(n+1)=24$. Dividing the first equation by the second, we get $n=5$. The largest of the three is $n+1=\boxed{6}$.