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

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

Alternatively, maybe using more numbers. Let's try with 4 numbers.

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

If I replace 7 with something smaller? But 7 is a prime factor.

Alternatively, 2, 3, 2, 7. Duplicates.

Alternatively, 2, 3, 1, 14. But that's with 1. Which we already considered.

Wait, but in case 2, we are not using 1. So maybe 2, 3, 4, 7 is too big.

Alternatively, 2, 3, 7, and then... but that's three numbers.

Wait, maybe 2, 3, 7, and 1? But that's case 1.

So in case 2, without 1, the best we can do is 3, 4, 7 with sum 14.

But let me check if there are other combinations.

What if we split the factors into more numbers? For example, 2, 2, 3, 7. But duplicates.

If we can split into 2, 3, 2, 7. Still duplicates.

Alternatively, 2, 3, 7, and then... but we need to have distinct numbers.

Alternatively, maybe 2, 3, 7, and 1. But that's case 1.

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

Alternatively, think of other factor combinations.

For example, 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. 7 and 12: sum 19. 4 and 21: 25. 3 and 28: 31. 2 and 42: 44. So the best is 7 and 12 with sum 19. But we already have a better sum with three numbers: 3, 4, 7 (sum 14).

Alternatively, if we take three numbers. Let me check all possible triplets.

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.

- 2, 3, 14: same as first.

- 2, 7, 6: same as second.

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

What about four numbers? Let me check.

For example, 2, 3, 4, 7: product 168. Too big.

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.

So the best with four numbers is 15.

So comparing with the triplet sum of 14, which is better.

Is there a way to get a sum less than 14? Let's see.

What about five numbers? Let me try.

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

If I replace 7 with something smaller? But 7 is a prime factor.

Alternatively, 1, 2, 3, 2, 7: duplicates.

Alternatively, 1, 2, 3, 4, something? Let me see.

If I have 1, 2, 3, 4, then the product is 24. To get 84, need 84 / 24 = 3.5. Not integer.

Alternatively, 1, 2, 3, 7, and then... but that's four numbers.

Alternatively, 1, 2, 3, 4, and then divide one of the numbers? Not sure.

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

So seems like five numbers is not helpful.

What about using more numbers? Let me think.

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

Alternatively, maybe using fractions? No, the problem says positive integers.

So maybe the minimal sum is 14?

But let me check if there are other triplets.

Wait, what about 2, 3, 14? Sum 19.

Or 2, 7, 6? Sum 15.

Or 3, 4, 7? Sum 14.

Is there another triplet?

What about 1, 2, 42? Sum 1+2+42=45. Worse.

Or 1, 3, 28? 1+3+28=32. Worse.

Or 1, 4, 21? 1+4+21=26. Worse.

Or 1, 6, 14? 1+6+14=21. Worse.

Or 1, 7, 12? 1+7+12=20. Worse.

So no, the triplet with 3,4,7 is the best.

But let me check if there are other triplets without 1.

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

Or 2, 6, 7. Sum 15.

Or 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, maybe four numbers without 1? Let me check.

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

If I take 2, 3, 4, and then divide 7 by something? Not possible.

Alternatively, 2, 3, 7, and then... but that's three numbers.

Alternatively, 2, 3, 2, 7. Duplicates.

So no.

So seems like the minimal sum is 14.

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

For example, if I take 1, 2, 3, 4, 7. Product is 168. If I can adjust some numbers to get product 84.

Wait, if I take 1, 2, 3, 4, and then instead of 7, take something else. But 84 divided by (1×2×3×4)= 84 / 24 = 3.5. Not integer.

Alternatively, take 1, 2, 3, 7, and then... but that's four numbers.

Alternatively, take 1, 2, 3, 4, and then replace 4 with something?

If I remove 4 and add something else. Let me think.

Suppose I have 1, 2, 3, and then instead of 4 and 7, maybe  something else.

Wait, this is getting too convoluted.

Alternatively, think of breaking down the prime factors into more numbers.

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

We need to combine them into distinct integers.

If we take 2, 3, 7, and then combine the other 2 with something?

But if we combine the other 2 with 3, we get 6, so numbers are 6, 2, 7. Which is sum 15.

If we combine the other 2 with 7, we get 14, so numbers are 14, 2, 3. Sum 19.

If we combine the two 2s into 4, then numbers are 4, 3, 7. Sum 14.

Alternatively, combine one 2 with 3 to make 6, and the other 2 with 7 to make 14. Then numbers are 6, 14. Product is 6×14=84. Sum is 20.

Alternatively, combine one 2 with 3 to make 6, and leave the other 2 and 7. So numbers are 6, 2, 7. Sum 15.

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

Is there a way to split into more numbers? For example, 2, 2, 3, 7. But duplicates.

If we can split into 2, 3, 2, 7. Still duplicates.

Alternatively, split into 2, 3, 7, and 1. Which is case 1.

So seems like 14 is the minimal sum.

But let me check if there is another combination.

What if we use 1, 2, 3, 4, 7? Product is 168. If we can divide one of the numbers by 2? But we can't.

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

Alternatively, think of other factor combinations.

For example, 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 1×2×2×21, duplicates.

Or 1×3×2×14, same as 1×2×3×14.

So the minimal sum with 1 is 15.

But without 1, we have 14.

So 14 is better.

Is there a way to get even lower than 14? Let me think.

What if we use four numbers without 1?

For example, 2, 3, 4, 7. Product is 168. If we can adjust.

Wait, if we take 2, 3, 4, and then instead of 7, take something else. But 84 divided by (2×3×4) is 84 / 24 = 3.5. Not integer.

Alternatively, take 2, 3, 7, and then... but that's three numbers.

Alternatively, take 2, 3, 7, and 1. Which is case 1.

Alternatively, take 2, 3, 4, and 7 divided by something? Not helpful.

Alternatively, think of splitting the prime factors into more numbers.

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

If we split into 2, 2, 3, 7. But duplicates.

If we split into 2, 3, 2, 7. Still duplicates.

If we split into 2, 3, 7, and 1. Which is case 1.

Alternatively, split into 2, 3, 4, 7. But that's combining two 2s into 4.

So that's allowed.

So that gives us 4, 3, 7.

Is there a way to split into more numbers? For example, 2, 3, 2, 7. Duplicates.

Alternatively, 2, 3, 7, and 1.

Alternatively, 2, 3, 4, 7.

So seems like 3 numbers is the way to go.

Wait, what if we split into 2, 3, 7, and then split one of them further?

But they are primes.

For example, 2 can't be split further. 3 can't be split. 7 can't be split.

So no.

Alternatively, maybe use exponents? But the numbers have to be integers.

Alternatively, think of other factor combinations.

Wait, 84 can be written as 2×2×3×7.

If I take 2×3=6, and then 2×7=14. So numbers are 6 and 14. Sum is 20.

Or take 2×2=4, 3×7=21. Numbers are 4 and 21. Sum 25.

Or take 2×7=14, 2×3=6. Same as before.

Or take 2×2×3=12, and 7. Numbers are 12 and 7. Sum 19.

Or take 2×2×7=28, and 3. Sum 31.

Or take 2×3×7=42, and 2. Sum 44.

So the best among these is 12 and 7 with sum 19.

But we already have 3, 4, 7 with sum 14.

So that's better.

Alternatively, take 2×3=6, 2×7=14, and then... but that's two numbers.

Alternatively, take 2, 3, 7, and 1.

So seems like 14 is the minimal.

Wait, but let me check if there is a combination with four numbers without 1.

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

But we can't.

Alternatively, maybe 2, 3, 7, and then... but that's three numbers.

Alternatively, 2, 3, 4, and 7 divided by 2? Not allowed.

Alternatively, think of other numbers.

Wait, 84 divided by 2 is 42. So if I have 2 and 42. Sum 44.

Or 3 and 28. Sum 31.

Or 4 and 21. Sum 25.

Or 6 and 14. Sum 20.

Or 7 and 12. Sum 19.

So none of these are better than 14.

Alternatively, think of three numbers.

For example, 3, 4, 7. Sum 14.

Or 2, 6, 7. Sum 15.

Or 2, 3, 14. Sum 19.

Or 3, 2, 14. Same as above.

So 14 is the best.

Is there a way to get a sum less than 14? Let's think.

What if we use four numbers? Let me think.

Suppose we have numbers like 2, 3, 4, and something.

But product is 2×3×4×x = 24x = 84. So x = 84 / 24 = 3.5. Not integer.

If I take numbers 2, 3, 5, and something. But 5 is not a factor of 84.

Alternatively, 2, 3, 7, and something. Product is 2×3×7×x = 42x = 84. So x=2. But 2 is already there. So duplicates.

Alternatively, 2, 3, 7, and 1. Which is case 1.

Alternatively, 2, 3, 4, and 3.5. Not allowed.

So no.

What about five numbers?

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

But we can't.

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

Alternatively, 1, 2, 3, 7, and something.

Product is 1×2×3×7×x = 42x = 84. So x=2. But 2 is already there.

So duplicates.

Alternatively, 1, 2, 3, 7, and 1. Duplicates.

So no.

Alternatively, 1, 2, 3, 4, and 7 divided by 2. Not allowed.

So seems like no.

Therefore, the minimal sum 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 replace 7 with something smaller.

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

Alternatively, replace 4 with something.

If I take 1, 2, 3, 3.5, 7. Not integer.

No.

Alternatively, take 1, 2, 3, 2, 7. Duplicates.

No.

Alternatively, take 1, 2, 3, 4, and 7 divided by 2. Not allowed.

So no.

Therefore, I think the minimal sum is 14.

But let me check if there is another triplet.

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

Or 2, 7, 6. Sum 15.

Or 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, think of numbers like 1, 2, 3, 4, 7. But that's five numbers.

Alternatively, think of numbers like 1, 2, 3, 7, and 2. Duplicates.

No.

Therefore, I think the answer is 14.

But let me check once again.

Wait, what if I use numbers like 1, 2, 3, 4, 7? Product is 168. If I can divide one of the numbers by 2.

But we can't.

Alternatively, maybe use fractions? No, the problem says positive integers.

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 yes, 14 is the minimal.

Therefore, the least possible sum is 14.

But wait, let me check if there is a combination with more numbers.

For example, 1, 2, 3, 4, 7. Product is 168. If I can replace 7 with something else.

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

Alternatively, replace 4 with something.

If I take 1, 2, 3, 3.5, 7. Not integer.

No.

Alternatively, take 1, 2, 3, 2, 7. Duplicates.

No.

Alternatively, take 1, 2, 3, 4, and 7 divided by 2. Not allowed.

So no.

Therefore, I think the answer is 14.

But let me check once again if there is a combination with numbers like 1, 2, 3, 4, 7.

But that's five numbers.

Alternatively, maybe 1, 2, 3, 4, and then instead of 7, take something else.

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

Alternatively, take 1, 2, 3, 4, and 3.5. Not allowed.

No.

Alternatively, take 1, 2, 3, 4, and 7 divided by 2. Not allowed.

So no.

Therefore, I think the minimal sum is 14.

But let me check if there is a combination with numbers like 1, 2, 3, 4, 7.

No, that's not helpful.

Alternatively, think of numbers like 1, 2, 3, 7, and 2. Duplicates.

No.

Therefore, I think the answer is 14.

But wait, let me check if there is a combination with numbers like 1, 2, 3, 4, 7.

No, that's not helpful.

Alternatively, think of numbers like 1, 2, 3, 4, 7.

No.

Therefore, I think 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 4, 3, 7.

Sum is 4 + 3 + 7 = 14.

Yes.

Alternatively, if I combine one 2 with 3 into 6, then I have 6, 2, 7. Sum is 15.

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

Alternatively, combine both 2s with 3 into 6 and 2? Wait, no.

Alternatively, combine one 2 with 3 into 6, and the other 2 with 7 into 14. Then numbers are 6 and 14. Sum is 20.

So the best is 14.

Therefore, the least possible sum is 14.

But let me check if there is a way to split into more numbers.

For example, 2, 3, 7, and 1.

Sum is 1 + 2 + 3 + 7 = 13? Wait, no. Wait, 1 + 2 + 3 + 7 = 13? Wait, 1+2 is 3, +3 is 6, +7 is 13. Wait, but earlier I thought that 1, 3, 4, 7 had sum 15. Wait, but if I take 1, 2, 3, 7, that's sum 13? But product is 1×2×3×7=42. Not 84.

Oh! Wait, I think I made a mistake here.

Hold on! If I take 1, 2, 3, 7, the product is 42, not 84. So that's not valid.

But earlier, I thought that 1, 3, 4, 7 has product 1×3×4×7=84. Yes, that's correct.

But if I take 1, 2, 3, 14, product is 1×2×3×14=84. Sum is 1+2+3+14=20.

But if I take 1, 2, 6, 7, product is 1×2×6×7=84. Sum is 1+2+6+7=16.

If I take 1, 3, 4, 7, product is 84. Sum is 1+3+4+7=15.

But if I take 1, 2, 3, 7, product is 42. Not 84.

So that's invalid.

So earlier, I thought that 1, 2, 3, 7 is a valid set, but it's not.

So that was a mistake.

So the sum of 13 is invalid.

So that was a mistake.

So going back, the minimal sum with 1 is 15.

So the triplet without 1 is 3, 4, 7 with sum 14.

So that is valid.

Therefore, the answer is 14.

But let me check again.

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

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

Product is 4×3×7=84.

Sum is 4+3+7=14.

Yes, that's correct.

So that's valid.

Therefore, the minimal sum is 14.

But let me check if there is another combination with sum less than 14.

For example, can I have numbers like 2, 3, 14? Sum 19.

Or 2, 6, 7? Sum 15.

Or 3, 4, 7? Sum 14.

Or 1, 3, 4, 7? Sum 15.

Or 1, 2, 6, 7? Sum 16.

So no, 14 is the minimal.

Therefore, 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 we can't.

Alternatively, take 1, 2, 3, 4, and 7 divided by 2. Not allowed.

Alternatively, take 1, 2, 3, 4, and 7 divided by something. Not helpful.

Alternatively, take 1, 2, 3, 4, and 7 divided by 2. Not allowed.

So no.

Therefore, I think the answer is 14.

But let me check if there is a combination with numbers like 2, 3, 7, and 2. But duplicates.

No.

Alternatively, 2, 3, 7, and 1. Product is 42. Not 84.

So no.

Therefore, I think 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 **combine 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 need to find all combinations of **distinct positive integers** whose product is 84. Since the integers must be **distinct**, we cannot repeat any number. We also want to **minimize the sum**, so we aim to use **smaller numbers** where possible.

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

---

### Step 2: Try Grouping the Factors

We can try different groupings of the prime factors:

- **Group 1:** $2 \times 2 = 4$, $3$, $7$
  → Integers: $4, 3, 7$
  → Product: $4 \times 3 \times 7 = 84$
  → Sum: $4 + 3 + 7 = \boxed{14}$

- **Group 2:** $2, 2, 3, 7$
  → Duplicates (two 2s), so **invalid**

- **Group 3:** $2, 3, 14$
  → Product: $2 \times 3 \times 14 = 84$
  → Sum: $2 + 3 + 14 = 19$

- **Group 4:** $2, 6, 7$
  → Product: $2 \times 6 \times 7 = 84$
  → Sum: $2 + 6 + 7 = 15$

- **Group 5:** $3, 4, 7$
  → Product: $3 \times 4 \times 7 = 84$
  → Sum: $3 + 4 + 7 = 14$

- **Group 6:** $1, 2, 3, 14$
  → Product: $1 \times 2 \times 3 \times 14 = 84$
  → Sum: $1 + 2 + 3 + 14 = 20$

- **Group 7:** $1, 2, 6, 7$
  → Product: $1 \times 2 \times 6 \times 7 = 84$
  → Sum: $1 + 2 + 6 + 7 = 16$

- **Group 8:** $1, 3, 4, 7$
  → Product: $1 \times 3 \times 4 \times 7 = 84$
  → Sum: $1 + 3 + 4 + 7 = 15$

---

### Step 3: Compare All Valid Sets

From the above, the valid sets of distinct integers whose product is 84 are:

- $3, 4, 7$ → Sum = 14
- $2, 6, 7$ → Sum = 15
- $1, 3, 4, 7$ → Sum = 15
- $1, 2, 6, 7$ → Sum = 16
- $1, 2, 3, 14$ → Sum = 20
- $2, 3, 14$ → Sum = 19

Among all these, the **smallest sum** is:

$$
\boxed{14}
$$

---

### 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, 3, 2, 7? But duplicates. Not allowed.

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

Alternatively, maybe using more numbers. Let's try with 4 numbers.

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

If I replace 7 with something smaller? But 7 is a prime factor.

Alternatively, 2, 3, 2, 7. Duplicates.

Alternatively, 2, 3, 1, 14. But that's with 1. Which we already considered.

Wait, but in case 2, we are not using 1. So maybe 2, 3, 4, 7 is too big.

Alternatively, 2, 3, 7, and then... but that's three numbers.

Wait, maybe 2, 3, 7, and 1? But that's case 1.

So in case 2, without 1, the best we can do is 3, 4, 7 with sum 14.

But let me check if there are other combinations.

What if we split the factors into more numbers? For example, 2, 2, 3, 7. But duplicates.

If we can split into 2, 3, 2, 7. Still duplicates.

Alternatively, 2, 3, 7, and then... but we need to have distinct numbers.

Alternatively, maybe 2, 3, 7, and 1. But that's case 1.

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

Alternatively, think of other factor combinations.

For example, 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. 7 and 12: sum 19. 4 and 21: 25. 3 and 28: 31. 2 and 42: 44. So the best is 7 and 12 with sum 19. But we already have a better sum with three numbers: 3, 4, 7 (sum 14).

Alternatively, if we take three numbers. Let me check all possible triplets.

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.

- 2, 3, 14: same as first.

- 2, 7, 6: same as second.

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

What about four numbers? Let me check.

For example, 2, 3, 4, 7: product 168. Too big.

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.

So the best with four numbers is 15.

So comparing with the triplet sum of 14, which is better.

Is there a way to get a sum less than 14? Let's see.

What about five numbers? Let me try.

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

If I replace 7 with something smaller? But 7 is a prime factor.

Alternatively, 1, 2, 3, 2, 7: duplicates.

Alternatively, 1, 2, 3, 4, something? Let me see.

If I have 1, 2, 3, 4, then the product is 24. To get 84, need 84 / 24 = 3.5. Not integer.

Alternatively, 1, 2, 3, 7, and then... but that's four numbers.

Alternatively, 1, 2, 3, 4, and then divide one of the numbers? Not sure.

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

So seems like five numbers is not helpful.

What about using more numbers? Let me think.

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

Alternatively, maybe using fractions? No, the problem says positive integers.

So maybe the minimal sum is 14?

But let me check if there are other triplets.

Wait, what about 2, 3, 14? Sum 19.

Or 2, 7, 6? Sum 15.

Or 3, 4, 7? Sum 14.

Is there another triplet?

What about 1, 2, 42? Sum 1+2+42=45. Worse.

Or 1, 3, 28? 1+3+28=32. Worse.

Or 1, 4, 21? 1+4+21=26. Worse.

Or 1, 6, 14? 1+6+14=21. Worse.

Or 1, 7, 12? 1+7+12=20. Worse.

So no, the triplet with 3,4,7 is the best.

But let me check if there are other triplets without 1.

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

Or 2, 6, 7. Sum 15.

Or 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, maybe four numbers without 1? Let me check.

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

If I take 2, 3, 4, and then divide 7 by something? Not possible.

Alternatively, 2, 3, 7, and then... but that's three numbers.

Alternatively, 2, 3, 2, 7. Duplicates.

So no.

So seems like the minimal sum is 14.

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

For example, if I take 1, 2, 3, 4, 7. Product is 168. If I can adjust some numbers to get product 84.

Wait, if I take 1, 2, 3, 4, and then instead of 7, take something else. But 84 divided by (1×2×3×4)= 84 / 24 = 3.5. Not integer.

Alternatively, take 1, 2, 3, 7, and then... but that's four numbers.

Alternatively, take 1, 2, 3, 4, and then replace 4 with something?

If I remove 4 and add something else. Let me think.

Suppose I have 1, 2, 3, and then instead of 4 and 7, maybe  something else.

Wait, this is getting too convoluted.

Alternatively, think of breaking down the prime factors into more numbers.

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

We need to combine them into distinct integers.

If we take 2, 3, 7, and then combine the other 2 with something?

But if we combine the other 2 with 3, we get 6, so numbers are 6, 2, 7. Which is sum 15.

If we combine the other 2 with 7, we get 14, so numbers are 14, 2, 3. Sum 19.

If we combine the two 2s into 4, then numbers are 4, 3, 7. Sum 14.

Alternatively, combine one 2 with 3 to make 6, and the other 2 with 7 to make 14. Then numbers are 6, 14. Product is 6×14=84. Sum is 20.

Alternatively, combine one 2 with 3 to make 6, and leave the other 2 and 7. So numbers are 6, 2, 7. Sum 15.

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

Is there a way to split into more numbers? For example, 2, 2, 3, 7. But duplicates.

If we can split into 2, 3, 2, 7. Still duplicates.

Alternatively, split into 2, 3, 7, and 1. Which is case 1.

So seems like 14 is the minimal sum.

But let me check if there is another combination.

What if we use 1, 2, 3, 4, 7? Product is 168. If we can divide one of the numbers by 2? But we can't.

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

Alternatively, think of other factor combinations.

For example, 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 1×2×2×21, duplicates.

Or 1×3×2×14, same as 1×2×3×14.

So the minimal sum with 1 is 15.

But without 1, we have 14.

So 14 is better.

Is there a way to get even lower than 14? Let me think.

What if we use four numbers without 1?

For example, 2, 3, 4, 7. Product is 168. If we can adjust.

Wait, if we take 2, 3, 4, and then instead of 7, take something else. But 84 divided by (2×3×4) is 84 / 24 = 3.5. Not integer.

Alternatively, take 2, 3, 7, and then... but that's three numbers.

Alternatively, take 2, 3, 7, and 1. Which is case 1.

Alternatively, take 2, 3, 4, and 7 divided by something? Not helpful.

Alternatively, think of splitting the prime factors into more numbers.

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

If we split into 2, 2, 3, 7. But duplicates.

If we split into 2, 3, 2, 7. Still duplicates.

If we split into 2, 3, 7, and 1. Which is case 1.

Alternatively, split into 2, 3, 4, 7. But that's combining two 2s into 4.

So that's allowed.

So that gives us 4, 3, 7.

Is there a way to split into more numbers? For example, 2, 3, 2, 7. Duplicates.

Alternatively, 2, 3, 7, and 1.

Alternatively, 2, 3, 4, 7.

So seems like 3 numbers is the way to go.

Wait, what if we split into 2, 3, 7, and then split one of them further?

But they are primes.

For example, 2 can't be split further. 3 can't be split. 7 can't be split.

So no.

Alternatively, maybe use exponents? But the numbers have to be integers.

Alternatively, think of other factor combinations.

Wait, 84 can be written as 2×2×3×7.

If I take 2×3=6, and then 2×7=14. So numbers are 6 and 14. Sum is 20.

Or take 2×2=4, 3×7=21. Numbers are 4 and 21. Sum 25.

Or take 2×7=14, 2×3=6. Same as before.

Or take 2×2×3=12, and 7. Numbers are 12 and 7. Sum 19.

Or take 2×2×7=28, and 3. Sum 31.

Or take 2×3×7=42, and 2. Sum 44.

So the best among these is 12 and 7 with sum 19.

But we already have 3, 4, 7 with sum 14.

So that's better.

Alternatively, take 2×3=6, 2×7=14, and then... but that's two numbers.

Alternatively, take 2, 3, 7, and 1.

So seems like 14 is the minimal.

Wait, but let me check if there is a combination with four numbers without 1.

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

But we can't.

Alternatively, maybe 2, 3, 7, and then... but that's three numbers.

Alternatively, 2, 3, 4, and 7 divided by 2? Not allowed.

Alternatively, think of other numbers.

Wait, 84 divided by 2 is 42. So if I have 2 and 42. Sum 44.

Or 3 and 28. Sum 31.

Or 4 and 21. Sum 25.

Or 6 and 14. Sum 20.

Or 7 and 12. Sum 19.

So none of these are better than 14.

Alternatively, think of three numbers.

For example, 3, 4, 7. Sum 14.

Or 2, 6, 7. Sum 15.

Or 2, 3, 14. Sum 19.

Or 3, 2, 14. Same as above.

So 14 is the best.

Is there a way to get a sum less than 14? Let's think.

What if we use four numbers? Let me think.

Suppose we have numbers like 2, 3, 4, and something.

But product is 2×3×4×x = 24x = 84. So x = 84 / 24 = 3.5. Not integer.

If I take numbers 2, 3, 5, and something. But 5 is not a factor of 84.

Alternatively, 2, 3, 7, and something. Product is 2×3×7×x = 42x = 84. So x=2. But 2 is already there. So duplicates.

Alternatively, 2, 3, 7, and 1. Which is case 1.

Alternatively, 2, 3, 4, and 3.5. Not allowed.

So no.

What about five numbers?

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

But we can't.

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

Alternatively, 1, 2, 3, 7, and something.

Product is 1×2×3×7×x = 42x = 84. So x=2. But 2 is already there.

So duplicates.

Alternatively, 1, 2, 3, 7, and 1. Duplicates.

So no.

Alternatively, 1, 2, 3, 4, and 7 divided by 2. Not allowed.

So seems like no.

Therefore, the minimal sum 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 replace 7 with something smaller.

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

Alternatively, replace 4 with something.

If I take 1, 2, 3, 3.5, 7. Not integer.

No.

Alternatively, take 1, 2, 3, 2, 7. Duplicates.

No.

Alternatively, take 1, 2, 3, 4, and 7 divided by 2. Not allowed.

So no.

Therefore, I think the minimal sum is 14.

But let me check if there is another triplet.

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

Or 2, 7, 6. Sum 15.

Or 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, think of numbers like 1, 2, 3, 4, 7. But that's five numbers.

Alternatively, think of numbers like 1, 2, 3, 7, and 2. Duplicates.

No.

Therefore, I think the answer is 14.

But let me check once again.

Wait, what if I use numbers like 1, 2, 3, 4, 7? Product is 168. If I can divide one of the numbers by 2.

But we can't.

Alternatively, maybe use fractions? No, the problem says positive integers.

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 yes, 14 is the minimal.

Therefore, the least possible sum is 14.

But wait, let me check if there is a combination with more numbers.

For example, 1, 2, 3, 4, 7. Product is 168. If I can replace 7 with something else.

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

Alternatively, replace 4 with something.

If I take 1, 2, 3, 3.5, 7. Not integer.

No.

Alternatively, take 1, 2, 3, 2, 7. Duplicates.

No.

Alternatively, take 1, 2, 3, 4, and 7 divided by 2. Not allowed.

So no.

Therefore, I think the answer is 14.

But let me check once again if there is a combination with numbers like 1, 2, 3, 4, 7.

But that's five numbers.

Alternatively, maybe 1, 2, 3, 4, and then instead of 7, take something else.

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

Alternatively, take 1, 2, 3, 4, and 3.5. Not allowed.

No.

Alternatively, take 1, 2, 3, 4, and 7 divided by 2. Not allowed.

So no.

Therefore, I think the minimal sum is 14.

But let me check if there is a combination with numbers like 1, 2, 3, 4, 7.

No, that's not helpful.

Alternatively, think of numbers like 1, 2, 3, 7, and 2. Duplicates.

No.

Therefore, I think the answer is 14.

But wait, let me check if there is a combination with numbers like 1, 2, 3, 4, 7.

No, that's not helpful.

Alternatively, think of numbers like 1, 2, 3, 4, 7.

No.

Therefore, I think 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 4, 3, 7.

Sum is 4 + 3 + 7 = 14.

Yes.

Alternatively, if I combine one 2 with 3 into 6, then I have 6, 2, 7. Sum is 15.

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

Alternatively, combine both 2s with 3 into 6 and 2? Wait, no.

Alternatively, combine one 2 with 3 into 6, and the other 2 with 7 into 14. Then numbers are 6 and 14. Sum is 20.

So the best is 14.

Therefore, the least possible sum is 14.

But let me check if there is a way to split into more numbers.

For example, 2, 3, 7, and 1.

Sum is 1 + 2 + 3 + 7 = 13? Wait, no. Wait, 1 + 2 + 3 + 7 = 13? Wait, 1+2 is 3, +3 is 6, +7 is 13. Wait, but earlier I thought that 1, 3, 4, 7 had sum 15. Wait, but if I take 1, 2, 3, 7, that's sum 13? But product is 1×2×3×7=42. Not 84.

Oh! Wait, I think I made a mistake here.

Hold on! If I take 1, 2, 3, 7, the product is 42, not 84. So that's not valid.

But earlier, I thought that 1, 3, 4, 7 has product 1×3×4×7=84. Yes, that's correct.

But if I take 1, 2, 3, 14, product is 1×2×3×14=84. Sum is 1+2+3+14=20.

But if I take 1, 2, 6, 7, product is 1×2×6×7=84. Sum is 1+2+6+7=16.

If I take 1, 3, 4, 7, product is 84. Sum is 1+3+4+7=15.

But if I take 1, 2, 3, 7, product is 42. Not 84.

So that's invalid.

So earlier, I thought that 1, 2, 3, 7 is a valid set, but it's not.

So that was a mistake.

So the sum of 13 is invalid.

So that was a mistake.

So going back, the minimal sum with 1 is 15.

So the triplet without 1 is 3, 4, 7 with sum 14.

So that is valid.

Therefore, the answer is 14.

But let me check again.

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

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

Product is 4×3×7=84.

Sum is 4+3+7=14.

Yes, that's correct.

So that's valid.

Therefore, the minimal sum is 14.

But let me check if there is another combination with sum less than 14.

For example, can I have numbers like 2, 3, 14? Sum 19.

Or 2, 6, 7? Sum 15.

Or 3, 4, 7? Sum 14.

Or 1, 3, 4, 7? Sum 15.

Or 1, 2, 6, 7? Sum 16.

So no, 14 is the minimal.

Therefore, 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 we can't.

Alternatively, take 1, 2, 3, 4, and 7 divided by 2. Not allowed.

Alternatively, take 1, 2, 3, 4, and 7 divided by something. Not helpful.

Alternatively, take 1, 2, 3, 4, and 7 divided by 2. Not allowed.

So no.

Therefore, I think the answer is 14.

But let me check if there is a combination with numbers like 2, 3, 7, and 2. But duplicates.

No.

Alternatively, 2, 3, 7, and 1. Product is 42. Not 84.

So no.

Therefore, I think 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 **combine 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 need to find all combinations of **distinct positive integers** whose product is 84. Since the integers must be **distinct**, we cannot repeat any number. We also want to **minimize the sum**, so we aim to use **smaller numbers** where possible.

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

---

### Step 2: Try Grouping the Factors

We can try different groupings of the prime factors:

- **Group 1:** $2 \times 2 = 4$, $3$, $7$
  → Integers: $4, 3, 7$
  → Product: $4 \times 3 \times 7 = 84$
  → Sum: $4 + 3 + 7 = \boxed{14}$

- **Group 2:** $2, 2, 3, 7$
  → Duplicates (two 2s), so **invalid**

- **Group 3:** $2, 3, 14$
  → Product: $2 \times 3 \times 14 = 84$
  → Sum: $2 + 3 + 14 = 19$

- **Group 4:** $2, 6, 7$
  → Product: $2 \times 6 \times 7 = 84$
  → Sum: $2 + 6 + 7 = 15$

- **Group 5:** $3, 4, 7$
  → Product: $3 \times 4 \times 7 = 84$
  → Sum: $3 + 4 + 7 = 14$

- **Group 6:** $1, 2, 3, 14$
  → Product: $1 \times 2 \times 3 \times 14 = 84$
  → Sum: $1 + 2 + 3 + 14 = 20$

- **Group 7:** $1, 2, 6, 7$
  → Product: $1 \times 2 \times 6 \times 7 = 84$
  → Sum: $1 + 2 + 6 + 7 = 16$

- **Group 8:** $1, 3, 4, 7$
  → Product: $1 \times 3 \times 4 \times 7 = 84$
  → Sum: $1 + 3 + 4 + 7 = 15$

---

### Step 3: Compare All Valid Sets

From the above, the valid sets of distinct integers whose product is 84 are:

- $3, 4, 7$ → Sum = 14
- $2, 6, 7$ → Sum = 15
- $1, 3, 4, 7$ → Sum = 15
- $1, 2, 6, 7$ → Sum = 16
- $1, 2, 3, 14$ → Sum = 20
- $2, 3, 14$ → Sum = 19

Among all these, the **smallest sum** is:

$$
\boxed{14}
$$

---

### 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 with the negative sign, it's -128. Yep. So, the product of all possible values of x is -128.

Alternatively, is there a formula or method to find the product without solving for x? For quadratic equations, the product of the roots is c/a. But this is an absolute value equation, which can be converted into a quadratic equation. Let me try that approach to cross-verify.

Starting from |x - 4| = 12. If I square both sides to eliminate the absolute value, I get:

(x - 4)^2 = 12^2
(x - 4)^2 = 144

Expanding the left side:

x^2 - 8x + 16 = 144

Subtract 144 from both sides:

x^2 - 8x + 16 - 144 = 0
x^2 - 8x - 128 = 0

So, the quadratic equation is x² - 8x - 128 = 0. For a quadratic equation ax² + bx + c = 0, the product of the roots is c/a. Here, a = 1, c = -128. Therefore, the product is -128/1 = -128. Which matches the previous result. So that's another way to confirm that the product is indeed -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 converting the absolute value equation into a quadratic form. Start from:

$$
|x - 4| = 12
$$

Squaring both sides:

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

Expanding:

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

Subtract 144 from both sides:

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

This is a quadratic equation in standard form $ ax^2 + bx + c = 0 $, where $ a = 1 $, $ b = -8 $, and $ c = -128 $. The product of the roots of a quadratic equation is given by $ \frac{c}{a} $, so:

$$
\text{Product of roots} = \frac{-128}{1} = -128
$$

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 with the negative sign, it's -128. Yep. So, the product of all possible values of x is -128.

Alternatively, is there a formula or method to find the product without solving for x? For quadratic equations, the product of the roots is c/a. But this is an absolute value equation, which can be converted into a quadratic equation. Let me try that approach to cross-verify.

Starting from |x - 4| = 12. If I square both sides to eliminate the absolute value, I get:

(x - 4)^2 = 12^2
(x - 4)^2 = 144

Expanding the left side:

x^2 - 8x + 16 = 144

Subtract 144 from both sides:

x^2 - 8x + 16 - 144 = 0
x^2 - 8x - 128 = 0

So, the quadratic equation is x² - 8x - 128 = 0. For a quadratic equation ax² + bx + c = 0, the product of the roots is c/a. Here, a = 1, c = -128. Therefore, the product is -128/1 = -128. Which matches the previous result. So that's another way to confirm that the product is indeed -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 converting the absolute value equation into a quadratic form. Start from:

$$
|x - 4| = 12
$$

Squaring both sides:

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

Expanding:

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

Subtract 144 from both sides:

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

This is a quadratic equation in standard form $ ax^2 + bx + c = 0 $, where $ a = 1 $, $ b = -8 $, and $ c = -128 $. The product of the roots of a quadratic equation is given by $ \frac{c}{a} $, so:

$$
\text{Product of roots} = \frac{-128}{1} = -128
$$

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 the given value

We are told that:

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

So we set up the equation:

$$
b - 2 = -3
$$

---

### Step 4: Solve for $ b $

Add 2 to both sides:

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

---

### 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 the given value

We are told that:

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

So we set up the equation:

$$
b - 2 = -3
$$

---

### Step 4: Solve for $ b $

Add 2 to both sides:

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

---

### 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 of the equations:

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

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

This gives us a new equation:

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

---

### Step 2: Solve for Each Variable

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

#### Find $ c $:

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

Substitute into equation (4):

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

#### Find $ b $:

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

Substitute $ c = -8 $:

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

#### Find $ a $:

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

Substitute $ b = 5 $:

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

---

### 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 of the equations:

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

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

This gives us a new equation:

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

---

### Step 2: Solve for Each Variable

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

#### Find $ c $:

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

Substitute into equation (4):

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

#### Find $ b $:

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

Substitute $ c = -8 $:

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

#### Find $ a $:

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

Substitute $ b = 5 $:

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

---

### 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 $abc=3\cdot5\cdot-8=\boxed{-120}$.