<Question>
The product of a set of distinct positive integers is 84. What is the least possible sum of these integers?
</Question>

<think>
<Plan_1>
Okay, so I need to find a set of distinct positive integers whose product is 84, and I want the sum of these integers to be as small as possible. Let me start by breaking down the problem step by step.

First, I know that 84 is the product of the numbers. So, I should factorize 84 into its prime factors. Let me do that.
The prime factors of 84 are 2, 2, 3, and 7. Because 84 divided by 2 is 42, divided by 2 again is 21, then divided by 3 is 7, and 7 is a prime number. So, prime factorization is 2² × 3 × 7.

But the problem says the integers have to be distinct. So, if I just take the prime factors as they are, that would be 2, 2, 3, 7. However, the two 2s are the same, so they aren't distinct. Therefore, I need to combine some of these factors in a way that results in distinct integers.

My goal is to minimize the sum. Since smaller numbers generally lead to a smaller sum, but they also have to multiply to 84. So, maybe grouping the prime factors differently can help.

Let me list out possible combinations. Let's start with the prime factors: 2, 2, 3, 7.

If I group the two 2s together, that gives me 4, 3, 7. So the numbers would be 3, 4, 7.

<prompt_1.1> Calculate the sum of the numbers 3, 4, and 7, and verify if they multiply to 84. </prompt_1.1>
-----

But wait, maybe there's another combination with more numbers but smaller integers? Because sometimes splitting into more numbers can lead to a smaller sum, even if the numbers are smaller. For example, using 1s, but wait, 1 is a positive integer, but if we include 1, then the product would still need to be 84. However, including 1 would require the other numbers to multiply to 84, but since 1 doesn't affect the product, maybe that's possible.

<prompt_1.2> Explore combinations including the number 1 and determine if they can provide a smaller sum. </prompt_1.2>
-----

Alternatively, let's see if there's another way to factor 84 into distinct integers. For example, 84 can be written as 6 × 7 × 2, but that's the same as before. Alternatively, 84 = 4 × 3 × 7, which is the same set.

<prompt_1.3> Verify if there are any other distinct factor combinations of 84 and calculate their sums. </prompt_1.3>
-----

Alternatively, if we use four numbers: Let's see. Let's try 2, 3, 4, 7.

<prompt_1.4> Attempt to find combinations with four or five numbers and calculate their sums. </prompt_1.4>
</Plan_1>

<Execution_1>
<execution_1.1>
Wait, but they need to be distinct. Wait, hold on. The two 2s can't both be there. So, maybe I need to combine them with other factors to make different numbers. Let me think.

Alternatively, instead of having two 2s, I can combine them with another prime. For example, 2×3=6. Then the factors would be 6, 2, 7. But again, 2 and 6 are both there. Wait, but 2 is still a separate number. Wait, but 6, 2, 7 are all distinct. Let me check: 6×2×7=84. So that's three numbers: 2, 6, 7. Their sum is 2+6+7=15.
</execution_1.1>

<execution_1.2>
But the problem says "distinct positive integers," so 1 is allowed, but if I include 1, then the remaining numbers need to multiply to 84. However, 1 doesn't help with the product, so maybe adding 1 and then other numbers. Let's see.

For example, if I take 1, 2, 4, 5... Wait, 1×2×4×5=40, which is not 84. Hmm. Maybe 1, 3, 4, 7. Let's check: 1×3×4×7=84. Yes, that works. The numbers are 1, 3, 4, 7. Their sum is 1+3+4+7=15.
</execution_1.2>

<execution_1.3>
Wait, but if I use 1, 6, 14. Let's check: 1×6×14=84. The sum here is 1+6+14=21, which is higher than 14. What about 1, 2, 4, 21? 1×2×4×21=168, which is too much. Not 84. So that's not good. How about 1, 3, 28? Sum is 32. Not better. 1, 4, 21: sum 26. 1, 7, 12: sum 20. 1, 8, 10.5: but 10.5 isn't an integer. So maybe using 1 doesn't help here because the product becomes too large unless we include fractions, which aren't allowed. So perhaps using 1 isn't helpful here.
</execution_1.3>

<execution_1.4>
That's the same as before, sum is 16. If we try to include 1, then the numbers would be 1, 2, 6, 7. Let's check the product: 1×2×6×7 = 84. The sum here is 1+2+6+7=16. Wait, same sum. Hmm. So that's another set with the same sum.
</execution_1.4>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Is that the smallest possible sum? Let me check other combinations.

What if I group 2 and 3 first? 2×3=6, then we have 6, 2, 7. Wait, that's the same as before. Alternatively, 2×7=14. Then the factors would be 14, 2, 3. But 14×2×3=84. The numbers are 2, 3, 14. Sum is 2+3+14=19. That's worse.
-----

Based on execution_1.2:
What if I use more numbers? Let's try five numbers.

<prompt_2.1> Attempt to find a combination with five numbers and calculate their sum. </prompt_2.1>
-----

Based on execution_1.4:
Let me check if there's a combination with more numbers but even smaller sum. For example, 1, 2, 3, 14: product is 1×2×3×14=84.

<prompt_2.2> Verify if there are any other combinations with different numbers and calculate their sums. </prompt_2.2>
</Plan_2>

<Execution_2>
<execution_2.1>
How about 1, 2, 3, 4, 7. Their product is 1×2×3×4×7=168. That's too big. So if I want the product to be 84, maybe 1, 2, 2, 3, 7. But then there are duplicate 2s, which are not allowed. So that's invalid. Alternatively, 1, 2, 3, 14: 1×2×3×14=84. Sum is 1+2+3+14=20. Still higher than 15.
</execution_2.1>

<execution_2.2>
Sum is 1+2+3+14=20. Not better. 1,2,4,21: sum 26. 1,2,5, ... Wait, 84 divided by 1,2,5 is 42/5=8.4, which is not an integer. So that's not possible. 1,2,6,7: sum 16. 1,3,4,7: sum 15. Let's see if there's another combination.

How about 2, 2, 3, 7. But those are the prime factors, but duplicates. So we can't have that. If we group them differently, like 2×2=4, then 3,4,7. Sum 14. Wait, but 4,3,7. Wait, but 4,3,7 are distinct. So 3×4×7=84. Sum is 3+4+7=14. Wait, that's a lower sum. Wait, how is that possible?

Wait, hold on. If I take 3, 4, 7, their product is 3×4×7=84. Their sum is 3+4+7=14. That's better than 15. But how did I not think of that earlier? Let me check.
</execution_2.2>
</Execution_2>

<Plan_3>
Based on execution_2.2:
3, 4, 7. Product: 3×4×7=84. Sum: 14. Is that valid? All numbers are distinct positive integers. Yes. So why didn't I think of that earlier? Maybe because when I tried grouping the prime factors, I thought of 2, 6, 7 which also sums to 15, but here 3, 4, 7 sum to 14. So that's better. So which one is the minimal?
-----

Let me see. Let's check if there's a way to get a lower sum. For example, if I can find three numbers that multiply to 84 with a sum lower than 14.

Let's see. Let's think of factors of 84 and see possible triplet combinations.

<prompt_3.1> Verify if there are any other triplet combinations and confirm the minimal sum. </prompt_3.1>
-----

Based on execution_2.2, execution_1.4:
Then the minimal sum is 14. But let me check if there's any other combination with four numbers that gives a lower sum.

<prompt_3.2> Confirm the minimal sum and ensure no other combinations yield a lower sum. </prompt_3.2>
</Plan_3>

<Execution_3>
<execution_3.1>
Starting with 3,4,7: sum 14. Is there another triplet?

How about 2, 3, 14: sum 19. Not better. 2, 6, 7: sum 15. 1, 12, 7: but 1 is allowed, sum 20. 3, 2, 14: same as before. 4, 3, 7: same as the first triplet. How about 1, 6, 14: sum 21. 2, 2, 21: duplicates. 3, 7, 4: same as before. 6, 2, 7: sum 15. 84 is also 1×84, but that's two numbers, sum 85. 2×42: sum 44. 3×28: sum 31. 4×21: sum 25. 6×14: sum 20. 7×12: sum 19. So none of these pairs have a sum lower than 14. So the triplet 3,4,7 seems to be the minimal sum.
</execution_3.1>

<execution_3.2>
For example, 2, 3, 4, 7: sum 16. If I include 1, 2, 3, 14: sum 20. 1, 2, 4, 21: sum 26. 1, 3, 4, 7: sum 15. 2, 3, 7: sum 12, but product is 42. Not 84. If I take 2, 2, 21: sum 25. Not good. 3, 3, (84/(3×3)=84/9=9.333). Not integer. So that's not possible. 4, 4, 5.25: nope. So seems like the only triplet that works is 3,4,7. Let me check another triplet: 2, 6, 7. Sum 15. 3, 4, 7. Sum 14. 1, 6, 14: sum 21. So 14 is the smallest so far.

Wait, another triplet: 1, 1, 84. But duplicates. Not allowed. 1, 2, 42. Sum 45. 1, 3, 28. Sum 32. 1, 4, 21. Sum 26. 1, 5, 16.8. Not integer. 1, 6, 14. Sum 21. 1, 7, 12. Sum 20. 1, 8, 10.5. Not integer. 2, 2, 21. Sum 25. 2, 3, 14. Sum 19. 2, 4, 10.5. Not integer. 2, 5, 8.4. Not integer. 2, 6, 7. Sum 15. 3, 3, (84/9=9.333). Not integer. 3, 4, 7. Sum 14. 3, 5, 5.6. Not integer. 4, 4, 5.25. Not integer. So seems like the only possible triplets are 3,4,7 and 2,6,7. The sums are 14 and 15 respectively. So 14 is better.

Alternatively, maybe four numbers. Let's see. Let's try 2, 3, 4, 7. Sum 16. If we can find four distinct numbers that multiply to 84 with a smaller sum. For example, 1, 2, 6, 7. Sum 16. 1, 3, 4, 7. Sum 15. If we try 2, 2, 3, 7. But duplicates. Not allowed. How about 2, 3, 2, 7. Still duplicates. So no. Alternatively, 1, 1, 6, 14. Duplicates again. So seems like four numbers can't get better than 15. Five numbers would require more 1s, but as we saw earlier, that increases the sum. For example, 1, 2, 3, 4, 7. Sum 1+2+3+4+7=17. Which is higher than 15.

So the minimal sum seems to be 14 with the numbers 3, 4, 7. Let me confirm once more if there's any other triplet or combination that I might have missed. Let's list all possible triplets of distinct positive integers that multiply to 84.

Starting with 3:
</execution_3.2>
</Execution_3>

<Plan_4>
Based on execution_3.2:
3 × 4 × 7 = 84. Sum 14.

<prompt_4.1> Confirm the minimal sum and ensure no other combinations yield a lower sum. </prompt_4.1>
</Plan_4>

<Execution_4>
<execution_4.1>
3 × 2 × 14 = 84. Sum 3+2+14=19.

3 × 1 × 28 = 84. But 1 is allowed, but sum is 31. Not better.

4 × 3 × 7 = same as first. 5 × something: 5 × 84/5 = 16.8, not integer. 6 × 14 = 84, but needs three numbers. 6 × 7 × 2 = 84, sum 15.

So yes, 3,4,7 is the minimal sum triplet.
</execution_4.1>
</Execution_4>

<Plan_5>
Based on execution_4.1:
Therefore, the least possible sum is 14.
</Plan_5>

<Execution_5>
No prompt to execute.
</Execution_5>

<Final_answer>
- Thus, the least possible sum of these integers is $\boxed{14}$.
</Final_answer>
</think>