<Question>
Suppose the polynomial
$$f(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_2x^2 + a_1x + a_0$$has integer coefficients, and its roots are distinct integers.

Given that $a_n=2$ and $a_0=66$, what is the least possible value of $|a_{n-1}|$?
</Question>

<think>
<Plan_1>
Okay, let's see. I need to find the least possible value of |a_{n-1}| for a polynomial f(x) with integer coefficients, where the leading coefficient a_n is 2, the constant term a_0 is 66, and all the roots are distinct integers. Hmm, right. So, the polynomial is of the form 2(x - r1)(x - r2)...(x - rk), where each ri is a distinct integer, and when expanded, the constant term is 66. Since the leading coefficient is 2, the polynomial is monic multiplied by 2. The constant term a0 is 2*(-1)^k*(r1*r2*...*rk) = 66. So, the product of the roots times (-1)^k is 66/2 = 33. Therefore, the product of the roots is either 33 or -33, depending on the degree.

Wait, but the roots are distinct integers. So, first, I need to factor 33 into distinct integer factors, considering both positive and negative factors. Let me note that 33 factors into 1*3*11, and also -1*-3*-11, but since the roots can be positive or negative, the actual roots could be combinations of these factors with possible sign changes. But since the product of the roots is 33 or -33, depending on the number of roots. Wait, but the number of roots is equal to the degree n. So, if n is even, then (-1)^k is 1, so the product is 33.
If n is odd, then (-1)^k is -1, so the product is -33. But the actual roots can be arranged such that their product is either 33 or -33. However, since the roots are distinct integers, we need to consider all possible factorizations of 33 and -33 into distinct integers.

But maybe I should start by listing all possible sets of distinct integer roots whose product is 33 or -33. Then, for each such set, compute the sum of the roots (since a_{n-1} is -2 times the sum of the roots), and find the minimal |sum|.

Wait, let's recall Vieta's formula. For a polynomial 2x^n + ... + a1x + a0, the coefficient a_{n-1} is equal to 2*(-sum of roots). Wait, let me check.

<prompt_1.1> Use Vieta's formulas to determine how the coefficient \(a_{n-1}\) relates to the sum of the roots of the polynomial. </prompt_1.1>

Therefore, to minimize |a_{n-1}|, we need to minimize |sum of roots|. So, the problem reduces to finding a set of distinct integers whose product is ±33, and whose sum has the smallest possible absolute value.

So, first, let's factor 33 into distinct integers. The prime factors of 33 are 3 and 11. So, possible factors are 1, 3, 11, 33, and their negatives. But since we need distinct integers, we can't repeat factors. So, let's consider all possible combinations of distinct integers whose product is 33 or -33.

Let me start by listing all possible factorizations of 33 into distinct integers:

<prompt_1.2> List all possible factorizations of 33 into distinct integers. </prompt_1.2>
</Plan_1>

<Execution_1>
<execution_1.1>
If the polynomial is f(x) = 2x^n + a_{n-1}x^{n-1} + ... + a0, then in terms of roots, it can be written as 2(x - r1)(x - r2)...(x - rn). Expanding this, the coefficient of x^{n-1} is 2*(-r1 - r2 - ... - rn) = a_{n-1}. Therefore, a_{n-1} = -2*(sum of roots). Therefore, |a_{n-1}| = 2*|sum of roots|. So, to minimize |a_{n-1}|, we need to minimize |sum of roots|.
</execution_1.1>

<execution_1.2>
1) 33 itself: but that's a single factor. However, if we have only one root, then the polynomial would be degree 1, but the product would be 33 or -33. But since the constant term is 66, which is 2*(33 or -33), so if the polynomial is linear, then f(x) = 2x + 66, which would have a root at x = -33. But in that case, the product of the roots is -33, which would mean a0 = 2*(-33) = -66, but given a0 is 66, so that's not possible. Wait, maybe I need to check this again.

Wait, the polynomial is f(x) = 2*(x - r1)*(x - r2)*...*(x - rk). Then, the constant term a0 is 2*(-1)^k*(r1*r2*...*rk) = 66. Therefore, (-1)^k*(r1*r2*...*rk) = 33. So, the product of the roots is 33*(-1)^k. Therefore, if k is even, product is 33; if k is odd, product is -33. So, the roots can be positive or negative, but their product must be 33 or -33. So, for example, if there's one root, then the product is r1 = 33 or -33. Then, a0 would be 2*(-1)^1*(-33) = 2*33 = 66. Wait, hold on. Let's check this.

If there's one root, then f(x) = 2*(x - r1). Then, a0 = 2*(-r1). So, 2*(-r1) = 66 => -r1 = 33 => r1 = -33. So, the root is -33. Then, the sum of the roots is -33, so a_{n-1} = -2*(-33) = 66. Therefore, |a_{n-1}| = 66. But maybe we can get a smaller sum by having more roots.
</execution_1.2>
</Execution_1>

<Plan_2>
Based on execution_1.2:
For example, if we have two roots, then the product of the roots is 33 or -33. Let's see.

<prompt_2.1> Analyze the case of two distinct integer roots whose product is 33 or -33. </prompt_2.1>

For three roots, the product of the roots is 33 or -33. Let's see. 33 can be factored into three distinct integers.

<prompt_2.2> Analyze the case of three distinct integer roots whose product is 33 or -33. </prompt_2.2>
</Plan_2>

<Execution_2>
<execution_2.1>
Case 1: Two roots. Then, product is 33 or -33. So, possible pairs of distinct integers whose product is 33: (1, 33), (3, 11), (-1, -33), (-3, -11). Similarly, for product -33: (1, -33), (-1, 33), (3, -11), (-3, 11). Wait, but since the roots are distinct, these are the possible pairs.

For each pair, the sum would be:

For (1, 33): sum = 34

For (3, 11): sum = 14

For (-1, -33): sum = -34

For (-3, -11): sum = -14

For (1, -33): sum = -32

For (-1, 33): sum = 32

For (3, -11): sum = -8

For (-3, 11): sum = 8

So, the sums are 34,14,-34,-14,-32,32,-8,8. Then, the absolute values are 34,14,34,14,32,32,8,8. Then, the minimal |sum| here is 8. So, if we take roots 3 and -11, sum is -8, or roots -3 and 11, sum is 8. Either way, |sum| =8. Then, |a_{n-1}| = 2*8 = 16. Which is better than the linear case of 66. So, 16 is a candidate.
</execution_2.1>

<execution_2.2>
Possible triplet factors of 33: 1, 3, 11. But 1*3*11 = 33. So, if we take roots as 1, 3, 11, then the product is 33. Then, the sum is 1 + 3 + 11 = 15. Then, a_{n-1} = -2*15 = -30, so |a_{n-1}| = 30. Alternatively, if we take roots as -1, -3, -11, product is (-1)*(-3)*(-11) = -33, so to get product 33, we need an even number of negative roots. Wait, but if we have three roots, to get product 33, which is positive, we need an even number of negative roots. So, possible sets: two negative and one positive. For example, (-1, -3, 11). Then product is (-1)*(-3)*11 = 33. Sum is (-1) + (-3) + 11 = 7. Then, a_{n-1} = -2*7 = -14, |a_{n-1}| =14. That's better. Alternatively, (-1, 3, -11). Product is (-1)*3*(-11) = 33. Sum is (-1) + 3 + (-11) = -9. Then, |sum| =9, so |a_{n-1}|=18. Not as good. Another combination: (-3, 3, -11). Product is (-3)*3*(-11)=99. Not 33. So, that's not good. Wait, maybe 1, -3, -11. Wait, we already did that. So, the best sum here is 7, leading to |a_{n-1}|=14.
</execution_2.2>
</Execution_2>

<Plan_3>
Based on execution_2.1, execution_2.2:
Wait, let me check again. For three roots, to get product 33, we need an even number of negative roots.

<prompt_3.1> Verify the calculations for triplet roots and correct any errors. </prompt_3.1>

Based on execution_2.2:
Alternatively, if we take roots as 1, -1, and -33. Wait, but then the product would be 1*(-1)*(-33)=33. Sum would be 1 + (-1) + (-33) = -33. Then, |sum|=33, which is worse.

<prompt_3.2> Explore other possible triplet roots and confirm the minimal |sum|. </prompt_3.2>
</Plan_3>

<Execution_3>
<execution_3.1>
So, either 0 or 2 negative roots. If 0 negative roots, then all positive: 1,3,11. Sum 15. If 2 negative roots: say -1, -3, 11. Sum is -1 -3 +11 =7. Alternatively, -1, -11, 3. Sum is -1 -11 +3 = -9. Or -3, -11, 1. Sum is -3 -11 +1 = -13. So, the minimal |sum| here is 7, leading to |a_{n-1}|=14.
</execution_3.1>

<execution_3.2>
Or 1, -3, 11. Wait, product is 1*(-3)*11 = -33. So, to get product 33, we need to have two negative roots. For example, (-1, -3, 11) as before, sum 7. Alternatively, (-1, -11, 3). Product is (-1)*(-11)*3=33. Sum is (-1) + (-11) +3 = -9. So, worse. So, the minimal |sum| in three roots is 7, leading to |a_{n-1}|=14.

But wait, can we get even a smaller sum? Let's see. For example, if we can find three distinct integers whose product is 33 or -33, and sum is smaller. Let's see. Let's consider all possible triplet combinations.

Possible triplet factors of 33: 1, 3, 11. Their product is 33. Sum is 15. If we take two negatives and one positive: (-1, -3, 11) sum 7; (-1, -11, 3) sum -9; (-3, -11, 1) sum -13. So, the minimal |sum| here is 7. So, 7 is the smallest for three roots.

Alternatively, if we have four roots. Then, the product of the roots is 33 or -33. Let's see if we can factor 33 into four distinct integers. For example, 1, -1, 3, -11. Then, product is 1*(-1)*3*(-11)=33. Sum is 1 -1 +3 -11= -9. Alternatively, 1, -1, 11, -3. Sum is same. Or 1, 3, 11, and -1. Same sum. Alternatively, 1, -3, 11, -1. Same. What if we use 1, -1, 3, 11. Then product is 1*(-1)*3*11= -33. So, to get product 33, we need even number of negatives. So, maybe two negatives: (-1, -3, 1, 11). Product is (-1)*(-3)*1*11=33. Sum is (-1) + (-3) +1 +11=8. So, |sum|=8. Then, |a_{n-1}|=16. Alternatively, (-1, -11, 3, 1). Sum is same as above. So, 8. So, this gives |a_{n-1}|=16, which is higher than the three-root case. So, 14 is better.

Alternatively, maybe another four roots. Let's see. 1, -1, 3, -11. Product 33, sum -9. So, |sum|=9. Not better. So, seems like four roots give a higher |sum| than three roots.
</execution_3.2>
</Execution_3>

<Plan_4>
Based on execution_3.1, execution_3.2:
Wait, but perhaps there are more combinations. Let me check.

<prompt_4.1> Consider other possible factorizations and confirm the minimal |sum|. </prompt_4.1>
</Plan_4>

<Execution_4>
<execution_4.1>
For example, if we take four roots: 1, -1, 3, -11. As before, sum is -9. If we take 1, -1, 11, -3. Sum is same. Alternatively, maybe another set. For example, 1, -3, 11, -1. Sum is same. What if we take 1, 3, -1, -11. Sum is 1 +3 -1 -11= -8. So, |sum|=8. Then, a_{n-1}= -2*(-8)=16, |16|=16. Which is worse than the three-root case. So, 14 is better.

Alternatively, if we take 1, -1, 3, 11. Then, product is 1*(-1)*3*11= -33. So, to get product 33, need even number of negatives. So, maybe two negatives: (-1, -3, 1, 11). Product is 33, sum is 8. So, same as before. So, seems like 14 is the minimal so far.

But wait, let's check if there are other four roots. For example, 1, -1, 3, -11. Product 33, sum -9. Alternatively, 1, -1, -3, 11. Product is 1*(-1)*(-3)*11=33. Sum is 1 -1 -3 +11=8. So, same as before. So, |sum|=8. So, 16.

Alternatively, 1, -1, 3, -3. Wait, but roots must be distinct. 3 and -3 are distinct, but 1 and -1 are distinct. However, product is 1*(-1)*3*(-3)=9. Not 33. So, not helpful. So, perhaps not.

Alternatively, maybe 1, -1, 3, 11. Product is -33. So, to get 33, need two negatives. So, (-1, -3, 1, 11). As before. So, sum 8. So, same.

Alternatively, 1, -1, 3, -11. Product is 33, sum -9. So, no improvement.

Alternatively, maybe 1, -3, 11, -11. Wait, but then product is 1*(-3)*11*(-11)= 363. Not 33. So, no. So, seems like four roots can give sum 8 or -9, which is worse than the three-root case.

Alternatively, maybe five roots. Then, the product of the roots is 33 or -33. Let's see. For example, 1, -1, 3, -3, 11. Product is 1*(-1)*3*(-3)*11= 99. Not 33. Alternatively, 1, -1, 3, 11, -11. Product is 1*(-1)*3*11*(-11)= 363. Not 33. Alternatively, 1, -1, 3, -1, 11. But roots must be distinct, so can't have duplicate 1 and -1. So, maybe 1, -1, 3, 11, -33. But product is 1*(-1)*3*11*(-33)= 1089. Not 33. So, seems like five roots would require a product that's too large. So, maybe not useful.

Alternatively, maybe using 1, -1, 3, -11, and 11. Wait, but 11 is already there. So, duplicates. Not allowed. So, perhaps five roots is not feasible.

Alternatively, maybe using 1, -1, 3, -3, 11. But product is 1*(-1)*3*(-3)*11= 99. Not 33.
</execution_4.1>
</Execution_4>

<Plan_5>
Based on execution_4.1:
Hmm.

Wait, perhaps another approach. Let's list all possible factorizations of 33 into distinct integers, both positive and negative, and compute their sums.

<prompt_5.1> List all possible factorizations and calculate their sums to find the minimal |sum|. </prompt_5.1>
</Plan_5>

<Execution_5>
<execution_5.1>
For two roots:

Possible pairs:

(1, 33): sum 34

(3, 11): sum 14

(-1, -33): sum -34

(-3, -11): sum -14

(1, -33): sum -32

(-1, 33): sum 32

(3, -11): sum -8

(-3, 11): sum 8

So, as before, the minimal |sum| here is 8.

For three roots:

Possible triplets:

(1, 3, 11): sum 15

(-1, -3, 11): sum 7

(-1, 3, -11): sum -9

(1, -3, -11): sum -13

(-1, -1, 33): but duplicates, so invalid.

Similarly, other combinations would either repeat or not multiply to 33. So, the minimal |sum| here is 7.

For four roots:

Possible quadruplets:

(1, -1, 3, -11): sum -9

(1, -1, 11, -3): sum same

(-1, -3, 1, 11): sum 8

(1, -3, 11, -11): product is 1*(-3)*11*(-11)= 363

No, that's not 33. So, need to find quadruplets where product is 33. Let's see.

For example, 1, -1, 3, 11. Product is -33. To get 33, need two negatives. So, (-1, -3, 1, 11). Product is 33. Sum is 8.

Alternatively, 1, -1, -3, -11. Product is 33. Sum is 1 -1 -3 -11= -14. |sum|=14.

Alternatively, 1, -1, 3, -3. Product is 9. Not 33. So, no.

Alternatively, 1, -1, 3, 11. Product is -33. So, to get 33, need two negatives. So, (-1, -3, 1, 11). As before.

Alternatively, 1, -3, 11, -1. Same as before.

So, seems like four roots can give sum 8 or -14. So, 8 is better, but 8 is worse than the three-root case of sum 7.

For five roots:

As discussed earlier, seems challenging. Let's see. For example, 1, -1, 3, -3, 11. Product is 1*(-1)*3*(-3)*11= 99. Not 33. Alternatively, 1, -1, 3, 11, -33. Product is 1*(-1)*3*11*(-33)= 1089. Not 33. So, seems not feasible.

Alternatively, 1, -1, 1, -1, 33. But duplicates. Not allowed. So, no.

So, seems like the minimal |sum| occurs when we have three roots: -1, 3, -11. Sum is 7. Therefore, |a_{n-1}|=2*7=14.
</execution_5.1>
</Execution_5>

<Plan_6>
Based on execution_5.1:
Wait, but let's check if there are other three-root combinations that I might have missed. For example, 1, -1, 33. But product is 1*(-1)*33= -33. So, to get product 33, need two negatives. So, (-1, -33, 1). Product is (-1)*(-33)*1=33. Sum is (-1) + (-33) +1= -33. |sum|=33. Worse.

<prompt_6.1> Verify the calculations and confirm the minimal |sum| for three roots. </prompt_6.1>
</Plan_6>

<Execution_6>
<execution_6.1>
Alternatively, (-1, 3, -11). Sum is 7. That's the minimal so far.

Alternatively, (3, -1, -11). Same as above. So, seems like 7 is the minimal for three roots.
</execution_6.1>
</Execution_6>

<Plan_7>
Based on execution_6.1:
Wait, but let me check another possibility. What if we use 1, -1, 3, -11, but that's four roots. Wait, we already checked that. The sum is -9.

<prompt_7.1> Verify the calculations and confirm the minimal |sum| for three roots. </prompt_7.1>
</Plan_7>

<Execution_7>
<execution_7.1>
Alternatively, maybe another three-root combination. For example, 1, -3, 11. Wait, that's three roots. Wait, 1, -3, 11. Product is 1*(-3)*11= -33. So, to get product 33, need two negatives. So, (-1, 3, 11). Product is (-1)*3*11= -33. Still negative. So, to get positive 33, we need even number of negatives. So, two negatives. So, (-1, -3, 11). Product is 33. Sum is 7. That's the minimal. Alternatively, (-1, 3, -11). Sum is -9. (-3, 11, -1). Sum is -13. So, 7 is the smallest.

Alternatively, if we take roots as 1, 3, -11. Product is 1*3*(-11)= -33. So, to get 33, need two negatives. So, (-1, -3, 11). As before.

Alternatively, 1, -1, -33. But that's three roots. Product is 1*(-1)*(-33)=33. Sum is 1 -1 -33= -33. |sum|=33. Worse.

Alternatively, 1, -3, 11. Product is 1*(-3)*11= -33. So, need two negatives. So, (-1, 3, 11). Product is (-1)*3*11= -33. Still negative. So, need two negatives. So, (-1, -3, 11). As before.

Alternatively, 3, -1, -11. Same as before.

So, seems like the minimal |sum| is 7. Therefore, the minimal |a_{n-1}| is 14. But wait, let's check if there's a case with more roots where the sum can be smaller. For example, five roots? Wait, we tried that earlier, but maybe there's a way.
</execution_7.1>
</Execution_7>

<Plan_8>
Based on execution_7.1:
Wait, let's see. If we have six roots, but that would require each root being 1 or -1, but 1 and -1 can only be used once each. Wait, but if we have six roots, we need six distinct integers. But 1, -1, 3, -3, 11, -11. Then, product is 1*(-1)*3*(-3)*11*(-11)= (1*-1)*(3*-3)*(11*-11)= (-1)*(-9)*(-121)= (-1)*(-9)=9; 9*(-121)= -1089. Not 33. So, product is -1089. So, to get product 33, we need to adjust. But since all roots are used, the product is fixed. So, not helpful.

Alternatively, maybe using 1, -1, 3, 11, -33, but that's six roots. Product is 1*(-1)*3*11*(-33)= -1*3*11*(-33)= -33*33= -1089. Not 33. So, not helpful.

Alternatively, maybe using 1, -1, 3, -3, 11, -11. Product is (1*-1)*(3*-3)*(11*-11)= (-1)*(-9)*(-121)= (-1)*(-9)=9; 9*(-121)= -1089. Same as before.

<prompt_8.1> Verify the calculations and confirm the minimal |sum| for three roots. </prompt_8.1>
</Plan_8>

<Execution_8>
<execution_8.1>
So, the product is too large. So, not helpful.

Alternatively, maybe using smaller numbers. Wait, but 33 is 3*11. The factors are limited. So, perhaps the minimal |sum| is indeed 7. But let's check another possibility. For example, roots like 2, but wait, the leading coefficient is 2. Wait, no, the roots can be any integers, but the polynomial is 2n degree. Wait, but the roots are integers, but the problem didn't specify the degree. Wait, actually, the problem states that the polynomial has integer coefficients, leading coefficient 2, and constant term 66. So, the degree n is the number of roots. So, for example, if we have three roots, the polynomial is quadratic? Wait, no. Wait, if there are three roots, then the polynomial is cubic. But the leading coefficient is 2, so f(x)=2(x - r1)(x - r2)(x - r3). Then, the constant term is 2*(-r1)(-r2)(-r3)=2*(-1)^3 r1 r2 r3= -2 r1 r2 r3=66. Therefore, r1 r2 r3= -33. So, the same as before. So, the sum of roots is r1 + r2 + r3, and a_{n-1}= -2*(sum of roots). Therefore, |a_{n-1}|=2*|sum|. So, to minimize |a_{n-1}|, we need the minimal |sum|.

So, in the three-root case, the minimal |sum| is 7, leading to |a_{n-1}|=14. For two roots, it's 8, leading to 16. For four roots, 8 or 14. Wait, no, four roots can lead to sum 8, but with |a_{n-1}|=16. Wait, but if with four roots, the sum is 8, then |a_{n-1}|=16, same as two roots. But if with three roots, sum is 7, which is smaller, so 14 is better.

But wait, is there a way to get a sum of 6 or less? Let's see. For three roots, can we have a sum of 6? Let's see. Let's see if there's a triplet of distinct integers whose product is 33 or -33 and sum is 6 or -6.
</execution_8.1>
</Execution_8>

<Plan_9>
Based on execution_8.1:
Let's see. For product 33, possible triplet with sum 6. Let's see. Let's think of three distinct integers multiplying to 33. 1, 3, 11. Sum 15. If we can find three numbers closer together. For example, 33 can be written as 3*11*1. But those are the only positive factors. If we use negative numbers. For example, 33= (-1)*(-3)*11. Sum is (-1) + (-3) +11=7. Which is what we had. If we use 33= (-1)*3*(-11). Sum is (-1) +3 + (-11)= -9. Alternatively, 1, -3, -11. Sum -13.

<prompt_9.1> Verify the calculations and confirm the minimal |sum| for three roots. </prompt_9.1>
</Plan_9>

<Execution_9>
<execution_9.1>
Alternatively, 1, 1, 33. But duplicates. Not allowed. 1, 2, 16.5. Not integers. So, seems like no. So, perhaps 7 is the minimal for three roots.

Alternatively, maybe using more roots. For example, five roots. Let's see. For example, 1, -1, 3, -3, 11. Product is 1*(-1)*3*(-3)*11= 99. Not 33. To get 33, need to adjust. For example, 1, -1, 3, 11, -1. Duplicates. Not allowed. So, perhaps 1, -1, 3, 11, -33. Product is 1*(-1)*3*11*(-33)=1089. Not 33. So, not helpful.

Alternatively, 1, -1, 3, -11, 33. Product is 1*(-1)*3*(-11)*33= 1* -1= -1; -1*3= -3; -3*-11=33; 33*33=1089. Not 33. So, not helpful. So, seems like with more roots, it's hard to get the product down to 33.

Alternatively, maybe using 1, -1, 3, -1, 11. But duplicates again. Not allowed.

Alternatively, 1, -1, 3, 11, -33. Product is 1*(-1)*3*11*(-33)= 1* -1= -1; -1*3= -3; -3*11= -33; -33*(-33)=1089. Not 33. So, same as before.

Therefore, seems like the minimal |sum| is 7, leading to |a_{n-1}|=14. But wait, let's check if there's another combination. For example, 3, -3, 11. Product is 3*(-3)*11= -99. So, to get product 33, need two negatives. So, (-3, 3, 11). Product is (-3)*3*11= -99. Not 33. Alternatively, (-3, -11, 1). Product is (-3)*(-11)*1=33. Sum is (-3) + (-11) +1= -13. |sum|=13. Worse than 7.

Alternatively, (-1, 3, -11). Sum is -9. So, worse. So, seems like 7 is the minimal.

But let me check another angle. Suppose we have more roots. For example, using 1, -1, 3, -1, 11. But duplicates. Not allowed. So, no.
</execution_9.1>
</Execution_9>

<Plan_10>
Based on execution_9.1:
Alternatively, maybe 1, -1, 3, -3, 11. But product is 1*(-1)*3*(-3)*11= 99. Not 33. So, no.

<prompt_10.1> Verify the calculations and confirm the minimal |sum| for three roots. </prompt_10.1>
</Plan_10>

<Execution_10>
<execution_10.1>
Alternatively, 1, -1, 3, -3, -11. Product is 1*(-1)*3*(-3)*(-11)= -99. Still not 33. So, seems like not possible.

Alternatively, maybe 1, -1, 3, 11, -33. Product is 1*(-1)*3*11*(-33)= 1089. Not 33. So, no.

Alternatively, 1, -1, 3, -11, 33. Product is 1*(-1)*3*(-11)*33= 1* -1= -1; -1*3= -3; -3*-11=33; 33*33=1089. Not 33. So, same.

Therefore, seems like the minimal |sum| is 7. So, the minimal |a_{n-1}| is 14. But wait, let me check another case. For example, four roots with sum 6. Let's see. Suppose we have four roots: 1, -1, 2, -3. Then, product is 1*(-1)*2*(-3)=6. So, to get product 33, need 33/6=5.5. Not integer. So, no. Alternatively, 1, -1, 3, -11. Product is 1*(-1)*3*(-11)=33. Sum is 1 -1 +3 -11= -8. |sum|=8. Which we already considered. So, worse than 7.

Alternatively, 1, -1, 11, -3. Same as above.

Alternatively, 3, -3, 11, -1. Product is 3*(-3)*11*(-1)=99. Not 33. So, no.

Alternatively, 1, -1, 3, -1. Duplicates. Not allowed.

Alternatively, 1, -1, 3, 11. Product is -33. So, need two negatives. So, (-1, -3, 1, 11). Sum 7. So, same as before.

So, seems like there's no combination of distinct integers that gives a smaller sum than 7. Therefore, the minimal |sum| is 7, leading to |a_{n-1}|=14. Therefore, the least possible value of |a_{n-1}| is 14.
</execution_10.1>
</Execution_10>

<Plan_11>
Based on execution_10.1:
But wait, let me check if there's another set of roots. For example, 1, -2, -16.5. But -16.5 is not integer. So, no.

<prompt_11.1> Verify if there are any other possible sets of roots that can yield a smaller sum. </prompt_11.1>
</Plan_11>

<Execution_11>
<execution_11.1>
Alternatively, maybe using 1, -1, 3, -11. But that's four roots. Sum is -8. So, already considered.

Alternatively, 1, -1, 33. Sum 33. |sum|=33. Worse.

Alternatively, 1, 3, 11. Sum 15. Worse.

Alternatively, 3, 11, -1. Sum 13. Worse.

Alternatively, 1, -3, -11. Sum -13. Worse.

Alternatively, -1, -3, 11. Sum 7. That's the same as before. So, same.

Alternatively, 1, -1, 3, -11. Sum -8. Worse.

Alternatively, 1, -3, 11, -1. Sum -4. Wait, 1 -3 +11 -1=8. Wait, no. 1 -3 is -2, plus 11 is 9, minus 1 is 8. So, sum 8. So, |sum|=8. So, worse than 7.

Alternatively, 3, -1, -11. Sum 3 -1 -11= -9. |sum|=9. Worse.

Alternatively, 11, -1, -3. Sum 11 -1 -3=7. So, same as before.

Therefore, the minimal |sum| is 7. Therefore, the least possible value of |a_{n-1}| is 14.
</execution_11.1>
</Execution_11>

<Final_answer>
This step-by-step analysis systematically explores the problem, considering different numbers of roots and their possible combinations, ensuring that all cases are exhausted to confirm that the minimum value of $|a_{n-1}|$ is indeed $\boxed{14}$.
</Final_answer>
</think>