Explanation:
Okay, let me try to figure out this problem. So, we have a polynomial with integer coefficients, distinct integer roots. The leading coefficient is 2, and the constant term is 66. We need to find the least possible value of |a_{n-1}|.

First, I remember that for a polynomial with integer coefficients and distinct integer roots, we can express it as f(x) = 2(x - r₁)(x - r₂)...(x - rₙ), where each rᵢ is an integer, and all roots are distinct. Since the leading coefficient is 2, that's why we have that 2 in front.

The constant term is given by a₀ = 66. The constant term of a polynomial is the product of the roots multiplied by the leading coefficient, but with a sign depending on the degree. Wait, more precisely, for a monic polynomial (leading coefficient 1), the constant term is (-1)^n * product of roots. But here the leading coefficient is 2, so the constant term would be 2 * (-1)^n * product of roots. Wait, let me check.

If we have f(x) = a_n(x - r₁)(x - r₂)...(x - rₙ), then expanding this, the constant term is a_n * (-1)^n * r₁r₂...rₙ. Therefore, in our case, since a_n = 2, the constant term a₀ = 2 * (-1)^n * (product of roots). But the problem says a₀ = 66, so:

2 * (-1)^n * (r₁r₂...rₙ) = 66 ⇒ (-1)^n * (r₁r₂...rₙ) = 33.

Therefore, the product of the roots is 33 or -33, depending on whether n is even or odd. Since 33 is positive, if n is even, then the product of roots is 33; if n is odd, the product of roots is -33.

But since all roots are distinct integers, we need to find integers r₁, r₂,..., rₙ such that their product is either 33 or -33, and then find the polynomial with leading coefficient 2 and find the minimal |a_{n-1}|.

Wait, but how does a_{n-1} relate to the roots? Let me recall. The coefficient a_{n-1} is the coefficient of x^{n-1} in the polynomial. For a monic polynomial, the coefficient of x^{n-1} is - (sum of roots). But in our case, the polynomial is 2*(x - r₁)...(x - rₙ). Therefore, expanding this, we get:

f(x) = 2x^n - 2*(r₁ + r₂ + ... + rₙ)x^{n-1} + ...

Therefore, the coefficient a_{n-1} is -2*(sum of roots). Wait, so a_{n-1} = -2*(sum of roots). Therefore, |a_{n-1}| = 2*|sum of roots|. Therefore, to minimize |a_{n-1}|, we need to minimize |sum of roots|.

Therefore, our problem reduces to finding distinct integers r₁, r₂,..., rₙ such that their product is 33 or -33 (depending on n's parity), and the absolute value of their sum is as small as possible, then multiply by 2 to get |a_{n-1}|.

But since we can choose the number of roots (n) as long as the roots are distinct integers, we need to consider different possibilities for n. However, the problem says the polynomial has degree n, so n can be any integer ≥ 1, but given that the roots are distinct integers and the product is 33 or -33, the number of roots can't be too large.

But since 33 factors into primes as 3 * 11, so the prime factors are 3 and 11. Also, the product can be positive or negative. Let me consider possible factorizations.

First, note that since the roots are integers, they must divide 33 or -33. The divisors of 33 are ±1, ±3, ±11, ±33. Therefore, the roots must be among these numbers. Since the roots are distinct integers, we can't repeat any of these.

So, to find possible sets of distinct integers whose product is 33 or -33. Also, note that if n is the degree, then n can be 1, 2, 3, ..., up to... Well, since the product of roots is 33 or -33, and each root is distinct, so the maximum number of roots would be the number of distinct divisors. Since there are 8 divisors (±1, ±3, ±11, ±33), but we need distinct ones. However, we need to consider that if we take more roots, we might need to include 1 or -1, but we also have to include other roots.

But since the product is fixed (±33), the more roots we have, the smaller the absolute value of each root would need to be, but they have to multiply to 33 or -33. For example, if we take more roots, we have to include 1 or -1, but then the other roots would have to be factors of 33 divided by those. However, we need to ensure that all roots are distinct integers. Let me think.

But since we need to minimize |sum of roots|, maybe having more roots can help? Because if we have more roots, maybe we can have some positive and negative numbers so that their sum is small. Let me try.

First, let me think about possible factorizations of 33 and -33 into distinct integers.

Case 1: The product is 33. Then the roots are integers whose product is 33. Similarly for -33.

First, for positive product 33:

Possible factorizations with distinct integers. Since 33 is positive, the number of negative roots must be even. Because negative times negative is positive. So, if we have an even number of negative roots, the product is positive.

Similarly, if the product is -33, the number of negative roots must be odd.

Let me first consider possible sets of roots for both cases.

Starting with product 33:

Possible combinations:

Option 1: Single root: 33. Then the polynomial is 2*(x - 33). Then sum of roots is 33, |a_{n-1}| = 2*33=66.

Option 2: Two roots: product 33. Since 33 factors as 1*33, 3*11, (-1)*(-33), (-3)*(-11). But since roots must be distinct integers, so possible pairs:

(1, 33), (3, 11), (-1, -33), (-3, -11). So sum of roots for (1, 33) is 34, for (3, 11) is 14, for (-1, -33) is -34, for (-3, -11) is -14. Therefore |sum| is 34, 14, 34, 14. So minimal |sum| is 14. Then |a_{n-1}| would be 2*14=28.

Option 3: Three roots. Product 33. Let's see. Since 33 is positive, need even number of negative numbers. Wait, if there are three roots, which is odd, so to have even number of negative roots? No, no. Wait, the product is positive, so number of negative roots must be even. But three roots can have 0 or 2 negative roots. Let's try.

Possible triplets. Let me think. Let me consider positive factors. Since 33 can be factored into three distinct integers? Well, 33 = 1 * 3 * 11. So (1, 3, 11), product is 33. Sum is 1 + 3 + 11 = 15. If we consider negative numbers, perhaps. Let me try. Suppose two negative numbers and one positive. Let me think. Let me take (-1, -3, 11). Their product is (-1)*(-3)*11 = 33. The sum is -1 -3 + 11 = 7. That's better. Similarly, (-1, -11, 3): sum is -1 -11 + 3 = -9. Absolute value 9. (-3, -11, 1): sum is -3 -11 + 1 = -13. Absolute value 13. Also, (-1, -3, 11), sum is 7. Alternatively, maybe other combinations. What about (-1, -1, 33)? But roots must be distinct, so can't have two -1's. Similarly, can't have duplicates. How about (-1, 1, -33)? Wait, product is (-1)(1)(-33) = 33. Sum is -1 + 1 - 33 = -33. Absolute value 33. Not better. How about (-1, 3, -11)? Product is (-1)(3)(-11) = 33. Sum is -1 + 3 -11 = -9. Absolute value 9. So the minimal |sum| for three roots is 7 or 9, so 7. Therefore |a_{n-1}| would be 2*7=14.

Wait, that's better than the two roots case. So for three roots, we have |sum| as 7, which gives |a_{n-1}| = 14. So that's better than the two roots case which was 14? Wait no, the two roots case gave |sum| of 14, so |a_{n-1}| 28, but here for three roots, |sum| is 7, so |a_{n-1}| is 14. So better. So 14 is better.

But wait, maybe even better with four roots? Let me check.

Case 4: Four roots. Product is 33. Since product is positive, number of negative roots must be even (0, 2, or 4). Let me see. Let me try to find four distinct integers with product 33. Since 33 factors into 1, 3, 11, 33. To have four distinct integers, we need to include 1, -1, 3, -3, 11, -11, etc. But their product must be 33.

Wait, for example, if we take 1, -1, 3, -11. Product is (1)(-1)(3)(-11) = 33. Yes. Sum is 1 -1 + 3 -11 = (1 -1) + (3 -11) = 0 - 8 = -8. Absolute value 8. So |sum| is 8. Therefore |a_{n-1}| would be 2*8=16. Wait, that's worse than the previous 14. Hmm.

Alternatively, maybe another combination. Let's see. What if we take 1, 3, -1, -11. Same thing. Alternatively, maybe 1, -1, 11, -3. Same sum.

Alternatively, perhaps include more numbers. Let's see. Let me think. To get four distinct integers, maybe include 1, -1, 3, -11. But maybe another way. If we take 1, -1, 3, 11. Product is 1*(-1)*3*11 = -33. Not 33. So need even number of negatives. If we take 1, -1, -3, -11. Product is 1*(-1)*(-3)*(-11) = -33. Not good.

Wait, need product 33. So two negative numbers. Let's try 1, -1, 3, -11 (product 33), sum -8. What about other combinations? For example, 1, 3, -1, -11. Same. Alternatively, 1, -3, -1, 11. Product is 1*(-3)*(-1)*11 = 33. Sum is 1 - 3 -1 + 11 = 8. Absolute value 8. So same as before.

Alternatively, can we have 1, -1, -3, 11? Product is 1*(-1)*(-3)*11 = 33. Sum is 1 -1 -3 + 11 = 8. Same.

Alternatively, maybe include 33? Let's try. For example, 1, -1, 3, -11 as before. If we take 1, -1, 33, ... wait, but if we have four roots, and one of them is 33, then the product would be 33 * ... but need to get 33. So for example, 1, -1, 33, 1/... Wait, but all roots must be integers. So if we have 33 as one root, then the product of the other three roots must be 1. Since 33*product=33 ⇒ product=1. Therefore, other three roots must multiply to 1. The only integer divisors of 1 are 1 and -1. So three distinct integers? But 1 and -1 are only two distinct integers. So you can't have three distinct integers multiplying to 1. Therefore, not possible. Similarly for other possibilities. So perhaps four roots can't have 33 or -33 as a root unless we have some 1s and -1s.

Therefore, maybe the four roots case gives |sum| of 8. So worse than three roots case.

What about five roots? Let me check. Product 33. Five distinct integers with product 33. Since 33 is small, we need to have some 1s, -1s, and other numbers. But the product is 33. Let's think. Since 33 has only two prime factors, 3 and 11, and we need five distinct integers. So we need to include 1s and -1s to make up the numbers.

For example, suppose we have 1, -1, 3, -1, 11. But wait, duplicate -1s. Not allowed. So need distinct integers. Let me try: 1, -1, 3, -3, 11? Wait, product is 1*(-1)*3*(-3)*11 = 99. Not 33.

Alternatively, 1, -1, 3, 11, and 1? No, duplicate 1. Not allowed.

Alternatively, 1, -1, 1, -1, 33? Again duplicates.

Alternatively, maybe some other combinations. Let me think. If we need five distinct integers, and their product is 33. Let's note that 33 is 3*11, so to have five distinct integers, we need to have 1, -1, and then... Well, if we have 1, -1, and need three more integers. Let me think. Let me try:

Let me think of 1, -1, 3, -11, and something. The product of these four numbers is 1*(-1)*3*(-11) = 33. So to get five numbers, we need another number which is 1. But duplicate 1 is not allowed. So can't.

Alternatively, maybe 1, -1, 3, -1, something. But duplicate -1. Not allowed.

Alternatively, maybe include 1, -1, 3, 11, and -1? No, duplicate again. It seems impossible to have five distinct integers whose product is 33. Because we need to have at least the numbers 1, -1, 3, 11, and then another number which would have to be 1 or -1 or something else, but that would duplicate. Therefore, five roots is impossible. Similarly for higher degrees. So, the maximum number of roots is four?

Wait, but what about if we have 1, -1, 3, -11, and then another root? Wait, if we have five roots, then the product is 1*(-1)*3*(-11)*x = 33x. We need 33x=33 ⇒ x=1. But 1 is already in the list. So duplicate.

Therefore, five roots is impossible. Therefore, the maximum number of roots is four.

But let's check for product -33. So if the product is -33. Then the number of negative roots must be odd. Let me check if that gives us a better |sum|.

So, first, product -33. Let me consider different cases.

Case 1: Single root. Then the root is -33. Sum is -33, |sum| = 33. Therefore |a_{n-1}| = 2*33=66.

Case 2: Two roots. Product is -33. So possible pairs: (1, -33), (-1, 33), (3, -11), (-3, 11). Their sums are 1 + (-33) = -32, -1 + 33 = 32, 3 + (-11) = -8, -3 + 11 = 8. So |sum| is 32, 32, 8, 8. So minimal is 8, so |a_{n-1}| is 2*8=16.

Case 3: Three roots. Product is -33. So need odd number of negative roots. So possible combinations. Let me think.

For example, one negative root and two positive roots. Let me see. Let's take (-1, 3, 11). Their product is -1*3*11 = -33. Sum is -1 + 3 + 11 = 13. Absolute value 13. Or (-3, 1, 11). Sum is -3 + 1 + 11 = 9. Or (-1, 1, 33). Sum is -1 + 1 + 33 = 33. Or (-1, 3, 11) sum 13, as above.

Alternatively, three negative roots. For example, (-1, -3, 11). Product is (-1)(-3)(11) = 33. Not -33. To get -33, need odd number of negatives. So three negatives would be product negative if odd number. Wait, three negatives: (-1)(-3)(-11) = -33. Yes. So that's a valid set. The roots are -1, -3, -11. Sum is -1 -3 -11 = -15, |sum| = 15.

Alternatively, another combination: (-1, -1, 33), but duplicates not allowed. So (-1, -3, 11) gives sum -15, (-1, -11, 3) gives sum -1 -11 + 3 = -9. Absolute value 9.

Wait, let me check (-1, -3, 11): product is (-1)(-3)(11)=33. Not -33. Oh, no, that's positive. Wait, to get product -33 with three roots, need an odd number of negatives. So three negatives would be (-1)(-3)(-11) = -33. So sum is -1 -3 -11 = -15. Absolute value 15. Or one negative and two positives. Let's try (-1, 3, 11): product -33, sum -1 + 3 + 11 = 13. Or (-1, 1, 33): sum -1 + 1 + 33 = 33. Or (-3, 1, 11): sum -3 + 1 + 11 = 9. Or (-11, 1, 3): same as above.

Alternatively, (-1, 3, -11): product is (-1)*3*(-11)=33. Not -33. So not valid.

So the minimal |sum| for three roots is 9. Therefore |a_{n-1}| would be 2*9=18. Which is worse than the two roots case of 16.

Case 4: Four roots. Product is -33. Need odd number of negative roots. Let's see. Let me think. Let's try to have three negative roots and one positive. Since 3 is odd. For example, (-1, -1, -3, 11). Wait, but duplicates. Not allowed. How about (-1, -3, -11, 1). Product is (-1)(-3)(-11)(1) = -33. Sum is -1 -3 -11 + 1 = -14. Absolute value 14.

Alternatively, (-1, -1, 3, 11) duplicates. No.

What about two negative roots and two positive roots? Wait, that would be even number of negatives, so product positive. But we need product -33. So need odd number of negative roots. So three or one.

Let me try three negatives and one positive. For example, (-1, -3, -11, 1). Sum -1 -3 -11 +1 = -14. Absolute value 14.

Alternatively, (-1, -3, -1, 11) duplicate -1. Not allowed.

Another possibility: (-1, -1, 3, 11) duplicate -1.

Alternatively, maybe one negative and three positives. Let's try. Let me take (-1, 1, 3, 11). Product is (-1)*1*3*11 = -33. Sum is -1 + 1 + 3 + 11 = 14. Absolute value 14. Same as before.

Alternatively, (-1, 1, 3, 11) sum 14. Or (-1, 1, 3, 11) same. Or (-1, 1, 11, 3). Same.

What if we take other factors? Like (-1, 1, 33, 1)? No, duplicate 1.

Alternatively, (-1, 3, 11, 1) same as above.

Alternatively, (-3, 1, 11, 1). Duplicate 1.

Alternatively, (-1, 3, 1, -11). Wait, product is (-1)(3)(1)(-11)=33. Not -33.

What if we take (-1, 3, -11, 1). Wait, two negatives: -1 and -11. So two negatives. Then product is positive? (-1)(-11)(3)(1)=33. Not -33. So need to have three negatives. So three negatives and one positive.

Another example: (-1, -3, -1, 11) duplicates. No good.

How about (-1, -3, 1, 11). Wait, that's two negatives, so product positive. Not good.

Alternatively, (-1, -3, -11, x). Product is (-1)(-3)(-11)x = -33x. We need this equal to -33. Therefore, -33x = -33 ⇒ x = 1. So roots are -1, -3, -11, 1. Which we already considered, sum -1 -3 -11 +1 = -14. Absolute value 14.

Alternatively, can we get lower |sum|? Let me think. Suppose we take (-1, -1, 3, 11) but duplicate. Not allowed. What about (-1, 1, -3, 11). Wait, two negatives: -1 and -3. Then product is (-1)(-3)(1)(11)=33. Not -33.

Alternatively, (-1, 1, 3, -11). Then product is (-1)(1)(3)(-11)=33. Not -33.

Alternatively, (-1, -1, -1, 33) duplicates. Not allowed.

So seems like with four roots, the minimal |sum| is 14, leading to |a_{n-1}|=28. Which is worse than two roots case of 16.

What about five roots? For product -33. Hmm. Let me see. Five distinct integers with product -33. Let me think. Again, similar to before, need to have duplicates? Maybe not. For example, include 1, -1, 3, -3, and 11. But product is 1*(-1)*3*(-3)*11 = 99. Not -33.

If I take 1, -1, 3, -1, 11, duplicate again. Not allowed.

Alternatively, 1, -1, 3, -11, and x. Product is 1*(-1)*3*(-11)*x = 33x. Need 33x = -33 ⇒ x = -1. But duplicate.

Alternatively, 1, -1, -3, 11, x. Product is 1*(-1)*(-3)*11*x = 33x. Set to -33: 33x=-33 ⇒ x=-1. Duplicate again.

Alternatively, 1, -1, 3, 11, x. Product is 1*(-1)*3*11*x = -33x. Set equal to -33: -33x = -33 ⇒ x=1. Duplicate.

So seems impossible. Therefore, five roots is not possible.

So for product -33, the minimal |sum| is 8 (from two roots case), leading to |a_{n-1}|=16.

So comparing all cases:

For product 33: minimal |a_{n-1}| is 14 (from three roots case: sum 7 or -7).

For product -33: minimal |a_{n-1}| is 16 (from two roots case: sum ±8).

Therefore, the overall minimal is 14. But wait, wait! Wait, but earlier when considering product 33 with three roots, we had |sum| of 7? Let me check again.

For product 33 with three roots: roots are -1, -3, 11. Wait, product is (-1)(-3)(11)=33. Sum is -1 -3 +11 = 7. So yes, |sum| is 7. So |a_{n-1}| is 2*7=14.

But let me check if there are other factorizations with even smaller |sum|. For example, four roots?

Wait, earlier for four roots, product 33, we had sum -8, which gives |a_{n-1}|=16. So that's worse. What about other combinations with three roots? Let me think. For example, (-1, 1, -33). Product is (-1)(1)(-33)=33. Sum is -1 + 1 -33 = -33. Absolute value 33. Worse.

Another three roots: (-1, 3, -11). Product is (-1)*3*(-11)=33. Sum is -1 + 3 -11 = -9. Absolute value 9. Worse than 7.

How about (1, 3, 11). Sum is 1 + 3 + 11 = 15. Worse.

How about (-1, -1, 33). Duplicate. Not allowed.

How about (-1, 1, 3, 11) but that's four roots.

So, the minimal |sum| for product 33 is 7. For product -33, the minimal |sum| is 8. Therefore, overall, the minimal |a_{n-1}| is 14.

But wait, hold on. Is there a case where n is higher? For example, with more roots? Let me think. Wait, for product 33, with more roots, maybe we can have more cancellation? For example, if we have more roots with 1 and -1.

But earlier, for four roots, with product 33, we had sum of -8. But perhaps with more roots?

Wait, if we have five roots? But earlier analysis shows that five roots is impossible. Because we need to have product 33 with five distinct integers.

Wait, but suppose we have 1, -1, 1, -1, 33. But duplicates again. Not allowed.

Alternatively, maybe include fractions? No, roots must be integers.

Alternatively, maybe other numbers. For example, 1, -1, 3, -3, and something. Let me check. Product of these four numbers is (1)(-1)(3)(-3)=9. So to get product 33, need another root x such that 9x=33 ⇒ x=33/9=11/3. Not integer. Not allowed.

Alternatively, 1, -1, 3, 11, and x. Product is 1*(-1)*3*11*x = -33x. We need this to be 33 ⇒ -33x = 33 ⇒ x = -1. But duplicate.

So, nope. So five roots not possible. Therefore, maximum is four roots.

So then, 14 seems to be the minimal. But let me check another case. What if the polynomial has more roots with some 1s and -1s and other factors? Wait, for example, suppose we take roots: -1, 1, 3, -11. Product is (-1)(1)(3)(-11)=33. Sum is -1 +1 +3 -11 = -8. Which we already considered.

Alternatively, other combinations. Let me think. What about including 1, -1, -3, 11. Product is 1*(-1)*(-3)*11 = 33. Sum is 1 -1 -3 +11 = 8. Same as before.

Is there a way to have sum closer to zero? Let me think. For example, if we have roots that cancel each other.

Suppose for product 33, with three roots. Let me think. Let me take 3, -1, -11. Sum is 3 -1 -11 = -9. Absolute value 9. Not better.

What about 1, -1, -33. Sum is -33. Worse.

How about 1, -3, -11. Sum is 1 -3 -11 = -13. Worse.

Alternatively, 3, -1, 11. Sum is 3 -1 +11=13.

Alternatively, maybe 3, 11, -1. Same thing.

Wait, so no. So the minimal |sum| for product 33 is 7, from -1, -3, 11.

Alternatively, other three roots? Let's see. For example, 33, -1, -1. Not allowed.

Or 3, -1, 11. Sum 13.

So seems 7 is the smallest.

But wait, let me check another possibility. Suppose we have more than three roots but with some cancellation. For example, four roots: 1, -1, 3, -11. Sum -8. If I add another root, say 1, but duplicate. Not allowed.

Alternatively, if I have roots: 1, -1, 3, -11, and another root. Wait, but then the product would be 1*(-1)*3*(-11)*x = 33x. To make product 33, need x=1. But duplicate.

Alternatively, if I take roots 1, -1, 3, -11, and then x. Suppose I take x= something else, but then product becomes 33x. If I want product to still be 33, then x=1, which is duplicate. Not allowed.

Therefore, seems impossible.

So, then the minimal |a_{n-1}| is 14? Wait, but hold on. Wait, the problem says "distinct integers". So in the case of three roots: -1, -3, 11. Are they distinct? Yes.

But let me check whether there are other factorizations with three roots which give a smaller |sum|. For example, suppose we take roots -1, 3, and -11. Wait, product is (-1)(3)(-11)=33. Sum is -1 + 3 -11 = -9. So |sum|=9. Worse than 7.

How about -1, 1, 33? Product is -1*1*33 = -33. Not 33.

Wait, so only the combination with -1, -3, 11 gives sum 7.

But let me think again: what if the polynomial has roots that include other factors? For example, 33 can also be factored with other numbers? Wait, the factors of 33 are limited to ±1, ±3, ±11, ±33. So no other integers. Therefore, the roots can only be among these numbers. Therefore, there are no other possibilities.

So seems like 14 is achievable. However, wait, let me check if there's another combination with four roots where the sum is smaller.

For example, suppose I take roots 1, -1, 3, -11, sum is -8. If I take roots -1, 1, -3, 11, same sum. But what if I take roots 1, -1, -3, 11? Wait, that's the same as before. Sum is 1 -1 -3 +11=8. What if I take 1, -1, 3, -11? Sum is 1 -1 +3 -11=-8.

Is there a way to have sum closer to zero? For example, if I have roots 1, -1, 1, -1, ... but duplicates not allowed.

Alternatively, maybe something like 1, -1, 1, something? No.

Alternatively, can I take roots like 1, -1, 3, -3? Product is (1)(-1)(3)(-3)=9. Not 33. If I include another root, say 11: then product is 9*11=99. No.

Alternatively, 1, -1, 3, -11, and something? But as before, need another root that is 1, duplicate.

So no.

Alternatively, what if I take roots 1, -1, 3, -11, and then another root x. Wait, but then product is 33x. To have 33x=33, x=1. Duplicate. Not allowed.

So no.

Therefore, seems like the minimal |sum| is 7 for product 33, leading to |a_{n-1}|=14.

But wait, hold on. Wait, what about considering the polynomial with higher degree? For example, if the polynomial is of degree 4, with roots like 1, -1, 3, -11, which sum to -8, which gives |a_{n-1}|=16. But if we have a polynomial of degree 3 with roots -1, -3, 11, sum is 7, leading to |a_{n-1}|=14. So 14 is smaller.

But let me check if the problem allows for multiple roots? Wait, no, the problem says the roots are distinct integers. So that is okay.

But wait, another thought: the polynomial could have roots that are not just the factors of 33, but maybe with some other numbers if we use multiplicities... Wait, no. Because the product of roots is 33 or -33, and since they are integers, they must divide 33. Therefore, they have to be among the divisors.

Therefore, I think that 14 is the minimal. But let me check again.

Wait, let me think again about the case for product -33. The minimal |sum| was 8. So for that case, |a_{n-1}| is 16. But for product 33, we have 14. So 14 is smaller.

But wait, there is another case: if we have roots that include some numbers outside of the factors of 33? Wait, but since the product of the roots is 33 or -33, and they are integers, each root must divide 33. So for example, if I have a root like 2, then 2 divides 33? No. 33 divided by 2 is not integer. Therefore, roots must divide 33, so they must be among ±1, ±3, ±11, ±33. Therefore, no other possibilities.

Therefore, I think the minimal is indeed 14. But let me check if there are other three roots combinations.

For example, suppose I take roots -1, 1, and -33. Product is (-1)(1)(-33) = 33. Sum is -1 + 1 - 33 = -33. |sum| is 33. Worse.

Or roots -1, -1, 33. Not allowed.

Or roots 3, -1, -11. Product is 3*(-1)*(-11)=33. Sum is 3 -1 -11 = -9. |sum| is 9. Worse than 7.

Or roots -3, -1, 11. Product (-3)(-1)(11)=33. Sum -3 -1 +11=7. Same as previous case.

So that's the same as -1, -3, 11. So same |sum|.

Alternatively, roots -1, -3, 11.

So that's the same. So the minimal is indeed 7.

Therefore, |a_{n-1}|=2*7=14.

But wait, let me check the problem statement again. It says the polynomial has integer coefficients, distinct integer roots, a_n=2, and a_0=66.

So in this case, the polynomial would be 2*(x - (-1))(x - (-3))(x - 11) = 2*(x + 1)(x + 3)(x - 11). Let me expand that to check if a_0 is indeed 66.

First, compute (x + 1)(x + 3)(x - 11). Let's compute step by step.

First multiply (x + 1)(x + 3) = x² + 4x + 3. Then multiply by (x - 11):

(x² + 4x + 3)(x - 11) = x³ -11x² +4x² -44x +3x -33 = x³ -7x² -41x -33.

Then multiply by 2: 2x³ -14x² -82x -66.

So the constant term is -66. But the problem says a₀=66. Wait! What?

Wait a_0 is the constant term, which is (-1)^n * product of roots * leading coefficient. Wait, but here, n=3. So according to the formula: a₀ = 2*(-1)^3*(product of roots).

Product of roots is (-1)*(-3)*11=33. Therefore, a₀ = 2*(-1)^3*(33) = 2*(-1)*33= -66.

But the problem says a₀=66. So this contradicts. So there's an error here.

Oh no! I forgot the sign! Wait, let me check again.

Earlier, I had that for the polynomial f(x) = 2*(x - r₁)(x - r₂)...(x - rₙ), the constant term is 2*(-1)^n*(r₁r₂...rₙ). Therefore, if the product of roots is 33, then a₀ = 2*(-1)^n * 33.

In the case where n is odd (three roots), a₀ = 2*(-1)^3 * 33 = -66. But the problem says a₀=66.

Therefore, this case does not satisfy a₀=66. Therefore, this is not acceptable. Therefore, my previous conclusion was wrong.

Oh no! So I need to make sure that the product of roots times (-1)^n times leading coefficient is 66.

So, given that the leading coefficient is 2, and the constant term is 66, then:

a₀ = 2*(-1)^n*(product of roots) = 66 ⇒ (-1)^n*(product of roots) = 33.

Therefore, if the product of roots is 33, then (-1)^n = 1 ⇒ n is even.

If the product of roots is -33, then (-1)^n*(-33) = 33 ⇒ (-1)^n = -1 ⇒ n is odd.

Therefore, I need to consider in my previous analysis whether the number of roots, n, is even or odd accordingly.

In the case of product of roots being 33, n must be even.

In the case of product of roots being -33, n must be odd.

This is crucial. So my earlier analysis had an error in not considering the parity of n.

So let me go back.

First, for the case where product of roots is 33, which requires n even. Then, the sum of roots is S = r₁ + r₂ + ... + rₙ. So then a_{n-1} = -2*S. Therefore, |a_{n-1}| = 2|S|.

Similarly, for product of roots being -33, which requires n odd. Then, same thing: |a_{n-1}| = 2|S|.

Therefore, I need to take this into account. Let me redo my previous analysis with this in mind.

Starting again:

Case 1: product of roots is 33, n even.

Therefore, need to find distinct integers roots with product 33 and n even.

Case 2: product of roots is -33, n odd.

Need to find distinct integers roots with product -33 and n odd.

Therefore, for example, the case with three roots (n odd) and product -33 is allowed. While the case with three roots and product 33 is not allowed (because n odd, which would make (-1)^n * product = -33 ≠ 33). Therefore, my previous mistake was not considering the parity.

Therefore, the case of three roots with product 33 is invalid because n must be even.

Therefore, we need to recast our analysis.

So let me start again.

First, let me consider Case 1: product of roots is 33, n even.

Possible factorizations of 33 into even number of distinct integers.

Possible even number of roots: n=2,4,...

For n=2: product is 33, so possible pairs as before.

Possible pairs: (1,33), (3,11), (-1,-33), (-3,-11).

Sum for these pairs: 1+33=34, 3+11=14, -1+(-33)=-34, -3 + (-11)=-14. Absolute value of sum is 34,14,34,14. So minimal |sum| is 14. Therefore, |a_{n-1}|=2*14=28.

For n=4: need four distinct integers with product 33.

As before, possible roots: 1, -1, 3, -11. Product is (1)(-1)(3)(-11)=33. Sum is 1 -1 + 3 -11 = -8. |sum| is 8. Therefore |a_{n-1}|=16.

Is there another combination with four distinct integers? For example, 1, -1, 11, -3. Same sum.

What about 1, -1, 3, -11: same as above.

Is there a way to have a smaller |sum| for n=4?

Suppose we take roots 1, -1, 3, -11, sum is -8. If we take other combinations? For example, 1, -1, -3, 11. Sum is 1 -1 -3 +11=8.

If we take roots 1, 3, -1, -11. Same sum.

Is there a combination with roots that sum to something closer to zero? For example, 1, -1, 3, -11: sum -8. What if we take roots 1, -1, 1, -33? But duplicate 1. Not allowed.

Alternatively, 3, -1, 11, -1. Duplicate -1. Not allowed.

So seems minimal |sum| is 8. Therefore |a_{n-1}|=16.

n=4 gives 16. n=2 gives 28. What about n=6? Let me check.

For n=6, need six distinct integers whose product is 33. Since 33 is small, need to include 1, -1, and maybe other numbers. But with six distinct integers, need to have some factors. Let me think. Suppose we take 1, -1, 3, -3, 11, -11. Their product is (1)(-1)(3)(-3)(11)(-11) = (1*-1)*(3*-3)*(11*-11) = (-1)*(-9)*(-121) = (-1)(1089) = -1089 ≠ 33.

Alternatively, maybe take some 1s and -1s and others. For example, 1, -1, 3, -1, 11, -11... but duplicates. Not allowed.

Alternatively, 1, -1, 3, -3, 1, -1... duplicates.

No way to get six distinct integers whose product is 33. Because the prime factors are only 3 and 11, and 33 has limited number of divisors. Therefore, n=6 is impossible. Similarly for higher n. Therefore, only n=2 and n=4 possible for product 33.

Case 1: minimal |a_{n-1}| is 16 (from n=4).

Now, Case 2: product of roots is -33, n odd.

Possible n=1,3,5,...

For n=1: only one root, which is -33. Sum is -33. |a_{n-1}|=2*33=66.

For n=3: three distinct integers with product -33.

Earlier, we found possible triplets. Let me re-examine with consideration of n odd (which it is, n=3).

Possible triplets:

For example:

(-1, 3, 11): product is -1*3*11=-33. Sum is -1 + 3 + 11 = 13. |sum|=13.

(-1, 1, 33): product is -1*1*33=-33. Sum is -1 +1 +33=33. |sum|=33.

(-1, -3, 11): product is (-1)(-3)(11)=33. Not -33.

(-1, -3, -11): product is (-1)(-3)(-11)=-33. Sum is -1 -3 -11=-15. |sum|=15.

(-3, 1, 11): product is -3*1*11=-33. Sum is -3 +1 +11=9. |sum|=9.

(-1, 3, -11): product is (-1)*3*(-11)=33. Not -33.

(-3, -1, 11): product is (-3)(-1)(11)=33. Not -33.

(-3, -1, -11): product is (-3)(-1)(-11)=-33. Sum is -3 -1 -11=-15. |sum|=15.

(1, -3, 11): same as above.

(1, -1, 33): sum 33.

(3, -1, -11): product 3*(-1)*(-11)=33. Not -33.

So the minimal |sum| for n=3 is 9 (from triplet -3,1,11). Therefore |a_{n-1}|=2*9=18.

What about n=5? Let me check if possible. Product -33, n=5 odd.

As before, five distinct integers with product -33. Let me think. Let me try to include 1, -1, 3, -3, and another root. Product is (1)(-1)(3)(-3)(x) = 9x. Set equal to -33: 9x = -33 ⇒ x = -11/3. Not integer.

Alternatively, include 1, -1, 3, 11, and x. Product is 1*(-1)*3*11*x = -33x. Set equal to -33 ⇒ -33x = -33 ⇒ x=1. Duplicate.

Or 1, -1, 11, -3, and x. Product is same as above.

Therefore, seems impossible. Similarly for other combinations, duplicates or non-integer roots.

Therefore, n=5 not possible.

Therefore, for Case 2, minimal |a_{n-1}| is 18 (from n=3).

Comparing Case 1 (product 33, n even) minimal |a_{n-1}| is 16 and Case 2 (product -33, n odd) minimal is 18.

Therefore, overall minimal is 16.

But wait, earlier I thought there was a case with product 33 and n=4 giving |a_{n-1}|=16. Let me check if that is valid.

For n=4, product of roots is 33, n even. Let me take roots 1, -1, 3, -11. Their product is 1*(-1)*3*(-11) = 33. n=4, even, okay.

Compute the constant term: a₀ = 2*(-1)^4*(product) = 2*1*33=66. Which is correct.

The sum of roots is 1 + (-1) + 3 + (-11) = 1 -1 +3 -11 = -8.

Therefore, a_{n-1} = -2*(sum of roots) = -2*(-8) = 16. Therefore |a_{n-1}|=16.

So that's valid.

Earlier, I had a mistake when I considered n=3 with product 33, which was invalid because n had to be even. Therefore, the minimal |a_{n-1}| is 16.

But wait, is there a way to get lower than 16? Let me check for n=4, can we have a smaller |sum|? For example, can we find four distinct integers with product 33 and smaller |sum|?

Let me think. For example, if we take roots 1, -1, -3, 11. Product is 1*(-1)*(-3)*11 = 33. Sum is 1 -1 -3 +11= 8. |sum|=8. Therefore, |a_{n-1}|=2*8=16. Same as previous.

Or take roots -1,1,3,-11: same thing.

Is there another set of four distinct integers with product 33 and smaller |sum|? Let me think.

Suppose we take 1, -1, 11, -3. Same as above.

What about 1, -1, 33, -1/... wait, no, roots have to be integers.

Alternatively, 3, -3, 1, -11. Sum is 3 -3 +1 -11 = -10. |sum|=10. Worse than 8.

Or 3, -1, 11, -1. Sum is 3 -1 +11 -1 = 12. Worse.

Alternatively, 11, -1, -1, 3. Duplicates.

No.

Alternatively, is there a way to have roots 1, -1,  something else? For example, 1, -1, 3, -11. Which we have.

Alternatively, 1, -1, 1, -33. Duplicate.

No.

What about including more numbers? For example, 1, -1, 3, -11, and another root? But with n=4, we can't.

Therefore, the minimal |sum| for n=4 is 8, leading to |a_{n-1}|=16.

Is there another case with n=2 and product 33? For example, roots (3,11), sum 14. |a_{n-1}|=28. Worse.

So, with Case 1 minimal is 16.

Case 2 minimal is 18.

Therefore, overall minimal is 16.

Therefore, the answer is 16.

But let me check once more. Is there another combination for product -33 with n odd (n=3) with smaller |sum|?

Earlier, we found a triplet with sum 9.

Wait, what about triplet (-1, 3, -11). Product is (-1)*3*(-11)=33. Not -33.

(-1, -3, 11). Product is (-1)(-3)(11)=33. Not -33.

(-1, -1, 33): duplicate.

(-1, 1, 33): sum 33, product -33.

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

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

(-3, -1, 11): product 33.

(-3, -1, -11): product -33. Sum -15.

(-1, -3, -11): product -33. Sum is -1 -3 -11 = -15.

Alternatively, is there another triplet? For example, (1, -3, 11). Sum is 1 -3 +11=9.

Or (1, -1, 33). Sum 1 -1 +33=33.

Or (3, -1, 11). Sum 3 -1 +11=13.

So, the minimal |sum| is indeed 9.

Therefore, |a_{n-1}|=2*9=18.

Therefore, comparing Case 1 (16) and Case 2 (18), the minimal is 16.

Therefore, the answer should be 16.

But wait, let me make sure that there isn't a Case 1 with n=6 or something else, but we saw that n=6 is not possible.

Alternatively, n=4 is the maximum possible for Case 1.

Therefore, the minimal |a_{n-1}| is 16.

But let me check if there is another Case 1 with n=4 and different roots.

For example, roots 1, -1, -3, 11. Product is 1*(-1)*(-3)*11=33. Sum is 1 -1 -3 +11 = 8. |a_{n-1}|=16.

Yes, as before.

Wait, but what if we take roots -1, 1, -3, 11? Same sum.

Therefore, conclusion: the minimal |a_{n-1}| is 16.

But hold on, earlier when I thought there was a case with |a_{n-1}|=14, that was invalid because the product of roots and parity didn't match.

Therefore, the correct answer is 16.

But to be thorough, let me check if there is a Case 2 with n=5 or something else. For example, with five roots.

Wait, previously, I thought it's impossible, but let me think again.

Suppose we take roots 1, -1, 3, -3, and x. Product is (1)(-1)(3)(-3)(x) = 9x. We need this equal to -33. So 9x=-33 ⇒ x=-11/3. Not integer.

Alternatively, roots 1, -1, 3, -11, x. Product is (1)(-1)(3)(-11)x = 33x. Need this equal to -33 ⇒ x = -1. Duplicate.

Therefore, no.

Alternatively, roots 1, -1, 11, -1, x. Duplicate again.

No.

Therefore, it's impossible.

Therefore, minimal is indeed 16.

Wait, but let me check for Case 1, if there's another set of four roots.

For example, if I take roots 1, -1, 11, -3. Sum is 1 -1 +11 -3=8. Same as before.

Is there a way to get sum 0? For example, roots 1, -1, 3, -3. Product is (1)(-1)(3)(-3)=9. Not 33.

If I take roots 1, -1, 3, -11. Product 33. Sum 8.

If I take roots 1, -1, 11, -3. Same.

Alternatively, roots 1, -1, something. No others.

Therefore, no.

Therefore, answer is 16.

But wait, let me think once again. The problem says the polynomial has "distinct integers" as roots. So in the case of four roots, they must be distinct. Which they are: 1, -1, 3, -11 are distinct.

Yes.

So then, the minimal |a_{n-1}| is 16.

However, I need to check whether there exists a polynomial with even degree and product 33 with smaller |sum|.

Wait, for n=4, we have |sum|=8, leading to |a_{n-1}|=16.

Is there a way to have n=4 with |sum| smaller than 8?

For example, if we take roots like 1, -1, 3, -11 (sum 8), or 1, -1, -3, 11 (sum 8).

What if we take roots like 1, -1, something else? Let me check.

Suppose I take roots 1, -1, x, y. Their product is 1*(-1)*x*y = -xy. Need -xy=33 ⇒ xy=-33.

So x and y are distinct integers, not 1 or -1, and different from each other.

So possible pairs (x,y): (3, -11), (-3,11), (11, -3), (-11,3), (33, -1), (-33,1), etc. But since x and y can't be 1 or -1, as already used.

So possible pairs (3, -11), (-3, 11), etc.

Therefore, sum would be 1 -1 + x + y = x + y.

So for x=3 and y=-11, sum is 3 + (-11)= -8.

For x=-3 and y=11, sum is -3 +11=8.

Therefore, same as before.

So no way to get smaller than 8.

Alternatively, other pairs? For example, x= 11 and y=-3. Same as before.

Therefore, minimal |sum| is 8. So |a_{n-1}|=16.

Therefore, I think this is indeed the minimal.

But let me check if there is another way to get product 33 with n=4 and different roots.

For example, using 33 as one of the roots.

Let me try. Suppose roots are 33, 1, -1, -1. But duplicate -1. Not allowed.

Or 33, 1, -1, something. Product is 33*1*(-1)*x = -33x. Need this equal to 33 ⇒ -33x=33 ⇒ x=-1. Duplicate.

So no.

Alternatively, 33, -1, 1, -1. Duplicate.

So no.

Therefore, no other combinations.

Therefore, indeed, the minimal is 16.

Therefore, the answer is \boxed{16}.

**Final Answer**
\boxed{16}

To solve the problem, we start with the given polynomial:

$$
f(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 = 2(x - r_1)(x - r_2)\cdots(x - r_n)
$$

Given that all roots $ r_1, r_2, ..., r_n $ are **distinct integers**, and the polynomial has **integer coefficients**, the expression is valid. We are told that:

- $ a_0 = 66 $
- $ a_n = 2 $

We aim to **minimize** the absolute value of the coefficient $ a_{n-1} $, which we determine from the expression of the polynomial.

---

### Step 1: Constant Term and Product of Roots

The constant term of the polynomial is:

$$
a_0 = 2 \cdot (-1)^n \cdot (r_1 r_2 \cdots r_n)
$$

Given $ a_0 = 66 $, we get:

$$
(-1)^n \cdot (r_1 r_2 \cdots r_n) = 33
$$

So the *product of the roots* (with sign) is 33. That is, either:

- $ \prod r_i = 33 $ and $ n $ is even, or
- $ \prod r_i = -33 $ and $ n $ is odd

Because the product of all roots is an integer, each root must divide 33. The divisors of 33 are:

$$
\pm1, \pm3, \pm11, \pm33
$$

We can only use these to construct the roots.

---

### Step 2: Objective

We need to find distinct integer roots such that their product is ±33 and the **absolute value of the coefficient $ a_{n-1} $** is minimized.

Recall that in a monic polynomial with root sum $ S = r_1 + r_2 + \cdots + r_n $, the coefficient of $ x^{n-1} $ is $ -S $. However, since our polynomial is not monic (it has leading coefficient 2), the coefficient $ a_{n-1} $ is:

$$
a_{n-1} = -2 \cdot (r_1 + r_2 + \cdots + r_n)
$$

Hence,

$$
|a_{n-1}| = 2 \cdot |\text{sum of roots}|
$$

So the problem reduces to **minimizing the absolute value of the sum of the roots**, subject to the constraints:

- All roots are distinct integers
- Product of roots is 33 if $ n $ is even, or -33 if $ n $ is odd

---

### Step 3: Try Possible Root Sets

We will consider sets of distinct integers whose product is ±33 and check which gives the **smallest absolute sum**.

#### Case 1: Product = 33, $ n $ even

This allows us to use even numbers of negative roots to preserve a positive product.

One such good set is:

- Roots: $ 1, -1, 3, -11 $ (product: $ 1 \cdot (-1) \cdot 3 \cdot (-11) = 33 $)
- Sum: $ 1 - 1 + 3 - 11 = -8 $
- Absolute sum: $ 8 $
- Hence, $ |a_{n-1}| = 2 \cdot 8 = 16 $

This is valid (distinct, correct product, even number of roots) and gives:

$$
f(x) = 2(x - 1)(x + 1)(x - 3)(x + 11)
$$

Expanding confirms that the constant term is $ 66 $, and this polynomial has distinct integer roots.

#### Case 2: Product = -33, $ n $ odd

Here, $ n $ is odd, and the product is negative, so an odd number of roots are negative.

Consider the roots: $ -3, 1, 11 $

- Product: $ (-3) \cdot 1 \cdot 11 = -33 $
- Sum: $ -3 + 1 + 11 = 9 $
- Absolute sum: $ 9 $
- Hence, $ |a_{n-1}| = 2 \cdot 9 = 18 $

This is also valid, and the sum is greater than in the previous case.

#### Other Cases

Trying any other possible root combinations with distinct integers from $ \{-33, -11, -3, -1, 1, 3, 11, 33\} $ and trying other combinations of even or odd number of roots results in **either duplicates or higher absolute sum**.

Therefore, the **smallest possible value** of $ |a_{n-1}| $ is achieved with the root set $ \{1, -1, 3, -11\} $, giving:

$$
\boxed{16}
$$

[Assistant]
Explanation:
Okay, let me try to figure out this problem. So, we have a polynomial with integer coefficients, distinct integer roots. The leading coefficient is 2, and the constant term is 66. We need to find the least possible value of |a_{n-1}|.

First, I remember that for a polynomial with integer coefficients and distinct integer roots, we can express it as f(x) = 2(x - r₁)(x - r₂)...(x - rₙ), where each rᵢ is an integer, and all roots are distinct. Since the leading coefficient is 2, that's why we have that 2 in front.

The constant term is given by a₀ = 66. The constant term of a polynomial is the product of the roots multiplied by the leading coefficient, but with a sign depending on the degree. Wait, more precisely, for a monic polynomial (leading coefficient 1), the constant term is (-1)^n * product of roots. But here the leading coefficient is 2, so the constant term would be 2 * (-1)^n * product of roots. Wait, let me check.

If we have f(x) = a_n(x - r₁)(x - r₂)...(x - rₙ), then expanding this, the constant term is a_n * (-1)^n * r₁r₂...rₙ. Therefore, in our case, since a_n = 2, the constant term a₀ = 2 * (-1)^n * (product of roots). But the problem says a₀ = 66, so:

2 * (-1)^n * (r₁r₂...rₙ) = 66 ⇒ (-1)^n * (r₁r₂...rₙ) = 33.

Therefore, the product of the roots is 33 or -33, depending on whether n is even or odd. Since 33 is positive, if n is even, then the product of roots is 33; if n is odd, the product of roots is -33.

But since all roots are distinct integers, we need to find integers r₁, r₂,..., rₙ such that their product is either 33 or -33, and then find the polynomial with leading coefficient 2 and find the minimal |a_{n-1}|.

Wait, but how does a_{n-1} relate to the roots? Let me recall. The coefficient a_{n-1} is the coefficient of x^{n-1} in the polynomial. For a monic polynomial, the coefficient of x^{n-1} is - (sum of roots). But in our case, the polynomial is 2*(x - r₁)...(x - rₙ). Therefore, expanding this, we get:

f(x) = 2x^n - 2*(r₁ + r₂ + ... + rₙ)x^{n-1} + ...

Therefore, the coefficient a_{n-1} is -2*(sum of roots). Wait, so a_{n-1} = -2*(sum of roots). Therefore, |a_{n-1}| = 2*|sum of roots|. Therefore, to minimize |a_{n-1}|, we need to minimize |sum of roots|.

Therefore, our problem reduces to finding distinct integers r₁, r₂,..., rₙ such that their product is 33 or -33 (depending on n's parity), and the absolute value of their sum is as small as possible, then multiply by 2 to get |a_{n-1}|.

But since we can choose the number of roots (n) as long as the roots are distinct integers, we need to consider different possibilities for n. However, the problem says the polynomial has degree n, so n can be any integer ≥ 1, but given that the roots are distinct integers and the product is 33 or -33, the number of roots can't be too large.

But since 33 factors into primes as 3 * 11, so the prime factors are 3 and 11. Also, the product can be positive or negative. Let me consider possible factorizations.

First, note that since the roots are integers, they must divide 33 or -33. The divisors of 33 are ±1, ±3, ±11, ±33. Therefore, the roots must be among these numbers. Since the roots are distinct integers, we can't repeat any of these.

So, to find possible sets of distinct integers whose product is 33 or -33. Also, note that if n is the degree, then n can be 1, 2, 3, ..., up to... Well, since the product of roots is 33 or -33, and each root is distinct, so the maximum number of roots would be the number of distinct divisors. Since there are 8 divisors (±1, ±3, ±11, ±33), but we need distinct ones. However, we need to consider that if we take more roots, we might need to include 1 or -1, but we also have to include other roots.

But since the product is fixed (±33), the more roots we have, the smaller the absolute value of each root would need to be, but they have to multiply to 33 or -33. For example, if we take more roots, we have to include 1 or -1, but then the other roots would have to be factors of 33 divided by those. However, we need to ensure that all roots are distinct integers. Let me think.

But since we need to minimize |sum of roots|, maybe having more roots can help? Because if we have more roots, maybe we can have some positive and negative numbers so that their sum is small. Let me try.

First, let me think about possible factorizations of 33 and -33 into distinct integers.

Case 1: The product is 33. Then the roots are integers whose product is 33. Similarly for -33.

First, for positive product 33:

Possible factorizations with distinct integers. Since 33 is positive, the number of negative roots must be even. Because negative times negative is positive. So, if we have an even number of negative roots, the product is positive.

Similarly, if the product is -33, the number of negative roots must be odd.

Let me first consider possible sets of roots for both cases.

Starting with product 33:

Possible combinations:

Option 1: Single root: 33. Then the polynomial is 2*(x - 33). Then sum of roots is 33, |a_{n-1}| = 2*33=66.

Option 2: Two roots: product 33. Since 33 factors as 1*33, 3*11, (-1)*(-33), (-3)*(-11). But since roots must be distinct integers, so possible pairs:

(1, 33), (3, 11), (-1, -33), (-3, -11). So sum of roots for (1, 33) is 34, for (3, 11) is 14, for (-1, -33) is -34, for (-3, -11) is -14. Therefore |sum| is 34, 14, 34, 14. So minimal |sum| is 14. Then |a_{n-1}| would be 2*14=28.

Option 3: Three roots. Product 33. Let's see. Since 33 is positive, need even number of negative numbers. Wait, if there are three roots, which is odd, so to have even number of negative roots? No, no. Wait, the product is positive, so number of negative roots must be even. But three roots can have 0 or 2 negative roots. Let's try.

Possible triplets. Let me think. Let me consider positive factors. Since 33 can be factored into three distinct integers? Well, 33 = 1 * 3 * 11. So (1, 3, 11), product is 33. Sum is 1 + 3 + 11 = 15. If we consider negative numbers, perhaps. Let me try. Suppose two negative numbers and one positive. Let me think. Let me take (-1, -3, 11). Their product is (-1)*(-3)*11 = 33. The sum is -1 -3 + 11 = 7. That's better. Similarly, (-1, -11, 3): sum is -1 -11 + 3 = -9. Absolute value 9. (-3, -11, 1): sum is -3 -11 + 1 = -13. Absolute value 13. Also, (-1, -3, 11), sum is 7. Alternatively, maybe other combinations. What about (-1, -1, 33)? But roots must be distinct, so can't have two -1's. Similarly, can't have duplicates. How about (-1, 1, -33)? Wait, product is (-1)(1)(-33) = 33. Sum is -1 + 1 - 33 = -33. Absolute value 33. Not better. How about (-1, 3, -11)? Product is (-1)(3)(-11) = 33. Sum is -1 + 3 -11 = -9. Absolute value 9. So the minimal |sum| for three roots is 7 or 9, so 7. Therefore |a_{n-1}| would be 2*7=14.

Wait, that's better than the two roots case. So for three roots, we have |sum| as 7, which gives |a_{n-1}| = 14. So that's better than the two roots case which was 14? Wait no, the two roots case gave |sum| of 14, so |a_{n-1}| 28, but here for three roots, |sum| is 7, so |a_{n-1}| is 14. So better. So 14 is better.

But wait, maybe even better with four roots? Let me check.

Case 4: Four roots. Product is 33. Since product is positive, number of negative roots must be even (0, 2, or 4). Let me see. Let me try to find four distinct integers with product 33. Since 33 factors into 1, 3, 11, 33. To have four distinct integers, we need to include 1, -1, 3, -3, 11, -11, etc. But their product must be 33.

Wait, for example, if we take 1, -1, 3, -11. Product is (1)(-1)(3)(-11) = 33. Yes. Sum is 1 -1 + 3 -11 = (1 -1) + (3 -11) = 0 - 8 = -8. Absolute value 8. So |sum| is 8. Therefore |a_{n-1}| would be 2*8=16. Wait, that's worse than the previous 14. Hmm.

Alternatively, maybe another combination. Let's see. What if we take 1, 3, -1, -11. Same thing. Alternatively, maybe 1, -1, 11, -3. Same sum.

Alternatively, perhaps include more numbers. Let's see. Let me think. To get four distinct integers, maybe include 1, -1, 3, -11. But maybe another way. If we take 1, -1, 3, 11. Product is 1*(-1)*3*11 = -33. Not 33. So need even number of negatives. If we take 1, -1, -3, -11. Product is 1*(-1)*(-3)*(-11) = -33. Not good.

Wait, need product 33. So two negative numbers. Let's try 1, -1, 3, -11 (product 33), sum -8. What about other combinations? For example, 1, 3, -1, -11. Same. Alternatively, 1, -3, -1, 11. Product is 1*(-3)*(-1)*11 = 33. Sum is 1 - 3 -1 + 11 = 8. Absolute value 8. So same as before.

Alternatively, can we have 1, -1, -3, 11? Product is 1*(-1)*(-3)*11 = 33. Sum is 1 -1 -3 + 11 = 8. Same.

Alternatively, maybe include 33? Let's try. For example, 1, -1, 3, -11 as before. If we take 1, -1, 33, ... wait, but if we have four roots, and one of them is 33, then the product would be 33 * ... but need to get 33. So for example, 1, -1, 33, 1/... Wait, but all roots must be integers. So if we have 33 as one root, then the product of the other three roots must be 1. Since 33*product=33 ⇒ product=1. Therefore, other three roots must multiply to 1. The only integer divisors of 1 are 1 and -1. So three distinct integers? But 1 and -1 are only two distinct integers. So you can't have three distinct integers multiplying to 1. Therefore, not possible. Similarly for other possibilities. So perhaps four roots can't have 33 or -33 as a root unless we have some 1s and -1s.

Therefore, maybe the four roots case gives |sum| of 8. So worse than three roots case.

What about five roots? Let me check. Product 33. Five distinct integers with product 33. Since 33 is small, we need to have some 1s, -1s, and other numbers. But the product is 33. Let's think. Since 33 has only two prime factors, 3 and 11, and we need five distinct integers. So we need to include 1s and -1s to make up the numbers.

For example, suppose we have 1, -1, 3, -1, 11. But wait, duplicate -1s. Not allowed. So need distinct integers. Let me try: 1, -1, 3, -3, 11? Wait, product is 1*(-1)*3*(-3)*11 = 99. Not 33.

Alternatively, 1, -1, 3, 11, and 1? No, duplicate 1. Not allowed.

Alternatively, 1, -1, 1, -1, 33? Again duplicates.

Alternatively, maybe some other combinations. Let me think. If we need five distinct integers, and their product is 33. Let's note that 33 is 3*11, so to have five distinct integers, we need to have 1, -1, and then... Well, if we have 1, -1, and need three more integers. Let me think. Let me try:

Let me think of 1, -1, 3, -11, and something. The product of these four numbers is 1*(-1)*3*(-11) = 33. So to get five numbers, we need another number which is 1. But duplicate 1 is not allowed. So can't.

Alternatively, maybe 1, -1, 3, -1, something. But duplicate -1. Not allowed.

Alternatively, maybe include 1, -1, 3, 11, and -1? No, duplicate again. It seems impossible to have five distinct integers whose product is 33. Because we need to have at least the numbers 1, -1, 3, 11, and then another number which would have to be 1 or -1 or something else, but that would duplicate. Therefore, five roots is impossible. Similarly for higher degrees. So, the maximum number of roots is four?

Wait, but what about if we have 1, -1, 3, -11, and then another root? Wait, if we have five roots, then the product is 1*(-1)*3*(-11)*x = 33x. We need 33x=33 ⇒ x=1. But 1 is already in the list. So duplicate.

Therefore, five roots is impossible. Therefore, the maximum number of roots is four.

But let's check for product -33. So if the product is -33. Then the number of negative roots must be odd. Let me check if that gives us a better |sum|.

So, first, product -33. Let me consider different cases.

Case 1: Single root. Then the root is -33. Sum is -33, |sum| = 33. Therefore |a_{n-1}| = 2*33=66.

Case 2: Two roots. Product is -33. So possible pairs: (1, -33), (-1, 33), (3, -11), (-3, 11). Their sums are 1 + (-33) = -32, -1 + 33 = 32, 3 + (-11) = -8, -3 + 11 = 8. So |sum| is 32, 32, 8, 8. So minimal is 8, so |a_{n-1}| is 2*8=16.

Case 3: Three roots. Product is -33. So need odd number of negative roots. So possible combinations. Let me think.

For example, one negative root and two positive roots. Let me see. Let's take (-1, 3, 11). Their product is -1*3*11 = -33. Sum is -1 + 3 + 11 = 13. Absolute value 13. Or (-3, 1, 11). Sum is -3 + 1 + 11 = 9. Or (-1, 1, 33). Sum is -1 + 1 + 33 = 33. Or (-1, 3, 11) sum 13, as above.

Alternatively, three negative roots. For example, (-1, -3, 11). Product is (-1)(-3)(11) = 33. Not -33. To get -33, need odd number of negatives. So three negatives would be product negative if odd number. Wait, three negatives: (-1)(-3)(-11) = -33. Yes. So that's a valid set. The roots are -1, -3, -11. Sum is -1 -3 -11 = -15, |sum| = 15.

Alternatively, another combination: (-1, -1, 33), but duplicates not allowed. So (-1, -3, 11) gives sum -15, (-1, -11, 3) gives sum -1 -11 + 3 = -9. Absolute value 9.

Wait, let me check (-1, -3, 11): product is (-1)(-3)(11)=33. Not -33. Oh, no, that's positive. Wait, to get product -33 with three roots, need an odd number of negatives. So three negatives would be (-1)(-3)(-11) = -33. So sum is -1 -3 -11 = -15. Absolute value 15. Or one negative and two positives. Let's try (-1, 3, 11): product -33, sum -1 + 3 + 11 = 13. Or (-1, 1, 33): sum -1 + 1 + 33 = 33. Or (-3, 1, 11): sum -3 + 1 + 11 = 9. Or (-11, 1, 3): same as above.

Alternatively, (-1, 3, -11): product is (-1)*3*(-11)=33. Not -33. So not valid.

So the minimal |sum| for three roots is 9. Therefore |a_{n-1}| would be 2*9=18. Which is worse than the two roots case of 16.

Case 4: Four roots. Product is -33. Need odd number of negative roots. Let's see. Let me think. Let's try to have three negative roots and one positive. Since 3 is odd. For example, (-1, -1, -3, 11). Wait, but duplicates. Not allowed. How about (-1, -3, -11, 1). Product is (-1)(-3)(-11)(1) = -33. Sum is -1 -3 -11 + 1 = -14. Absolute value 14.

Alternatively, (-1, -1, 3, 11) duplicates. No.

What about two negative roots and two positive roots? Wait, that would be even number of negatives, so product positive. But we need product -33. So need odd number of negative roots. So three or one.

Let me try three negatives and one positive. For example, (-1, -3, -11, 1). Sum -1 -3 -11 +1 = -14. Absolute value 14.

Alternatively, (-1, -3, -1, 11) duplicate -1. Not allowed.

Another possibility: (-1, -1, 3, 11) duplicate -1.

Alternatively, maybe one negative and three positives. Let's try. Let me take (-1, 1, 3, 11). Product is (-1)*1*3*11 = -33. Sum is -1 + 1 + 3 + 11 = 14. Absolute value 14. Same as before.

Alternatively, (-1, 1, 3, 11) sum 14. Or (-1, 1, 3, 11) same. Or (-1, 1, 11, 3). Same.

What if we take other factors? Like (-1, 1, 33, 1)? No, duplicate 1.

Alternatively, (-1, 3, 11, 1) same as above.

Alternatively, (-3, 1, 11, 1). Duplicate 1.

Alternatively, (-1, 3, 1, -11). Wait, product is (-1)(3)(1)(-11)=33. Not -33.

What if we take (-1, 3, -11, 1). Wait, two negatives: -1 and -11. So two negatives. Then product is positive? (-1)(-11)(3)(1)=33. Not -33. So need to have three negatives. So three negatives and one positive.

Another example: (-1, -3, -1, 11) duplicates. No good.

How about (-1, -3, 1, 11). Wait, that's two negatives, so product positive. Not good.

Alternatively, (-1, -3, -11, x). Product is (-1)(-3)(-11)x = -33x. We need this equal to -33. Therefore, -33x = -33 ⇒ x = 1. So roots are -1, -3, -11, 1. Which we already considered, sum -1 -3 -11 +1 = -14. Absolute value 14.

Alternatively, can we get lower |sum|? Let me think. Suppose we take (-1, -1, 3, 11) but duplicate. Not allowed. What about (-1, 1, -3, 11). Wait, two negatives: -1 and -3. Then product is (-1)(-3)(1)(11)=33. Not -33.

Alternatively, (-1, 1, 3, -11). Then product is (-1)(1)(3)(-11)=33. Not -33.

Alternatively, (-1, -1, -1, 33) duplicates. Not allowed.

So seems like with four roots, the minimal |sum| is 14, leading to |a_{n-1}|=28. Which is worse than two roots case of 16.

What about five roots? For product -33. Hmm. Let me see. Five distinct integers with product -33. Let me think. Again, similar to before, need to have duplicates? Maybe not. For example, include 1, -1, 3, -3, and 11. But product is 1*(-1)*3*(-3)*11 = 99. Not -33.

If I take 1, -1, 3, -1, 11, duplicate again. Not allowed.

Alternatively, 1, -1, 3, -11, and x. Product is 1*(-1)*3*(-11)*x = 33x. Need 33x = -33 ⇒ x = -1. But duplicate.

Alternatively, 1, -1, -3, 11, x. Product is 1*(-1)*(-3)*11*x = 33x. Set to -33: 33x=-33 ⇒ x=-1. Duplicate again.

Alternatively, 1, -1, 3, 11, x. Product is 1*(-1)*3*11*x = -33x. Set equal to -33: -33x = -33 ⇒ x=1. Duplicate.

So seems impossible. Therefore, five roots is not possible.

So for product -33, the minimal |sum| is 8 (from two roots case), leading to |a_{n-1}|=16.

So comparing all cases:

For product 33: minimal |a_{n-1}| is 14 (from three roots case: sum 7 or -7).

For product -33: minimal |a_{n-1}| is 16 (from two roots case: sum ±8).

Therefore, the overall minimal is 14. But wait, wait! Wait, but earlier when considering product 33 with three roots, we had |sum| of 7? Let me check again.

For product 33 with three roots: roots are -1, -3, 11. Wait, product is (-1)(-3)(11)=33. Sum is -1 -3 +11 = 7. So yes, |sum| is 7. So |a_{n-1}| is 2*7=14.

But let me check if there are other factorizations with even smaller |sum|. For example, four roots?

Wait, earlier for four roots, product 33, we had sum -8, which gives |a_{n-1}|=16. So that's worse. What about other combinations with three roots? Let me think. For example, (-1, 1, -33). Product is (-1)(1)(-33)=33. Sum is -1 + 1 -33 = -33. Absolute value 33. Worse.

Another three roots: (-1, 3, -11). Product is (-1)*3*(-11)=33. Sum is -1 + 3 -11 = -9. Absolute value 9. Worse than 7.

How about (1, 3, 11). Sum is 1 + 3 + 11 = 15. Worse.

How about (-1, -1, 33). Duplicate. Not allowed.

How about (-1, 1, 3, 11) but that's four roots.

So, the minimal |sum| for product 33 is 7. For product -33, the minimal |sum| is 8. Therefore, overall, the minimal |a_{n-1}| is 14.

But wait, hold on. Is there a case where n is higher? For example, with more roots? Let me think. Wait, for product 33, with more roots, maybe we can have more cancellation? For example, if we have more roots with 1 and -1.

But earlier, for four roots, with product 33, we had sum of -8. But perhaps with more roots?

Wait, if we have five roots? But earlier analysis shows that five roots is impossible. Because we need to have product 33 with five distinct integers.

Wait, but suppose we have 1, -1, 1, -1, 33. But duplicates again. Not allowed.

Alternatively, maybe include fractions? No, roots must be integers.

Alternatively, maybe other numbers. For example, 1, -1, 3, -3, and something. Let me check. Product of these four numbers is (1)(-1)(3)(-3)=9. So to get product 33, need another root x such that 9x=33 ⇒ x=33/9=11/3. Not integer. Not allowed.

Alternatively, 1, -1, 3, 11, and x. Product is 1*(-1)*3*11*x = -33x. We need this to be 33 ⇒ -33x = 33 ⇒ x = -1. But duplicate.

So, nope. So five roots not possible. Therefore, maximum is four roots.

So then, 14 seems to be the minimal. But let me check another case. What if the polynomial has more roots with some 1s and -1s and other factors? Wait, for example, suppose we take roots: -1, 1, 3, -11. Product is (-1)(1)(3)(-11)=33. Sum is -1 +1 +3 -11 = -8. Which we already considered.

Alternatively, other combinations. Let me think. What about including 1, -1, -3, 11. Product is 1*(-1)*(-3)*11 = 33. Sum is 1 -1 -3 +11 = 8. Same as before.

Is there a way to have sum closer to zero? Let me think. For example, if we have roots that cancel each other.

Suppose for product 33, with three roots. Let me think. Let me take 3, -1, -11. Sum is 3 -1 -11 = -9. Absolute value 9. Not better.

What about 1, -1, -33. Sum is -33. Worse.

How about 1, -3, -11. Sum is 1 -3 -11 = -13. Worse.

Alternatively, 3, -1, 11. Sum is 3 -1 +11=13.

Alternatively, maybe 3, 11, -1. Same thing.

Wait, so no. So the minimal |sum| for product 33 is 7, from -1, -3, 11.

Alternatively, other three roots? Let's see. For example, 33, -1, -1. Not allowed.

Or 3, -1, 11. Sum 13.

So seems 7 is the smallest.

But wait, let me check another possibility. Suppose we have more than three roots but with some cancellation. For example, four roots: 1, -1, 3, -11. Sum -8. If I add another root, say 1, but duplicate. Not allowed.

Alternatively, if I have roots: 1, -1, 3, -11, and another root. Wait, but then the product would be 1*(-1)*3*(-11)*x = 33x. To make product 33, need x=1. But duplicate.

Alternatively, if I take roots 1, -1, 3, -11, and then x. Suppose I take x= something else, but then product becomes 33x. If I want product to still be 33, then x=1, which is duplicate. Not allowed.

Therefore, seems impossible.

So, then the minimal |a_{n-1}| is 14? Wait, but hold on. Wait, the problem says "distinct integers". So in the case of three roots: -1, -3, 11. Are they distinct? Yes.

But let me check whether there are other factorizations with three roots which give a smaller |sum|. For example, suppose we take roots -1, 3, and -11. Wait, product is (-1)(3)(-11)=33. Sum is -1 + 3 -11 = -9. So |sum|=9. Worse than 7.

How about -1, 1, 33? Product is -1*1*33 = -33. Not 33.

Wait, so only the combination with -1, -3, 11 gives sum 7.

But let me think again: what if the polynomial has roots that include other factors? For example, 33 can also be factored with other numbers? Wait, the factors of 33 are limited to ±1, ±3, ±11, ±33. So no other integers. Therefore, the roots can only be among these numbers. Therefore, there are no other possibilities.

So seems like 14 is achievable. However, wait, let me check if there's another combination with four roots where the sum is smaller.

For example, suppose I take roots 1, -1, 3, -11, sum is -8. If I take roots -1, 1, -3, 11, same sum. But what if I take roots 1, -1, -3, 11? Wait, that's the same as before. Sum is 1 -1 -3 +11=8. What if I take 1, -1, 3, -11? Sum is 1 -1 +3 -11=-8.

Is there a way to have sum closer to zero? For example, if I have roots 1, -1, 1, -1, ... but duplicates not allowed.

Alternatively, maybe something like 1, -1, 1, something? No.

Alternatively, can I take roots like 1, -1, 3, -3? Product is (1)(-1)(3)(-3)=9. Not 33. If I include another root, say 11: then product is 9*11=99. No.

Alternatively, 1, -1, 3, -11, and something? But as before, need another root that is 1, duplicate.

So no.

Alternatively, what if I take roots 1, -1, 3, -11, and then another root x. Wait, but then product is 33x. To have 33x=33, x=1. Duplicate. Not allowed.

So no.

Therefore, seems like the minimal |sum| is 7 for product 33, leading to |a_{n-1}|=14.

But wait, hold on. Wait, what about considering the polynomial with higher degree? For example, if the polynomial is of degree 4, with roots like 1, -1, 3, -11, which sum to -8, which gives |a_{n-1}|=16. But if we have a polynomial of degree 3 with roots -1, -3, 11, sum is 7, leading to |a_{n-1}|=14. So 14 is smaller.

But let me check if the problem allows for multiple roots? Wait, no, the problem says the roots are distinct integers. So that is okay.

But wait, another thought: the polynomial could have roots that are not just the factors of 33, but maybe with some other numbers if we use multiplicities... Wait, no. Because the product of roots is 33 or -33, and since they are integers, they must divide 33. Therefore, they have to be among the divisors.

Therefore, I think that 14 is the minimal. But let me check again.

Wait, let me think again about the case for product -33. The minimal |sum| was 8. So for that case, |a_{n-1}| is 16. But for product 33, we have 14. So 14 is smaller.

But wait, there is another case: if we have roots that include some numbers outside of the factors of 33? Wait, but since the product of the roots is 33 or -33, and they are integers, each root must divide 33. So for example, if I have a root like 2, then 2 divides 33? No. 33 divided by 2 is not integer. Therefore, roots must divide 33, so they must be among ±1, ±3, ±11, ±33. Therefore, no other possibilities.

Therefore, I think the minimal is indeed 14. But let me check if there are other three roots combinations.

For example, suppose I take roots -1, 1, and -33. Product is (-1)(1)(-33) = 33. Sum is -1 + 1 - 33 = -33. |sum| is 33. Worse.

Or roots -1, -1, 33. Not allowed.

Or roots 3, -1, -11. Product is 3*(-1)*(-11)=33. Sum is 3 -1 -11 = -9. |sum| is 9. Worse than 7.

Or roots -3, -1, 11. Product (-3)(-1)(11)=33. Sum -3 -1 +11=7. Same as previous case.

So that's the same as -1, -3, 11. So same |sum|.

Alternatively, roots -1, -3, 11.

So that's the same. So the minimal is indeed 7.

Therefore, |a_{n-1}|=2*7=14.

But wait, let me check the problem statement again. It says the polynomial has integer coefficients, distinct integer roots, a_n=2, and a_0=66.

So in this case, the polynomial would be 2*(x - (-1))(x - (-3))(x - 11) = 2*(x + 1)(x + 3)(x - 11). Let me expand that to check if a_0 is indeed 66.

First, compute (x + 1)(x + 3)(x - 11). Let's compute step by step.

First multiply (x + 1)(x + 3) = x² + 4x + 3. Then multiply by (x - 11):

(x² + 4x + 3)(x - 11) = x³ -11x² +4x² -44x +3x -33 = x³ -7x² -41x -33.

Then multiply by 2: 2x³ -14x² -82x -66.

So the constant term is -66. But the problem says a₀=66. Wait! What?

Wait a_0 is the constant term, which is (-1)^n * product of roots * leading coefficient. Wait, but here, n=3. So according to the formula: a₀ = 2*(-1)^3*(product of roots).

Product of roots is (-1)*(-3)*11=33. Therefore, a₀ = 2*(-1)^3*(33) = 2*(-1)*33= -66.

But the problem says a₀=66. So this contradicts. So there's an error here.

Oh no! I forgot the sign! Wait, let me check again.

Earlier, I had that for the polynomial f(x) = 2*(x - r₁)(x - r₂)...(x - rₙ), the constant term is 2*(-1)^n*(r₁r₂...rₙ). Therefore, if the product of roots is 33, then a₀ = 2*(-1)^n * 33.

In the case where n is odd (three roots), a₀ = 2*(-1)^3 * 33 = -66. But the problem says a₀=66.

Therefore, this case does not satisfy a₀=66. Therefore, this is not acceptable. Therefore, my previous conclusion was wrong.

Oh no! So I need to make sure that the product of roots times (-1)^n times leading coefficient is 66.

So, given that the leading coefficient is 2, and the constant term is 66, then:

a₀ = 2*(-1)^n*(product of roots) = 66 ⇒ (-1)^n*(product of roots) = 33.

Therefore, if the product of roots is 33, then (-1)^n = 1 ⇒ n is even.

If the product of roots is -33, then (-1)^n*(-33) = 33 ⇒ (-1)^n = -1 ⇒ n is odd.

Therefore, I need to consider in my previous analysis whether the number of roots, n, is even or odd accordingly.

In the case of product of roots being 33, n must be even.

In the case of product of roots being -33, n must be odd.

This is crucial. So my earlier analysis had an error in not considering the parity of n.

So let me go back.

First, for the case where product of roots is 33, which requires n even. Then, the sum of roots is S = r₁ + r₂ + ... + rₙ. So then a_{n-1} = -2*S. Therefore, |a_{n-1}| = 2|S|.

Similarly, for product of roots being -33, which requires n odd. Then, same thing: |a_{n-1}| = 2|S|.

Therefore, I need to take this into account. Let me redo my previous analysis with this in mind.

Starting again:

Case 1: product of roots is 33, n even.

Therefore, need to find distinct integers roots with product 33 and n even.

Case 2: product of roots is -33, n odd.

Need to find distinct integers roots with product -33 and n odd.

Therefore, for example, the case with three roots (n odd) and product -33 is allowed. While the case with three roots and product 33 is not allowed (because n odd, which would make (-1)^n * product = -33 ≠ 33). Therefore, my previous mistake was not considering the parity.

Therefore, the case of three roots with product 33 is invalid because n must be even.

Therefore, we need to recast our analysis.

So let me start again.

First, let me consider Case 1: product of roots is 33, n even.

Possible factorizations of 33 into even number of distinct integers.

Possible even number of roots: n=2,4,...

For n=2: product is 33, so possible pairs as before.

Possible pairs: (1,33), (3,11), (-1,-33), (-3,-11).

Sum for these pairs: 1+33=34, 3+11=14, -1+(-33)=-34, -3 + (-11)=-14. Absolute value of sum is 34,14,34,14. So minimal |sum| is 14. Therefore, |a_{n-1}|=2*14=28.

For n=4: need four distinct integers with product 33.

As before, possible roots: 1, -1, 3, -11. Product is (1)(-1)(3)(-11)=33. Sum is 1 -1 + 3 -11 = -8. |sum| is 8. Therefore |a_{n-1}|=16.

Is there another combination with four distinct integers? For example, 1, -1, 11, -3. Same sum.

What about 1, -1, 3, -11: same as above.

Is there a way to have a smaller |sum| for n=4?

Suppose we take roots 1, -1, 3, -11, sum is -8. If we take other combinations? For example, 1, -1, -3, 11. Sum is 1 -1 -3 +11=8.

If we take roots 1, 3, -1, -11. Same sum.

Is there a combination with roots that sum to something closer to zero? For example, 1, -1, 3, -11: sum -8. What if we take roots 1, -1, 1, -33? But duplicate 1. Not allowed.

Alternatively, 3, -1, 11, -1. Duplicate -1. Not allowed.

So seems minimal |sum| is 8. Therefore |a_{n-1}|=16.

n=4 gives 16. n=2 gives 28. What about n=6? Let me check.

For n=6, need six distinct integers whose product is 33. Since 33 is small, need to include 1, -1, and maybe other numbers. But with six distinct integers, need to have some factors. Let me think. Suppose we take 1, -1, 3, -3, 11, -11. Their product is (1)(-1)(3)(-3)(11)(-11) = (1*-1)*(3*-3)*(11*-11) = (-1)*(-9)*(-121) = (-1)(1089) = -1089 ≠ 33.

Alternatively, maybe take some 1s and -1s and others. For example, 1, -1, 3, -1, 11, -11... but duplicates. Not allowed.

Alternatively, 1, -1, 3, -3, 1, -1... duplicates.

No way to get six distinct integers whose product is 33. Because the prime factors are only 3 and 11, and 33 has limited number of divisors. Therefore, n=6 is impossible. Similarly for higher n. Therefore, only n=2 and n=4 possible for product 33.

Case 1: minimal |a_{n-1}| is 16 (from n=4).

Now, Case 2: product of roots is -33, n odd.

Possible n=1,3,5,...

For n=1: only one root, which is -33. Sum is -33. |a_{n-1}|=2*33=66.

For n=3: three distinct integers with product -33.

Earlier, we found possible triplets. Let me re-examine with consideration of n odd (which it is, n=3).

Possible triplets:

For example:

(-1, 3, 11): product is -1*3*11=-33. Sum is -1 + 3 + 11 = 13. |sum|=13.

(-1, 1, 33): product is -1*1*33=-33. Sum is -1 +1 +33=33. |sum|=33.

(-1, -3, 11): product is (-1)(-3)(11)=33. Not -33.

(-1, -3, -11): product is (-1)(-3)(-11)=-33. Sum is -1 -3 -11=-15. |sum|=15.

(-3, 1, 11): product is -3*1*11=-33. Sum is -3 +1 +11=9. |sum|=9.

(-1, 3, -11): product is (-1)*3*(-11)=33. Not -33.

(-3, -1, 11): product is (-3)(-1)(11)=33. Not -33.

(-3, -1, -11): product is (-3)(-1)(-11)=-33. Sum is -3 -1 -11=-15. |sum|=15.

(1, -3, 11): same as above.

(1, -1, 33): sum 33.

(3, -1, -11): product 3*(-1)*(-11)=33. Not -33.

So the minimal |sum| for n=3 is 9 (from triplet -3,1,11). Therefore |a_{n-1}|=2*9=18.

What about n=5? Let me check if possible. Product -33, n=5 odd.

As before, five distinct integers with product -33. Let me think. Let me try to include 1, -1, 3, -3, and another root. Product is (1)(-1)(3)(-3)(x) = 9x. Set equal to -33: 9x = -33 ⇒ x = -11/3. Not integer.

Alternatively, include 1, -1, 3, 11, and x. Product is 1*(-1)*3*11*x = -33x. Set equal to -33 ⇒ -33x = -33 ⇒ x=1. Duplicate.

Or 1, -1, 11, -3, and x. Product is same as above.

Therefore, seems impossible. Similarly for other combinations, duplicates or non-integer roots.

Therefore, n=5 not possible.

Therefore, for Case 2, minimal |a_{n-1}| is 18 (from n=3).

Comparing Case 1 (product 33, n even) minimal |a_{n-1}| is 16 and Case 2 (product -33, n odd) minimal is 18.

Therefore, overall minimal is 16.

But wait, earlier I thought there was a case with product 33 and n=4 giving |a_{n-1}|=16. Let me check if that is valid.

For n=4, product of roots is 33, n even. Let me take roots 1, -1, 3, -11. Their product is 1*(-1)*3*(-11) = 33. n=4, even, okay.

Compute the constant term: a₀ = 2*(-1)^4*(product) = 2*1*33=66. Which is correct.

The sum of roots is 1 + (-1) + 3 + (-11) = 1 -1 +3 -11 = -8.

Therefore, a_{n-1} = -2*(sum of roots) = -2*(-8) = 16. Therefore |a_{n-1}|=16.

So that's valid.

Earlier, I had a mistake when I considered n=3 with product 33, which was invalid because n had to be even. Therefore, the minimal |a_{n-1}| is 16.

But wait, is there a way to get lower than 16? Let me check for n=4, can we have a smaller |sum|? For example, can we find four distinct integers with product 33 and smaller |sum|?

Let me think. For example, if we take roots 1, -1, -3, 11. Product is 1*(-1)*(-3)*11 = 33. Sum is 1 -1 -3 +11= 8. |sum|=8. Therefore, |a_{n-1}|=2*8=16. Same as previous.

Or take roots -1,1,3,-11: same thing.

Is there another set of four distinct integers with product 33 and smaller |sum|? Let me think.

Suppose we take 1, -1, 11, -3. Same as above.

What about 1, -1, 33, -1/... wait, no, roots have to be integers.

Alternatively, 3, -3, 1, -11. Sum is 3 -3 +1 -11 = -10. |sum|=10. Worse than 8.

Or 3, -1, 11, -1. Sum is 3 -1 +11 -1 = 12. Worse.

Alternatively, 11, -1, -1, 3. Duplicates.

No.

Alternatively, is there a way to have roots 1, -1,  something else? For example, 1, -1, 3, -11. Which we have.

Alternatively, 1, -1, 1, -33. Duplicate.

No.

What about including more numbers? For example, 1, -1, 3, -11, and another root? But with n=4, we can't.

Therefore, the minimal |sum| for n=4 is 8, leading to |a_{n-1}|=16.

Is there another case with n=2 and product 33? For example, roots (3,11), sum 14. |a_{n-1}|=28. Worse.

So, with Case 1 minimal is 16.

Case 2 minimal is 18.

Therefore, overall minimal is 16.

Therefore, the answer is 16.

But let me check once more. Is there another combination for product -33 with n odd (n=3) with smaller |sum|?

Earlier, we found a triplet with sum 9.

Wait, what about triplet (-1, 3, -11). Product is (-1)*3*(-11)=33. Not -33.

(-1, -3, 11). Product is (-1)(-3)(11)=33. Not -33.

(-1, -1, 33): duplicate.

(-1, 1, 33): sum 33, product -33.

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

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

(-3, -1, 11): product 33.

(-3, -1, -11): product -33. Sum -15.

(-1, -3, -11): product -33. Sum is -1 -3 -11 = -15.

Alternatively, is there another triplet? For example, (1, -3, 11). Sum is 1 -3 +11=9.

Or (1, -1, 33). Sum 1 -1 +33=33.

Or (3, -1, 11). Sum 3 -1 +11=13.

So, the minimal |sum| is indeed 9.

Therefore, |a_{n-1}|=2*9=18.

Therefore, comparing Case 1 (16) and Case 2 (18), the minimal is 16.

Therefore, the answer should be 16.

But wait, let me make sure that there isn't a Case 1 with n=6 or something else, but we saw that n=6 is not possible.

Alternatively, n=4 is the maximum possible for Case 1.

Therefore, the minimal |a_{n-1}| is 16.

But let me check if there is another Case 1 with n=4 and different roots.

For example, roots 1, -1, -3, 11. Product is 1*(-1)*(-3)*11=33. Sum is 1 -1 -3 +11 = 8. |a_{n-1}|=16.

Yes, as before.

Wait, but what if we take roots -1, 1, -3, 11? Same sum.

Therefore, conclusion: the minimal |a_{n-1}| is 16.

But hold on, earlier when I thought there was a case with |a_{n-1}|=14, that was invalid because the product of roots and parity didn't match.

Therefore, the correct answer is 16.

But to be thorough, let me check if there is a Case 2 with n=5 or something else. For example, with five roots.

Wait, previously, I thought it's impossible, but let me think again.

Suppose we take roots 1, -1, 3, -3, and x. Product is (1)(-1)(3)(-3)(x) = 9x. We need this equal to -33. So 9x=-33 ⇒ x=-11/3. Not integer.

Alternatively, roots 1, -1, 3, -11, x. Product is (1)(-1)(3)(-11)x = 33x. Need this equal to -33 ⇒ x = -1. Duplicate.

Therefore, no.

Alternatively, roots 1, -1, 11, -1, x. Duplicate again.

No.

Therefore, it's impossible.

Therefore, minimal is indeed 16.

Wait, but let me check for Case 1, if there's another set of four roots.

For example, if I take roots 1, -1, 11, -3. Sum is 1 -1 +11 -3=8. Same as before.

Is there a way to get sum 0? For example, roots 1, -1, 3, -3. Product is (1)(-1)(3)(-3)=9. Not 33.

If I take roots 1, -1, 3, -11. Product 33. Sum 8.

If I take roots 1, -1, 11, -3. Same.

Alternatively, roots 1, -1, something. No others.

Therefore, no.

Therefore, answer is 16.

But wait, let me think once again. The problem says the polynomial has "distinct integers" as roots. So in the case of four roots, they must be distinct. Which they are: 1, -1, 3, -11 are distinct.

Yes.

So then, the minimal |a_{n-1}| is 16.

However, I need to check whether there exists a polynomial with even degree and product 33 with smaller |sum|.

Wait, for n=4, we have |sum|=8, leading to |a_{n-1}|=16.

Is there a way to have n=4 with |sum| smaller than 8?

For example, if we take roots like 1, -1, 3, -11 (sum 8), or 1, -1, -3, 11 (sum 8).

What if we take roots like 1, -1, something else? Let me check.

Suppose I take roots 1, -1, x, y. Their product is 1*(-1)*x*y = -xy. Need -xy=33 ⇒ xy=-33.

So x and y are distinct integers, not 1 or -1, and different from each other.

So possible pairs (x,y): (3, -11), (-3,11), (11, -3), (-11,3), (33, -1), (-33,1), etc. But since x and y can't be 1 or -1, as already used.

So possible pairs (3, -11), (-3, 11), etc.

Therefore, sum would be 1 -1 + x + y = x + y.

So for x=3 and y=-11, sum is 3 + (-11)= -8.

For x=-3 and y=11, sum is -3 +11=8.

Therefore, same as before.

So no way to get smaller than 8.

Alternatively, other pairs? For example, x= 11 and y=-3. Same as before.

Therefore, minimal |sum| is 8. So |a_{n-1}|=16.

Therefore, I think this is indeed the minimal.

But let me check if there is another way to get product 33 with n=4 and different roots.

For example, using 33 as one of the roots.

Let me try. Suppose roots are 33, 1, -1, -1. But duplicate -1. Not allowed.

Or 33, 1, -1, something. Product is 33*1*(-1)*x = -33x. Need this equal to 33 ⇒ -33x=33 ⇒ x=-1. Duplicate.

So no.

Alternatively, 33, -1, 1, -1. Duplicate.

So no.

Therefore, no other combinations.

Therefore, indeed, the minimal is 16.

Therefore, the answer is \boxed{16}.

**Final Answer**
\boxed{16}

To solve the problem, we start with the given polynomial:

$$
f(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 = 2(x - r_1)(x - r_2)\cdots(x - r_n)
$$

Given that all roots $ r_1, r_2, ..., r_n $ are **distinct integers**, and the polynomial has **integer coefficients**, the expression is valid. We are told that:

- $ a_0 = 66 $
- $ a_n = 2 $

We aim to **minimize** the absolute value of the coefficient $ a_{n-1} $, which we determine from the expression of the polynomial.

---

### Step 1: Constant Term and Product of Roots

The constant term of the polynomial is:

$$
a_0 = 2 \cdot (-1)^n \cdot (r_1 r_2 \cdots r_n)
$$

Given $ a_0 = 66 $, we get:

$$
(-1)^n \cdot (r_1 r_2 \cdots r_n) = 33
$$

So the *product of the roots* (with sign) is 33. That is, either:

- $ \prod r_i = 33 $ and $ n $ is even, or
- $ \prod r_i = -33 $ and $ n $ is odd

Because the product of all roots is an integer, each root must divide 33. The divisors of 33 are:

$$
\pm1, \pm3, \pm11, \pm33
$$

We can only use these to construct the roots.

---

### Step 2: Objective

We need to find distinct integer roots such that their product is ±33 and the **absolute value of the coefficient $ a_{n-1} $** is minimized.

Recall that in a monic polynomial with root sum $ S = r_1 + r_2 + \cdots + r_n $, the coefficient of $ x^{n-1} $ is $ -S $. However, since our polynomial is not monic (it has leading coefficient 2), the coefficient $ a_{n-1} $ is:

$$
a_{n-1} = -2 \cdot (r_1 + r_2 + \cdots + r_n)
$$

Hence,

$$
|a_{n-1}| = 2 \cdot |\text{sum of roots}|
$$

So the problem reduces to **minimizing the absolute value of the sum of the roots**, subject to the constraints:

- All roots are distinct integers
- Product of roots is 33 if $ n $ is even, or -33 if $ n $ is odd

---

### Step 3: Try Possible Root Sets

We will consider sets of distinct integers whose product is ±33 and check which gives the **smallest absolute sum**.

#### Case 1: Product = 33, $ n $ even

This allows us to use even numbers of negative roots to preserve a positive product.

One such good set is:

- Roots: $ 1, -1, 3, -11 $ (product: $ 1 \cdot (-1) \cdot 3 \cdot (-11) = 33 $)
- Sum: $ 1 - 1 + 3 - 11 = -8 $
- Absolute sum: $ 8 $
- Hence, $ |a_{n-1}| = 2 \cdot 8 = 16 $

This is valid (distinct, correct product, even number of roots) and gives:

$$
f(x) = 2(x - 1)(x + 1)(x - 3)(x + 11)
$$

Expanding confirms that the constant term is $ 66 $, and this polynomial has distinct integer roots.

#### Case 2: Product = -33, $ n $ odd

Here, $ n $ is odd, and the product is negative, so an odd number of roots are negative.

Consider the roots: $ -3, 1, 11 $

- Product: $ (-3) \cdot 1 \cdot 11 = -33 $
- Sum: $ -3 + 1 + 11 = 9 $
- Absolute sum: $ 9 $
- Hence, $ |a_{n-1}| = 2 \cdot 9 = 18 $

This is also valid, and the sum is greater than in the previous case.

#### Other Cases

Trying any other possible root combinations with distinct integers from $ \{-33, -11, -3, -1, 1, 3, 11, 33\} $ and trying other combinations of even or odd number of roots results in **either duplicates or higher absolute sum**.

Therefore, the **smallest possible value** of $ |a_{n-1}| $ is achieved with the root set $ \{1, -1, 3, -11\} $, giving:

$$
\boxed{16}
$$

Solution:
Since $f(x)$ has integer coefficients, the Integer Root Theorem tells us that all integer roots of $f(x)$ must divide the constant term $66=2\cdot 3\cdot 11$. Thus, the possible integer roots of $f(x)$ are
$$\pm 1,~\pm 2,~\pm 3,~\pm 6,~\pm 11,~\pm 22,~\pm 33,~\pm 66.$$Moreover, since we know that all roots of $f(x)$ are integers, we know that all roots of $f(x)$ appear in the list above.

Now we apply Vieta's formulas. The product of the roots of $f(x)$ is $(-1)^n\cdot\frac{a_0}{a_n}$, which is $33$ or $-33$. Also, the sum of the roots is $-\frac{a_{n-1}}{a_n}=-\frac{a_{n-1}}2$. Thus, in order to minimize $|a_{n-1}|$, we should make the absolute value of the sum of the roots as small as possible, working under the constraint that the product of the roots must be $33$ or $-33$.

We now consider two cases.

Case 1 is that one of $33,-33$ is a root, in which case the only other possible roots are $\pm 1$. In this case, the absolute value of the sum of the roots is at least $32$.

The alternative, Case 2, is that one of $11,-11$ is a root and one of $3,-3$ is a root. Again, the only other possible roots are $\pm 1$, so the absolute value of the sum of the roots is at least $11-3-1=7$, which is better than the result of Case 1. If the absolute value of the sum of the roots is $7$, then $|a_{n-1}|=7|a_n|=7\cdot 2=14$.

Therefore, we have shown that $|a_{n-1}|\ge 14$, and we can check that equality is achieved by
\begin{align*}
f(x) &= 2(x+11)(x-3)(x-1) \\
&= 2x^3+14x^2-82x+66,
\end{align*}which has integer coefficients and integer roots. So the least possible value of $|a_{n-1}|$ is $\boxed{14}$.