Okay, let's tackle this problem step by step. The question is about finding the least possible value of |a_{n-1}| for a polynomial with integer coefficients, distinct integer roots, leading coefficient 2, and constant term 66. 

First, let me recall some fundamental concepts. For a polynomial with integer coefficients and integer roots, the roots must divide the constant term. This is due to the Rational Root Theorem, which states that any rational root p/q (in lowest terms) of a polynomial with integer coefficients must have p dividing the constant term and q dividing the leading coefficient. Since our polynomial has integer roots, q must be 1, so each root is a divisor of the constant term, which here is 66.

Given that a_n = 2, the leading coefficient is 2. The constant term a_0 is 66. The polynomial can be expressed in factored form as f(x) = 2(x - r_1)(x - r_2)...(x - r_n), where each r_i is an integer root. 

Our goal is to find the minimal possible absolute value of the coefficient a_{n-1}. Let's recall that in a polynomial, the coefficient a_{n-1} is equal to -2 times the sum of the roots. That is, a_{n-1} = -2(r_1 + r_2 + ... + r_n). Therefore, |a_{n-1}| = 2|sum of roots|. So, to minimize |a_{n-1}|, we need to minimize the absolute value of the sum of the roots.

Therefore, the problem reduces to finding a set of distinct integers whose product is 66 (since the product of the roots, up to a sign, is 66 divided by the leading coefficient, which is 2. Wait, let's check that.

Wait, actually, for a polynomial f(x) = a_n x^n + ... + a_0, the product of the roots (with multiplicity) is (-1)^n * a_0 / a_n. Since all roots are integers and distinct, the product of the roots is (-1)^n * 66 / 2 = (-1)^n * 33. So, the product of the roots is ±33. Therefore, the roots must be distinct integers whose product is either 33 or -33. 

But since the roots are distinct integers, we need to factor 33 or -33 into distinct integer factors. So, first, let's factor 33. 33 factors into 1 * 3 * 11. But since we need distinct integers, we can also include negative factors. For example, 33 can be written as (-1) * 3 * (-11), but that would still be distinct integers. However, the product of the roots is 33 or -33, so the sign depends on the number of negative roots. 

Wait, let's get back to the factored form. The polynomial is f(x) = 2(x - r_1)(x - r_2)...(x - r_n). The constant term a_0 is 66, which is equal to 2*(-1)^n * r_1 r_2 ... r_n. Therefore, the product of the roots r_1 r_2 ... r_n = (-1)^n * 66 / 2 = (-1)^n * 33. So, the product of the roots is either 33 or -33, depending on whether n is even or odd.

Therefore, the roots are distinct integers whose product is ±33. Our task is to find such a set of integers with the minimal possible absolute sum. Then, |a_{n-1}| = 2|sum|. So, the problem is equivalent to finding a set of distinct integers whose product is 33 or -33, and the sum of which has the smallest possible absolute value.

So, first, let's list all possible factorizations of 33 and -33 into distinct integers, and compute their sums. Then, we can find the one with the minimal absolute sum.

But 33 is a positive number, and -33 is negative. Let's first consider the factorizations of 33:

33 can be factored into:

1 and 33

1, 3, 11

But since we need distinct integers, these are the possible factorizations into two factors. However, we can also have negative factors. For example:

-1 and -33

-1, -3, -11

But if we include negative factors, the product remains positive (for even number of negatives) or negative (for odd number of negatives). Wait, but since the product is 33 or -33, depending on the sign. Wait, perhaps I need to clarify.

Wait, the product of the roots is (-1)^n * 33. So, if n is even, the product is 33; if n is odd, the product is -33. Therefore, depending on whether n is even or odd, the product of the roots is either 33 or -33. So, if we factor 33 into distinct integers, their product is 33, and if n is odd, we need the product to be -33. So, perhaps the roots can be a mix of positive and negative numbers.

But first, let's note that 33 is 3 * 11, and 1 * 33. So, possible factorizations into distinct integers:

For two factors:

1 and 33

-1 and -33

1 and -33

But 1 and -33 would give product -33.

Similarly, 3 and 11, -3 and -11, 3 and -11, etc.

But we need to have all roots distinct. So, let's see:

Case 1: Two roots.

Possible pairs:

(1, 33): product 33, sum 34

(1, -33): product -33, sum -32

(3, 11): product 33, sum 14

(3, -11): product -33, sum -8

(-1, -33): product 33, sum -34

(-1, 33): product -33, sum 32

Wait, but the sum of roots is important. For two roots, the possible sums are 34, -32, 14, -8, -34, 32. The minimal absolute sum here is 8 (from 3 and -11, sum -8, absolute value 8).

Case 2: Three roots.

Factorizations of 33 into three distinct integers. Let's see:

33 can be written as 1 * 3 * 11. Similarly, -1 * -3 * -11, but that's product -33. Wait, 1 * 3 * 11 = 33, -1 * -3 * -11 = -33. So, for three factors, to get product 33, the factors must be 1, 3, 11. For product -33, the factors must be, say, -1, -3, 11 (but that's product 33). Wait, maybe not. Let's compute.

Wait, 1 * (-3) * (-11) = 33. So, three factors: 1, -3, -11. Their product is 33. Their sum is 1 -3 -11 = -13. Absolute value 13.

Alternatively, (-1) * 3 * (-11) = 33. Sum is -1 + 3 -11 = -9. Absolute value 9.

Alternatively, (-1) * (-3) * 11 = 33. Sum is -1 -3 +11 = 7. Absolute value 7.

Similarly, 1 * (-1) * (-33) = 33. Sum is 1 -1 -33 = -33. Absolute value 33.

So, the minimal absolute sum in three factors is 7 (from -1, -3, 11). 

Wait, but 1, -3, -11: product 33, sum -13. Absolute sum 13.

-1, 3, -11: product 33, sum -9. Absolute sum 9.

-1, -3, 11: product 33, sum 7. Absolute sum 7. That's the smallest so far.

Case 3: Four roots.

Factorizing 33 into four distinct integers. Let's see. 33 is 3*11, so to factor into four distinct integers, we need to include 1 and -1 as well.

For example: 1, -1, 3, 11. Product is 1*(-1)*3*11 = -33. But we need product 33 or -33 depending on the number of factors. Wait, n is the number of roots, which is the degree of the polynomial. So, if we have four roots, then the product of the roots is (-1)^4 * 33 = 33. So, the product must be 33. So, the four roots must multiply to 33.

So, how can we write 33 as a product of four distinct integers?

Possible options:

1, -1, 3, 11: product is -33. Not good.

1, -1, -3, 11: product is 33. Sum is 1 -1 -3 +11 = 8. Absolute value 8.

Alternatively, 1, -1, 3, -11: product is 33. Sum is 1 -1 +3 -11 = -8. Absolute value 8.

Or 1, 3, 11, -1: same as above.

Alternatively, 1, 3, -1, -11: product is 33. Sum is 1 +3 -1 -11 = -8.

Alternatively, 1, -3, -11, 1: but duplicates. Not allowed.

Alternatively, 3, 11, -1, -1: duplicates. Not allowed.

Wait, maybe another combination. For example, 2, 3, 11, -1: but 2 is not a factor of 33.

Wait, 33's prime factors are 3 and 11. So, to get four distinct integers, we need to include 1, -1, 3, 11, but that gives product -33. To get 33, we need even number of negatives. So, two negatives: 1, -1, -3, 11: product is 33. Sum is 1 -1 -3 +11 = 8. Similarly, 1, -1, 3, -11: product 33, sum -8. So, the minimal absolute sum here is 8.

Alternatively, maybe 1, 3, 11, -1: same as above. So, four roots give a sum of ±8.

Case 4: Five roots.

To factor 33 into five distinct integers. Let's see. 33 is 3*11. To have five distinct integers, we need to include 1, -1, 3, -3, 11. But 1*(-1)*3*(-3)*11 = 99. Not 33. Alternatively, 1, -1, 3, 11, -3: product is -99. Not 33. Alternatively, maybe 1, -1, 3, 11, and another number. Wait, but 33 is 3*11. So, to have five factors, we need to include 1, -1, 3, 11, and 1 or -1 again, but that would duplicate. Not allowed. Alternatively, maybe 1, -1, 3, 11, and 1: duplicates. So, seems impossible to factor 33 into five distinct integers. Therefore, five roots are not possible.

Therefore, the possible numbers of roots are 2, 3, or 4. For each, we can compute the minimal |sum|.

From above:

- Two roots: minimal |sum| is 8 (from 3 and -11 sum -8)

- Three roots: minimal |sum| is 7 (from -1, -3, 11 sum 7)

- Four roots: minimal |sum| is 8 (from 1, -1, 3, -11 sum -8 or 1, -1, -3, 11 sum 8)

Therefore, the minimal |sum| across all possible factorizations is 7 (from three roots: -1, -3, 11). Therefore, |a_{n-1}| = 2*7 = 14.

But wait, let's check if there's a way to get a smaller sum. For example, in three roots, is there a set of three distinct integers whose product is 33 and sum is smaller in absolute value?

Suppose we take 1, -3, 11: sum is 1 -3 +11 = 9. Absolute sum 9.

-1, 3, 11: sum -1 +3 +11 =13.

-1, -3, 11: sum -1 -3 +11 =7.

1, -1, -33: product -33, but n=3, so product should be -33. So sum is 1 -1 -33 = -33. Absolute sum 33.

Alternatively, 1, -1, 33: sum 1 -1 +33=33.

Alternatively, 3, 11, -1: same as above.

So, indeed, the minimal |sum| for three roots is 7.

Similarly, for four roots, the minimal |sum| is 8, which is higher than 7.

Therefore, the minimal |a_{n-1}| is 14.

Wait, but let's check if there's another way to factor 33 into three distinct integers with a smaller sum.

For example, 33 can be written as 1 * (-1) * (-33) = 33. Sum is 1 -1 -33 = -33. Absolute sum 33.

Alternatively, 1 * (-3) * 11 = -33. So, for three roots, product is -33, but n=3, so product should be (-1)^3 *33 = -33. So, that's okay. Sum is 1 -3 +11 =9. Absolute sum 9.

Alternatively, 3 * (-1) * 11 = -33. Sum is 3 -1 +11=13.

Alternatively, -1 * (-3) * 11 =33. Sum is -1 -3 +11=7. That's the same as before.

Alternatively, 1 * 3 * 11=33. Sum 1+3+11=15. Absolute sum 15.

Alternatively, 1 * (-1) * (-33)=33. Sum 1 -1 -33=-33. Absolute sum 33.

So, yeah, the minimal |sum| is 7 for three roots.

So, if we take roots -1, -3, 11, the product is (-1)*(-3)*11=33, which is correct for a polynomial of degree 3 (since n=3, product is (-1)^3 *33= -33, but wait: Wait, hold on. Wait, the product of the roots is r_1 r_2 r_3 = (-1)^3 * a_0 / a_n = (-1)^3 * 66 / 2 = -33. Wait, that's correct. So, the product of the roots is -33. But in the case of three roots: r1*r2*r3 = (-1)*(-3)*11 =33. But according to the formula, it should be -33. Therefore, this is a contradiction. Therefore, that set of roots would not satisfy the product condition.

Wait, hold on. Let's clarify.

Given that the polynomial is f(x) = 2(x - r1)(x - r2)...(x - rn). Then, the constant term a0 is 2*(-1)^n * r1 r2 ... rn. Therefore, the product of the roots r1 r2 ... rn = (-1)^n * a0 / 2 = (-1)^n * 66 / 2 = (-1)^n *33. So, if n is even, the product is 33; if n is odd, the product is -33.

Therefore, in our earlier analysis, if we have three roots (n=3), the product of roots must be -33. Therefore, the roots -1, -3, 11 multiply to 33, which is not equal to -33. Therefore, that set is invalid.

Wait, that changes things. So, in three roots, the product must be -33. So, we need three distinct integers whose product is -33. Let's recast that.

So, for three roots:

Possible factorizations into three distinct integers with product -33:

We need to write -33 as a product of three distinct integers.

Possible triplet factors:

1, 3, -11: product 1*3*(-11) = -33. Sum is 1 +3 -11 = -7. Absolute sum 7.

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

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

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

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

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

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

-3, 1, 11: same as above.

Therefore, the only triplet with product -33 is 1, 3, -11 (sum -7), and 1, -3, 11 (sum 9). Therefore, the minimal |sum| here is 7 (from 1, 3, -11 sum -7). 

But wait, the product is 1*3*(-11)=-33, which is correct. So, that's acceptable. So, |sum| is 7. Therefore, the sum is -7, so |sum|=7. Therefore, |a_{n-1}|=2*7=14.

But wait, is there another triplet with product -33 and smaller |sum|?

Suppose we take -1, 3, 11: product -1*3*11=-33. Sum -1 +3 +11=13. Absolute sum 13. Not better.

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

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

Another triplet: -1, -3, 11: product (-1)*(-3)*11=33. Not -33. So, invalid.

Alternatively, 1, 1, -33: but duplicates. Not allowed.

Therefore, the only valid triplet with product -33 is 1, 3, -11, sum -7, absolute sum 7. So, |a_{n-1}|=14.

Therefore, for n=3, the minimal |a_{n-1}| is 14.

For n=2, the minimal |sum| was 8, which gives |a_{n-1}|=16.

For n=4, the minimal |sum| was 8, which gives |a_{n-1}|=16.

Therefore, the minimal |a_{n-1}| is 14, achieved when n=3, with roots 1, 3, -11.

But wait, let's check if there's a way to have a sum of roots with a smaller absolute value. For example, if we can have sum 0, but probably not. Let's see.

Suppose we can find three distinct integers whose product is -33 and sum is 0. Let's see.

Looking for three distinct integers x, y, z such that x + y + z =0 and x y z = -33.

Possible candidates:

Let me check if there are such numbers. Let's see. Let's try small numbers.

Suppose one of them is 1. Then, the other two numbers must multiply to -33 and sum to -1.

So, solving a + b = -1 and a*b = -33.

The quadratic equation would be t^2 + t -33 =0. The discriminant is 1 + 132=133, which is not a perfect square. So, no integer solutions.

Similarly, if one number is -1. Then, the other two numbers multiply to 33 and sum to 1.

Solving a + b =1, a*b=33.

Quadratic equation: t^2 - t +33=0. Discriminant 1 -132=-131. No real solutions.

Similarly, take 3. Then, the other two numbers multiply to -11 and sum to -3.

So, a + b = -3, a*b = -11.

Quadratic equation: t^2 +3t -11=0. Discriminant 9 +44=53. Not a perfect square. No integer solutions.

Take -3. Then, other two numbers multiply to 11 and sum to 3.

Equation: t^2 -3t +11=0. Discriminant 9 -44=-35. No real solutions.

Similarly, take 11. Then, other two numbers multiply to -3 and sum to -11.

Equation: t^2 +11t -3=0. Discriminant 121 +12=133. Not a perfect square.

Take -11. Then, other two numbers multiply to 3 and sum to 11.

Equation: t^2 -11t +3=0. Discriminant 121 -12=109. Not a perfect square.

Therefore, there are no three distinct integers that multiply to -33 and sum to 0. Therefore, the minimal |sum| is indeed 7, achieved by 1, 3, -11.

Therefore, the minimal |a_{n-1}| is 14.

But let's confirm this with another approach. Suppose n=3, and the roots are 1, 3, -11. Then, the polynomial is 2(x -1)(x -3)(x +11). Let's expand this.

First, multiply (x -1)(x -3):

(x -1)(x -3) = x^2 -4x +3.

Then, multiply by (x +11):

(x^2 -4x +3)(x +11) = x^3 +11x^2 -4x^2 -44x +3x +33 = x^3 +7x^2 -41x +33.

Multiply by 2: 2x^3 +14x^2 -82x +66.

Therefore, a_2 =14, a_1=-82. So, |a_{n-1}|=| -82 | =82. Wait, that's not matching. Wait, wait, no. Wait, in the polynomial f(x) = a_n x^n + ... + a_0, so in this case, the polynomial is 2x^3 +14x^2 -82x +66. Therefore, a_{n-1} is a_2, which is14. So, |a_{n-1}|=14. So, that's correct. So, the coefficient a_{n-1} is 14. Therefore, the answer is 14.

But wait, wait, in the expansion above, the coefficient of x^2 is14. So, |a_{n-1}|=14. So, that's correct.

But let's check another factorization. Suppose we have roots -1, -3, 11. Then, the polynomial is 2(x +1)(x +3)(x -11). Let's expand this.

First, multiply (x +1)(x +3) =x^2 +4x +3.

Multiply by (x -11):

(x^2 +4x +3)(x -11) =x^3 -11x^2 +4x^2 -44x +3x -33 =x^3 -7x^2 -41x -33.

Multiply by 2: 2x^3 -14x^2 -82x -66.

Therefore, a_{n-1}= -14. So, |a_{n-1}|=14. Same result.

Therefore, regardless of the order of roots, |a_{n-1}|=14.

Therefore, the minimal possible value is14.

But wait, let's check if there's another set of roots with a smaller sum. For example, suppose we have roots 1, -1, -33. Then, the product is 1*(-1)*(-33)=33. But since n=3, product should be -33. Therefore, invalid. Similarly, roots 3, -1, -11: product is 3*(-1)*(-11)=33. Not -33. So, invalid.

Therefore, the only valid triplet with product -33 and minimal |sum| is 1,3,-11, sum -7, |sum|=7, giving |a_{n-1}|=14.

Therefore, the minimal possible value is14. So, the answer is \boxed{14}.

**Final Answer**
The least possible value of \(|a_{n-1}|\) is \boxed{14}.
Therefore, the least possible value of $|a_{n-1}|$ is $\boxed{14}$.
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