[THOUGHT] We are given natural numbers a,b,c satisfying:

1. a+b+c = 151
2. a^2+b^2+c^2 = 10939
3. a^3+b^3+c^3 = 957871

We need to find a,b,c. Since they are natural numbers (positive integers), we can try to solve.

Let s1 = a+b+c = 151. Let s2 = a^2+b^2+c^2 = 10939. Let s3 = a^3+b^3+c^3 = 957871.

We can use identities: Let p = ab+bc+ca. Then s1^2 = s2 + 2p => p = (s1^2 - s2)/2. Compute: s1^2 = 151^2 = 22801. Then p = (22801 - 10939)/2 = (11862)/2 = 5931.

Now, we have a+b+c=151, ab+bc+ca=5931.

Also, a^3+b^3+c^3 = (a+b+c)^3 - 3(a+b+c)(ab+bc+ca) + 3abc. So s3 = s1^3 - 3 s1 p + 3abc. Thus abc = (s3 - s1^3 + 3 s1 p)/3.

Compute s1^3 = 151^3 = 151151151. Let's compute stepwise: 151^2 = 22801, 15122801 = 15122801 = 15122800 + 1511 = 3442800 + 151 = 3442951. So s1^3 = 3442951. Now s3 = 957871. Compute 3 s1 p = 31515931 = 31515931 = 4535931? Let's compute carefully: 1515931 = 1515900 + 15131 = 890900 + 4681 = 895581. Then 3*895581 = 2686743. Now s3 - s1^3 + 3 s1 p = 957871 - 3442951 + 2686743 = (957871 + 2686743) - 3442951 = 3644614 - 3442951 = 201663. Then abc = 201663/3 = 67221.

So we have: a+b+c = 151 ab+bc+ca = 5931 abc = 67221.

Thus a,b,c are roots of polynomial t^3 - 151 t^2 + 5931 t - 67221 = 0.

We need natural numbers. Let's solve this cubic.

We can try integer roots. Divisors of 67221? Possibly small numbers. Let's try factorization.

We can use Python to find integer roots.

Then Q(x) = (x-a)(x-b)(x-c) = x^3 - 151 x^2 + 5931 x - 67221.

Now we need to compute I = ∫_{-∞}^{∞} Q(x) e^{ix} / (x - (205/69) i)^4 dx.

This is an integral of the form ∫_{-∞}^{∞} f(x) e^{ix} dx where f(x) = Q(x)/(x - iα)^4 with α = 205/69.

We can use contour integration in complex analysis. Consider the integral ∫_{-∞}^{∞} Q(z) e^{iz} / (z - iα)^4 dz, with α>0. Since e^{iz} decays in the upper half-plane for Im(z)>0, we can close the contour in the upper half-plane if the denominator has a pole there. The pole is at z = iα, which is in the upper half-plane because α>0. So we can compute using residue theorem.

But careful: Q(z) is a polynomial, entire. The integrand has a pole of order 4 at z = iα.

Thus I = 2πi * Residue at z = iα.

But we must consider orientation: For integral ∫_{-∞}^{∞} f(x) e^{ix} dx, with f(x) rational function decaying sufficiently fast, we can close in upper half-plane if f(z) decays as |z|→∞. Here Q(z) is degree 3 polynomial, denominator degree 4, so overall degree -1, which decays like 1/|z|, but that's not enough for Jordan's lemma? Actually Jordan's lemma requires integrand times exponential bounded by M/R and goes to zero. For e^{iz} in upper half-plane, |e^{iz}| = e^{-Im(z)} ≤ 1 for Im(z)≥0. So if |f(z)| ~ 1/|z| for large |z|, then integral over large semicircle tends to zero because length ~ πR, integrand ~ 1/R, so product ~ π → constant? Wait: |∫ f(z) e^{iz} dz| ≤ max|f(z)| * πR. If |f(z)| ~ C/|z|, then max on semicircle ~ C/R, so bound ~ C/R * πR = Cπ, constant, not zero. That's not enough to guarantee zero limit. But maybe because Q(z) is degree 3, denominator degree 4, so f(z) ~ constant/z for large z, indeed integral over semicircle tends to constant? Actually need to check: For large |z|, Q(z) ~ z^3, denominator ~ z^4, so f(z) ~ 1/z. Then |f(z)| ~ 1/R on semicircle of radius R. Length of semicircle = πR. So integral magnitude ≤ (1/R) * πR = π, constant. So limit may not be zero. However, Jordan's lemma for e^{iz} requires |f(z)| → 0 as |z|→∞ to guarantee integral over large arc goes to zero. But here f(z) ~ 1/z, which goes to zero, but the product with length R is constant, not zero. Actually Jordan's lemma states: If f(z) is continuous on the arc and |f(z)| ≤ M/R^α for α>0, then integral goes to zero. Here α=1, but we need α>0? Wait: For Jordan's lemma, if |f(z)| ≤ M/R for large R, then integral over semicircle ≤ πM/R * R? Let's recall: Jordan's lemma: ∫_C e^{imz} f(z) dz → 0 as R→∞ if m>0 and f(z) → 0 uniformly as |z|→∞. Here f(z) ~ 1/z → 0, but the lemma requires that max|f(z)| on the arc goes to zero, which it does (1/R). Then the integral bound is (π/R) * max|f|? Actually standard estimate: |∫C e^{imz} f(z) dz| ≤ (π/m) max{θ} |f(Re^{iθ})|. So if max|f| → 0, then integral → 0. Since max|f| ~ 1/R → 0, indeed integral → 0. So we can close contour in upper half-plane.

Thus I = 2πi * Res_{z=iα} [ Q(z) e^{iz} / (z - iα)^4 ].

Since pole of order 4, residue = (1/3!) lim_{z→iα} d^3/dz^3 [ Q(z) e^{iz} ].

Because residue for pole order n: Res = 1/(n-1)! lim_{z→z0} d^{n-1}/dz^{n-1} [ (z-z0)^n f(z) ].

Here f(z) = Q(z) e^{iz} / (z - iα)^4, so (z - iα)^4 f(z) = Q(z) e^{iz}. Then residue = (1/3!) * g'''(iα) where g(z) = Q(z) e^{iz}.

Thus we need to compute g'''(z) at z = iα.

Let g(z) = Q(z) e^{iz}. Then g'(z) = (Q'(z) + i Q(z)) e^{iz}. g''(z) = (Q''(z) + 2i Q'(z) - Q(z)) e^{iz}. g'''(z) = (Q'''(z) + 3i Q''(z) - 3 Q'(z) - i Q(z)) e^{iz}.

But Q(z) is cubic: Q(z) = z^3 - 151 z^2 + 5931 z - 67221. So Q'(z) = 3z^2 - 302 z + 5931. Q''(z) = 6z - 302. Q'''(z) = 6.

Thus g'''(z) = [6 + 3i (6z - 302) - 3(3z^2 - 302z + 5931) - i (z^3 - 151 z^2 + 5931 z - 67221)] e^{iz}.

Simplify: g'''(z) = [6 + 18i z - 906i - 9z^2 + 906z - 17793 - i z^3 + 151 i z^2 - 5931 i z + 67221 i] e^{iz}.

Combine real and imaginary parts: Real part: 6 - 9z^2 + 906z - 17793 = -9z^2 + 906z - 17787. Imaginary part: 18i z - 906i - i z^3 + 151 i z^2 - 5931 i z + 67221 i = i( -z^3 + 151 z^2 + 18z - 5931z - 906 + 67221 ) = i( -z^3 + 151 z^2 - 5913z + 66315).

Check: 18z - 5931z = -5913z. Constant: -906 + 67221 = 66315.

Thus g'''(z) = [ (-9z^2 + 906z - 17787) + i(-z^3 + 151 z^2 - 5913z + 66315) ] e^{iz}.

Now evaluate at z = iα, where α = 205/69.

Let α = 205/69. Compute iα.

We'll compute using Python.

Then residue = (1/6) * g'''(iα).

Then I = 2πi * residue = 2πi * (1/6) g'''(iα) = (πi/3) g'''(iα).

Thus I = (πi/3) * g'''(iα).

We need I = α_real + β i. Then compute floor(10 α_real) + floor(β).

But careful: α_real is the real part of I, β is imaginary part.

Let's compute numerically with Python.

First find a,b,c to verify polynomial.

[/THOUGHT] [PYTHON] import sympy as sp

Given sums

s1 = 151 s2 = 10939 s3 = 957871

Compute p = ab+bc+ca

p = (s1**2 - s2) / 2 print("p =", p)

Compute abc

abc = (s3 - s1**3 + 3s1p) / 3 print("abc =", abc)

Polynomial coefficients

t^3 - s1 t^2 + p t - abc = 0

coeff = [1, -s1, p, -abc] print("Coefficients:", coeff)

Solve cubic

t = sp.symbols('t') poly = t3 - s1*t2 + p*t - abc roots = sp.solve(poly, t) print("Roots:", roots)

Check if they are natural numbers

for r in roots: print(r, "is integer?", r.is_integer) [/PYTHON]

[THOUGHT] The roots are a=21, b=33, c=97 (or any permutation). So Q(x) = (x-21)(x-33)(x-97) = x^3 - 151x^2 + 5931x - 67221 as derived.

Now compute I = ∫_{-∞}^{∞} Q(x) e^{ix} / (x - iα)^4 dx with α = 205/69.

Using residue theorem: I = 2πi * Res_{z=iα} f(z) where f(z) = Q(z) e^{iz} / (z - iα)^4.

Residue = (1/3!) g'''(iα) with g(z) = Q(z) e^{iz}.

We have g'''(z) expression: g'''(z) = [ (-9z^2 + 906z - 17787) + i(-z^3 + 151 z^2 - 5913z + 66315) ] e^{iz}.

Let α_num = 205/69. Compute z0 = i * α_num.

Compute g'''(z0) numerically, then I = (πi/3) g'''(z0). Then extract real and imaginary parts.

Finally compute floor(10 * Re(I)) + floor(Im(I)).

Let's compute using Python.

[/THOUGHT] [PYTHON] import sympy as sp

Define symbols

z = sp.symbols('z')

Define α

alpha = sp.Rational(205, 69) z0 = sp.I * alpha

Define Q(z)

Q = z3 - 151*z2 + 5931*z - 67221

Compute g(z) = Q(z) * exp(I*z)

g = Q * sp.exp(sp.I * z)

Compute third derivative

g3 = sp.diff(g, z, 3) print("g'''(z) =", g3)

Evaluate at z = iα

g3_at_z0 = sp.simplify(g3.subs(z, z0)) print("g'''(iα) =", g3_at_z0)

Compute I = (π*i/3) * g'''(iα)

I = (sp.pi * sp.I / 3) * g3_at_z0 print("I =", I)

Simplify

I_simplified = sp.simplify(I) print("I simplified =", I_simplified)

Get real and imaginary parts

real_part = sp.re(I_simplified) imag_part = sp.im(I_simplified) print("Re(I) =", real_part) print("Im(I) =", imag_part)

Compute numeric approximations

real_val = sp.N(real_part) imag_val = sp.N(imag_part) print("Re(I) numeric =", real_val) print("Im(I) numeric =", imag_val)

Compute floor(10*Re(I)) + floor(Im(I))

floor_10real = int(sp.floor(10 * real_val)) floor_imag = int(sp.floor(imag_val)) result = floor_10real + floor_imag print("floor(10*Re(I)) =", floor_10real) print("floor(Im(I)) =", floor_imag) print("Sum =", result) [/PYTHON]

[THOUGHT] The computed sum is -36330. The problem asks for (\lfloor 10\alpha \rfloor + \lfloor \beta \rfloor) where (I = \alpha + \beta i). From the output, (\alpha \approx -3632.07201578072) and (\beta \approx -8.91859003926276). Then (\lfloor 10\alpha \rfloor = \lfloor -36320.7201578072 \rfloor = -36321) (since floor goes to the next lower integer), and (\lfloor \beta \rfloor = \lfloor -8.91859 \rfloor = -9). Their sum is (-36321 + (-9) = -36330). This matches the Python result.

Thus final answer is (-36330).

[/THOUGHT] \boxed{-36330}