[THOUGHT] We are given seven natural numbers a,b,c,d,e,f,g satisfying: p1 = sum = 58 p2 = sum of squares = 666 p3 = sum of cubes = 8944 p4 = sum of 4th powers = 129846 p5 = sum of 5th powers = 1972408 p6 = sum of 6th powers = 30870006 p7 = sum of 7th powers = 493478344

We also have a faulty mutation appends h = 10. We need to compute P12 = sum of 12th powers of all eight numbers (a through g and h).

We can treat this as a system of symmetric power sums. Let s_k = sum_{i=1}^7 x_i^k for k=1..7. Then s1=58, s2=666, s3=8944, s4=129846, s5=1972408, s6=30870006, s7=493478344.

We need to find s12 for 8 numbers: the original 7 plus h=10. So P12 = s12(7) + 10^12.

Thus we need to compute s12 of the original 7 numbers from given s1..s7.

We can use Newton's identities to find elementary symmetric sums e1..e7 from s1..s7, then compute s12 recursively.

Alternatively, we can use the fact that the numbers are natural numbers (positive integers). Possibly they are small integers? Let's check if there are integer solutions.

We can attempt to find the seven numbers by solving a polynomial whose roots are these numbers. Let p(x) = (x - a)(x - b)...(x - g) = x^7 - e1 x^6 + e2 x^5 - e3 x^4 + e4 x^3 - e5 x^2 + e6 x - e7.

We have e1 = s1 = 58. Newton's identities: k e_k = sum_{i=1}^k (-1)^{i-1} e_{k-i} s_i.

We can compute e1..e7 step by step.

Let's compute using Python.

We'll compute e1..e7, then compute s8..s12 recursively using s_k = e1 s_{k-1} - e2 s_{k-2} + ... + (-1)^{k-1} e_{k-1} s1 + (-1)^{k-1} k e_k? Wait careful.

Newton's identities for power sums p_k = sum x_i^k and elementary symmetric sums e_k: p_k - e1 p_{k-1} + e2 p_{k-2} - ... + (-1)^{k-1} e_{k-1} p_1 + (-1)^k k e_k = 0 for k <= n. For k > n, the formula is: p_k - e1 p_{k-1} + e2 p_{k-2} - ... + (-1)^n e_n p_{k-n} = 0.

So for n=7, for k > 7, we have recurrence: p_k = e1 p_{k-1} - e2 p_{k-2} + e3 p_{k-3} - e4 p_{k-4} + e5 p_{k-5} - e6 p_{k-6} + e7 p_{k-7}.

Thus we can compute p8, p9, p10, p11, p12 using e1..e7.

First compute e1..e7 from p1..p7.

We'll implement Newton's identities for k=1..7.

Let's do in Python. [/THOUGHT] [PYTHON] import itertools

Given power sums p_k for k=1..7

p = [0, 58, 666, 8944, 129846, 1972408, 30870006, 493478344] # p[1]..p[7]

Newton's identities: for k from 1 to 7, compute e_k

n = 7 e = [0]*(n+1) # e[0]=1

e[0] = 1

e[0] = 1

for k in range(1, n+1): # compute sum_{i=1}^{k} (-1)^{i-1} e_{k-i} * p_i total = 0 for i in range(1, k+1): sign = 1 if (i-1) % 2 == 0 else -1 # (-1)^{i-1} total += sign * e[k-i] * p[i] e[k] = total / k

print("Elementary symmetric sums e1..e7:") for i in range(1, n+1): print(f"e{i} = {e[i]}")

Now compute p8..p12 using recurrence for k>7

p_k = e1 p_{k-1} - e2 p_{k-2} + e3 p_{k-3} - e4 p_{k-4} + e5 p_{k-5} - e6 p_{k-6} + e7 p_{k-7}

p_extended = p[:] # copy p[0..7] for k in range(8, 13): # compute p_k val = 0 for j in range(1, n+1): sign = 1 if (j-1) % 2 == 0 else -1 # (-1)^{j-1} val += sign * e[j] * p_extended[k - j] p_extended.append(val) print(f"p{k} = {val}")

Now P12 = p12 + 10**12

p12_original = p_extended[12] h12 = 10**12 P12 = p12_original + h12 print(f"\nP12 = {P12}")

Check if the numbers are natural numbers: we can try to find integer roots.

Build polynomial x^7 - e1 x^6 + e2 x^5 - e3 x^4 + e4 x^3 - e5 x^2 + e6 x - e7

from sympy import symbols, Poly, factor, roots

x = symbols('x') coeffs = [1, -e[1], e[2], -e[3], e[4], -e[5], e[6], -e[7]] poly = Poly(coeffs, x) print("\nPolynomial:") print(poly.as_expr())

Try to factor

factored = factor(poly.as_expr()) print("\nFactored form:") print(factored)

Find roots

try: root_list = roots(poly.as_expr(), x) print("\nRoots (as dictionary):") print(root_list) except: # maybe roots returns dict with multiplicities pass

Let's compute numerical roots

import numpy as np coeff_np = [float(c) for c in coeffs] roots_np = np.roots(coeff_np) print("\nNumerical roots:") for r in roots_np: print(r)

Check if they are integers

int_roots = [r for r in roots_np if abs(r - round(r)) < 1e-6] print("\nInteger roots (candidates):") for r in int_roots: print(int(r))

Check sum, sum squares, etc.

We'll compute power sums for these roots to verify

def power_sum(roots, k): return sum([r**k for r in roots])

Use the exact integer roots

int_roots_exact = [int(round(r)) for r in int_roots] print(f"\nCandidate integer roots: {int_roots_exact}")

Check p1..p7

for k in range(1, 8): calc = sum([r**k for r in int_roots_exact]) given = p[k] print(f"p{k}: calc={calc}, given={given}, diff={calc-given}")

If they match exactly, we have the set.

But note: there might be repeated roots? The problem says natural numbers, not necessarily distinct.

Let's check if the polynomial factors nicely.

We can also try to factor over integers.

But from our earlier calculation, we can trust the recurrence.

However, we should verify that the numbers are natural numbers.

Let's compute the power sums for the candidate roots up to 7.

If they match, then we can use those roots to compute p12 directly.

But careful: the recurrence we used is based on the same e's, so p12 should be consistent.

However, we can compute p12 directly from the roots if they are integers.

Let's do that.

[/PYTHON]

[THOUGHT] From the Python output, we have:

• The original seven numbers are {2,3,5,7,11,13,17} (natural numbers).
• Their 12th power sum p12 = 609072836692326.
• h = 10, so h^12 = 10^12 = 1000000000000.
• Therefore P12 = p12 + h^12 = 609072836692326 + 1000000000000 = 610072836692326.

Thus the final answer is 610072836692326.

\boxed{610072836692326}