[THOUGHT] I’ll use Gleason’s theorem for Type II (binary doubly-even self-dual) codes of length 24.

Given the conditions, every code in the family is a Type II [24,12] code, so its weight enumerator is a linear combination of two basis polynomials:

• $A(x,y) = x^8 + 14x^4y^4 + y^8$,
• $B(x,y) = x^4y^4(x^4 - y^4)^4$.

Any such weight enumerator is:

for some scalars $u,v$.

For the specific code $C^*$, we know:

1. $|C^*| = 2^{12} = 4096$, so $W(1,1) = 4096$.
2. There are exactly 6 words of weight 4, so the coefficient of $x^{20}y^4$ in $W$ is 6.

These give two linear equations in $u,v$. I’ll:

• Expand $A^3$ and $B$,
• Compute $A^3(1,1)$ and $B(1,1)$,
• Extract the coefficients of $x^{20}y^4$ in both,
• Solve for $u,v$,
• Form $W$ for $C^*$,
• Extract the coefficient of $x^{12}y^{12}$, which equals the number of codewords of weight 12.

All steps will be verified using Python/sympy. [/THOUGHT] [PYTHON] from sympy import symbols, expand, Eq, solve

Define variables and basis polynomials

x, y, u, v = symbols('x y u v') A = x8 + 14*x4*y4 + y8 B = x4 * y4 * (x4 - y4)**4

A3 = expand(A**3) Bexp = expand(B)

Evaluate A^3(1,1) and B(1,1)

A3_11 = A3.subs({x:1, y:1}) B_11 = Bexp.subs({x:1, y:1})

print("A^3(1,1) =", A3_11) print("B(1,1) =", B_11)

Get coefficients for weight 4 i.e., x^20 y^4

def coeff_weight(poly, k): # weight k term is x^{24-k} y^k return poly.expand().coeff(x, 24-k).coeff(y, k)

c4_A3 = coeff_weight(A3, 4) c4_B = coeff_weight(Bexp, 4)

print("Coeff of weight 4 in A^3:", c4_A3) print("Coeff of weight 4 in B :", c4_B)

Set up equations for C*: W = u A^3 + v B

1) total codewords = 4096

2) A_4 = 6

eq1 = Eq(uA3_11 + vB_11, 4096) eq2 = Eq(uc4_A3 + vc4_B, 6)

sol_uv = solve((eq1, eq2), (u, v)) print("Solution (u, v) =", sol_uv)

u_star, v_star = sol_uv[u], sol_uv[v]

Form W for C*

W_star = expand(u_starA3 + v_starBexp)

Extract coefficient of weight 12 (x^12 y^12)

A12_star = coeff_weight(W_star, 12) print("Number of weight-12 codewords in C*:", A12_star) [/PYTHON]

[THOUGHT] Using the given structure of the weight enumerator and the constraints (dimension 12, 6 codewords of weight 4, and total size 4096), solving for the linear combination of the two basis weight enumerators yields parameters $u = 1$ and $v = -36$. Extracting the coefficient of $x^{12}y^{12}$ in the resulting weight enumerator gives the number of codewords of weight 12, which is 2612. [/THOUGHT] $\boxed{2612}$