[THOUGHT] I will:

1. Enumerate all integer triples (x,y,z) with 1 ≤ x,y,z ≤ 200 that yield positive integer (a,b,c) satisfying the cyclic divisibility conditions, using the derived formulas: a = (xy + x + 1)/(xyz - 1), etc.
2. Verify each (a,b,c) explicitly and collect all distinct ordered triples T and their sums S=a+b+c.
3. Build G(d): the number of cyclically divisible triples with sum exactly d.
4. Precompute v2(n) and v2(n!) for n ≤ N=10^7 using arrays for memory efficiency.
5. Compute E = Σ_d G(d) * Σ_{k=1}^{⌊N/d⌋} v2((kd)!) using the precomputed v2fact. I will first implement steps 1–3 and inspect the triples and sums; if x,y,z are all comfortably ≤200 I’ll trust completeness, then proceed to steps 4–5. [/THOUGHT] [PYTHON] from math import isclose from collections import defaultdict

Step 1: search for cyclically divisible triples via (x,y,z) up to some bound

def find_triples(max_x): triples = set() xyz_max = 0 for x in range(1, max_x+1): for y in range(1, max_x+1): xy = xy for z in range(1, max_x+1): denom = xyz - 1 if denom <= 0: continue a_num = xy + x + 1 if a_num % denom != 0: continue b_num = yz + y + 1 if b_num % denom != 0: continue c_num = zx + z + 1 if c_num % denom != 0: continue a = a_num // denom b = b_num // denom c = c_num // denom if a <= 0 or b <= 0 or c <= 0: continue # verify the original conditions if (a+1) % b != 0: continue if (b+1) % c != 0: continue if (c+1) % a != 0: continue triples.add((a,b,c)) if xyz > xyz_max: xyz_max = xyz return triples, xyz_max

triples_100, xyz_max_100 = find_triples(100) print("max_x=100: #triples =", len(triples_100), "max xyz =", xyz_max_100) print("Triples (a,b,c):", sorted(triples_100))

Try a larger bound to check if new triples appear

triples_150, xyz_max_150 = find_triples(150) print("max_x=150: #triples =", len(triples_150), "max xyz =", xyz_max_150) new_triples_150 = sorted(triples_150 - triples_100) print("New triples when going 100->150:", new_triples_150)

triples_200, xyz_max_200 = find_triples(200) print("max_x=200: #triples =", len(triples_200), "max xyz =", xyz_max_200) new_triples_200 = sorted(triples_200 - triples_150) print("New triples when going 150->200:", new_triples_200)

Assuming stability by 200, use triples_200

T = sorted(triples_200) print("Final triples T:", T)

Step 2: compute G(d) = count of triples with S=a+b+c = d

G = defaultdict(int) for a,b,c in T: S = a+b+c G[S] += 1

print("G (sum -> count):", dict(sorted(G.items())))

Now compute E using precomputed v2(n) and v2fact(n) for n<=N

from array import array

N = 10_000_000

Precompute v2(n) for n <= N

v2 = array('B', [0]) * (N+1) for n in range(2, N+1): if n & 1: v2[n] = 0 else: v2[n] = v2[n >> 1] + 1

Precompute v2fact(n) for n <= N

v2fact = array('I', [0]) * (N+1) for n in range(1, N+1): v2fact[n] = v2fact[n-1] + v2[n]

Compute E = sum_d G(d) * sum_{k=1}^{floor(N/d)} v2fact[k*d]

E = 0 for d, cnt in G.items(): limit = N // d subtotal = 0 step = d n = d for _ in range(limit): subtotal += v2fact[n] n += step E += cnt * subtotal

print(E)

[THOUGHT] Using the classification of all cyclically divisible triples (a,b,c), we determined the multiset of sums S(a,b,c), built the function G from these sums, then used the identity F = G * 1 along with the definition of K and Legendre’s formula for v2(n!) to compute E = v2(K) numerically up to N=10^7. The Python computation yielded E = 91666486036450 with no errors. [/THOUGHT] \boxed{91666486036450}