Problem Analysis #0

[THOUGHT] To solve this, first identify all cyclically divisible triples (a,b,c) by brute force over small ranges since T is finite and values are small (max around 5-10). List them, compute S=a+b+c for each, and create a dictionary g where g[s] is the number of triples with that S. Then, F(n) = sum_{d|n} g(d). For E = v_2(K) = sum_{n=1}^N F(n) v_2(n!), compute it as sum_k sum_n F(n) floor(n/2^k) for k=1 to 30 say. To compute efficiently, for each k, m=2**k, sum_n F(n) floor(n/m) = sum_s g(s) * sum_{t=1}^{floor(N/s)} floor( (st)/m ). Since number of s small, and for each pair (s,m), compute sum_{t=1}^T floor(s t / m) efficiently. There is a standard O(1) or O(log) way, but since N=1e7, and few s (~10?), few k=25, I can loop over t if T small, but for small s, T=1e7/s large. For s=1, T=1e7, summing 1e7 terms 25 times too slow? No, python can do 251e7 ~2.5e8 operations fast <1s. But there are multiple s, say 10 s, 1025 avg 1e7/10=1e6, total ~2.5e8 still ok, but for s=3 T~3e6, etc. Worst is smallest s. But python loop sum for 1e7 is fine. To be safe, implement a fast floor sum function.

Actually, there is a formula for sum_{t=1}^T floor(t * a / b), with a=s, b=m.

A standard way is O(sqrt(T)) but for speed, since N=1e7, even naive loop is ok, as tool is python.

But to make faster, I can precompute all v_2(n!) cumulatives? No.

Since g supported on few s, F(n) = sum_{s|gcd?} no.

Another way: since F(n) = sum_{s in S_set} g(s) * 1_{s divides n}, so sum_n F(n) f(n) = sum_s g(s) sum_{n: s|n} f(n), with n=1 to N.

Yes, and sum_{multiples of s up to N} f(n), with f(n)=floor(n/m)

Yes.

To compute sum_{j=1}^{floor(N/s)} f(s j) = sum_j floor(s j / m)

To compute this fast, I can use the fact that floor(s j / m) is arithmetic, and use the divisor sum trick for floor functions.

There is a known O(log N) per query or better.

But for simplicity, since N=1e7, and total operations small, I'll write a function to compute sum_{t=1}^T floor( (s*t) // m ) using a loop, python3 is fast enough.

First, find the triples.

Code to find triples: loop a=1 to 100, b divisors of a+1, c divisors of b+1, check a divides c+1.

This will be instant.

Yes.

So first code: find all triples, print list of (a,b,c,S), and dict g = collections.Counter([a+b+c for ...])

Then print g.

Once I have the g, I can hardcode it in next thought.

Since multi-turn, first run this. [/THOUGHT] [PYTHON] from collections import Counter, defaultdict

triples = [] maxv = 100 # ample

for a in range(1, maxv + 1): divisors_b = [d for d in range(1, a+2) if (a+1) % d == 0] # b divides a+1, b <=a+1 for b in divisors_b: divisors_c = [d for d in range(1, b+2) if (b+1) % d == 0] for c in divisors_c: if (c + 1) % a == 0: triples.append((a, b, c)) print(f"Triple: ({a},{b},{c}), S={a+b+c}")

print(f"\nTotal triples: {len(triples)}")

sums = [a+b+c for a,b,c in triples] g = Counter(sums) print("\ng[s] multiplicities:") for s in sorted(g): print(f"s={s}: {g[s]}")

print("\nAll sums:", sorted(set(sums))) [/PYTHON]

[THOUGHT] Using the output, G(3)=1, G(4)=3, G(6)=3, G(12)=3. Now compute E using numpy for fast summation of floor(s j / m) over j=1 to T. [/THOUGHT] [PYTHON] import numpy as np

N = 10000000 g_dict = {3:1, 4:3, 6:3, 12:3} E = 0 for k in range(1, 30): m = 1 << k if m > N: break contrib = 0 for s, mult in g_dict.items(): T = N // s if T == 0: continue j = np.arange(1, T+1, dtype=np.int64) floorsum = np.sum((s * j) // m) contrib += mult * floorsum E += contrib print(E) [/PYTHON]

[THOUGHT] The computation is correct based on the triples found and the summation using numpy for efficiency. The output 91666486036450 is the exact value of E. [/THOUGHT] \boxed{91666486036450}

Explanation

The reasoning follows a structured mathematical and computational process to compute E, the 2-adic valuation of K.

1. Identify all cyclically divisible triples: Iterate over possible values of a from 1 to 100 (sufficient since T is finite and triples are small). For each a, find divisors b of a+1, then divisors c of b+1, and check if a divides c+1. This yields the 10 triples: (1,1,1), (1,1,2), (1,2,1), (1,2,3), (2,1,1), (2,3,1), (3,1,2), (3,4,5), (4,5,3), (5,3,4), with corresponding S values: one triple for S=3, three for S=4, three for S=6, and three for S=12. Thus, G(3)=1, G(4)=3, G(6)=3, G(12)=3.

2. Express F(n): F(n) = ∑_{d|n} G(d), so F(n) counts triples where S divides n.

3. Compute E = v_2(K): K = ∏{n=1}^N (n!)^{F(n)}, so v_2(K) = ∑{n=1}^N F(n) ⋅ v_2(n!). Since v_2(n!) = ∑{k≥1} floor(n / 2^k), interchange sums: E = ∑{k≥1} ∑_{n=1}^N F(n) ⋅ floor(n / 2^k).

4. Evaluate inner sum: For each k, let m = 2^k (up to k=29 where m > N=10^7). Then ∑_{n=1}^N F(n) ⋅ floor(n/m) = ∑s G(s) ⋅ ∑{t=1}^{floor(N/s)} floor((s ⋅ t) / m), where s are the support values {3,4,6,12}.

5. Numerical computation: Use NumPy for efficient summation. For each k and s, compute T = floor(N/s), generate array j from 1 to T, and sum floor((s ⋅ j) / m) multiplied by G(s). Aggregate over all k and s to obtain E = 91666486036450.

This verifies the result through exact enumeration and verified Python execution.