Detailed mathematical approach

Problem Summary

For Problem 522, the central observation is that the target quantity can be evaluated directly from a finite formula instead of simulating the underlying blackout process. The C++, Python, and Java implementations compute the value for \(n=12344321\) modulo

$M=135707531.$

The practical task is therefore to evaluate one leading term and one indexed sum efficiently in modular arithmetic.

Mathematical Approach

The implementations use the identity

$F(n)\equiv n(n-1)(n-2)^{n-1}+\sum_{m=2}^{n-2}\frac{n!}{m!(n-m)}(m-1)^m \pmod{M}.$

The mathematics of the solution is about reorganizing this expression so that every summand can be produced quickly and safely modulo \(M\).

Step 1: Separate the leading term from the indexed family

Write

$Z(n)=n(n-1)(n-2)^{n-1},\qquad T_m(n)=\frac{n!}{m!(n-m)}(m-1)^m.$

Then

$F(n)\equiv Z(n)+\sum_{m=2}^{n-2}T_m(n)\pmod{M}.$

This split is useful because the first part is a single modular power, while the second part is a uniform family of terms indexed by \(m\). It also makes the small edge cases transparent: if \(n<4\), the interval \(2\le m\le n-2\) is empty, so only \(Z(n)\) remains.

The coefficient is genuinely integral, because

$\frac{n!}{m!(n-m)}=\binom{n}{m}(n-m-1)!.$

So the formula introduces no fractional ambiguity before the reduction modulo \(M\).

Step 2: Rewrite every denominator as a modular inverse

For the target range, \(n<M\), and the modulus \(M\) is prime. Hence every integer \(1,2,\dots,n\) has a multiplicative inverse modulo \(M\). Define

$u_i\equiv i^{-1}\pmod{M},\qquad v_m\equiv (m!)^{-1}\pmod{M},\qquad P\equiv n!\pmod{M}.$

Then each summand becomes

$T_m(n)\equiv P\,v_m\,u_{n-m}(m-1)^m\pmod{M}.$

After this rewrite, the entire computation is reduced to modular multiplications and modular exponentiation. No explicit division is needed during the main loop.

Step 3: Precompute the inverse tables in linear time

The inverse table is built with the standard recurrence

$u_1=1,\qquad u_i\equiv M-\left\lfloor\frac{M}{i}\right\rfloor u_{M\bmod i}\pmod{M}$

for \(2\le i\le n\). Once the \(u_i\) values are known, the inverse factorials follow from a forward product:

$v_0=1,\qquad v_m\equiv v_{m-1}u_m\pmod{M}.$

A separate forward pass computes \(P=n!\bmod M\). After these linear precomputations, the coefficient part \(P\,v_m\,u_{n-m}\) of every term is available in \(O(1)\) time.

Step 4: Compute the power factor by repeated squaring

The remaining factor in every indexed term is

$(m-1)^m\pmod{M}.$

The implementations evaluate it with binary exponentiation. Repeated squaring reduces one power evaluation from linear time in \(m\) to logarithmic time in \(m\), so each term needs only a constant number of modular multiplications plus one \(O(\log m)\) exponentiation.

This is why the sum is feasible even for a very large target \(n\): the expensive part is the sequence of fast power calls, not the factorial coefficient.

Worked Example: \(n=8\)

For \(n=8\), the leading term is

$Z(8)=8\cdot 7\cdot 6^7=15676416.$

The summation runs over \(m=2,3,4,5,6\). The corresponding coefficients and contributions are

$\frac{8!}{2!(8-2)}=3360,\qquad 3360\cdot 1^2=3360,$

$\frac{8!}{3!(8-3)}=1344,\qquad 1344\cdot 2^3=10752,$

$\frac{8!}{4!(8-4)}=420,\qquad 420\cdot 3^4=34020,$

$\frac{8!}{5!(8-5)}=112,\qquad 112\cdot 4^5=114688,$

$\frac{8!}{6!(8-6)}=28,\qquad 28\cdot 5^6=437500.$

Adding everything gives

$F(8)=15676416+3360+10752+34020+114688+437500=16276736.$

In this example the total is still below \(M\), so the modular answer equals the ordinary integer answer. This matches the checkpoint used by the implementation.

How the Code Works

The C++, Python, and Java implementations all follow the same arithmetic plan. They first precompute modular inverses up to \(n\), then compute \(n!\bmod M\), and then build a prefix table for inverse factorials. After that they evaluate the leading term \(n(n-1)(n-2)^{n-1}\) once.

Next they iterate through \(m=2,\dots,n-2\). For each \(m\), the factorial coefficient is assembled from the precomputed tables, the power \((m-1)^m\) is obtained by binary exponentiation, and the contribution is added modulo \(M\). The C++ implementation additionally splits the \(m\)-interval into chunks and combines partial sums from multiple workers; the Python and Java implementations use the same formula sequentially.

Complexity Analysis

Precomputing inverses, the factorial value, and the inverse factorial table takes \(O(n)\) time and \(O(n)\) memory. The main sum has \(n-3\) terms, and each term requires one modular exponentiation of cost \(O(\log m)\). Therefore the total time complexity is \(O(n\log n)\), with \(O(n)\) memory usage. In the multithreaded C++ version, the arithmetic workload is the same, but the wall-clock time is reduced by distributing the summation interval across several workers.

Footnotes and References

1. Problem page: https://projecteuler.net/problem=522
2. Modular multiplicative inverse: Wikipedia — Modular multiplicative inverse
3. Exponentiation by squaring: Wikipedia — Exponentiation by squaring
4. Factorial: Wikipedia — Factorial
5. Binomial coefficient: Wikipedia — Binomial coefficient