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SKILL·06290A

solve-modular-arithmetic

pjt222
Actualizado 1 month ago
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Esta Skill de Claude resuelve problemas de aritmética modular, incluyendo congruencias lineales, inversos modulares, exponenciación modular con números grandes y sistemas de congruencias utilizando el Teorema Chino del Resto. Combina razonamiento manual paso a paso con verificación computacional mediante el algoritmo de Euclides extendido y el teorema de Euler. Los desarrolladores deben usarla para tareas de teoría de números que involucren grupos cíclicos, contextos de logaritmo discreto o simplificación de expresiones modulares complejas.

Instalación rápida

Claude Code

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Principal
npx skills add pjt222/agent-almanac -a claude-code
Comando PluginAlternativo
/plugin add https://github.com/pjt222/agent-almanac
Git CloneAlternativo
git clone https://github.com/pjt222/agent-almanac.git ~/.claude/skills/solve-modular-arithmetic

Copia y pega este comando en Claude Code para instalar esta habilidad

Documentación

Solve Modular Arithmetic

Parse congruence systems → extended Euclidean for inverses → CRT for systems → Euler for exponentiation. Verify all by substitution.

Use When

  • Single linear congruence ax = b (mod m)
  • System of simultaneous congruences (CRT)
  • Mod inverse a^{-1} (mod m)
  • Large mod exp a^k (mod m)
  • Order of element in Z/mZ
  • Cyclic groups, primitive roots, discrete log

In

  • Required: Congruence(s) | mod eqn
  • Optional: Show extended Euclidean steps explicit?
  • Optional: Apply Euler | Fermat?
  • Optional: Find primitive roots | element orders?
  • Optional: Output (step-by-step, compact, proof)

Do

Step 1: Parse System / Eqn

Extract math structure.

  1. ID type:

    • Single linear: ax = b (mod m)
    • System: x = a1 (mod m1), x = a2 (mod m2), ...
    • Mod exp: a^k (mod m)
    • Mod inverse: find a^{-1} (mod m)
  2. Normalize: Reduce coefs mod respective moduli. a, b, m non-neg, m > 0.

  3. Record parsed problem in standard notation.

Got: Clearly parsed + normalized problem, all values reduced.

If err: Ambiguous notation ("solve 3x + 5 = 2 mod 7" → equation or 5=2 mod 7?) → clarify w/ user. Default: mod applies to entire eqn.

Step 2: Solve Single Congruence

ax = b (mod m) via extended Euclidean.

  1. g = gcd(a, m) via Euclidean:

    • Repeated div: m = q1·a + r1, a = q2·r1 + r2, ... until rem = 0
    • Last non-zero rem = gcd(a, m)
  2. Solvability: solution iff g | b. Else no solution. Stop.

  3. Reduce: ÷ g → (a/g)x = (b/g) (mod m/g). Now gcd(a/g, m/g) = 1.

  4. Find inverse of a/g mod m/g via extended Euclidean:

    • Back-sub through Euclidean → 1 = (a/g)·s + (m/g)·t
    • Coef s (reduced mod m/g) = inverse
  5. Particular sol: x0 = s·(b/g) mod (m/g).

  6. General sol: x = x0 + (m/g)·k for k = 0, 1, ..., g-1 → all g incongruent solutions mod m.

Extended Euclidean ex (find 17^{-1} mod 43):

43 = 2*17 + 9
17 = 1*9  + 8
 9 = 1*8  + 1
 8 = 8*1  + 0

Back-substitute:
1 = 9 - 1*8
  = 9 - 1*(17 - 1*9) = 2*9 - 17
  = 2*(43 - 2*17) - 17 = 2*43 - 5*17

So 17*(-5) = 1 (mod 43), i.e., 17^{-1} = -5 = 38 (mod 43).

Got: Complete solution set, or proof no solution exists.

If err: Back-sub wrong result → verify each div step. Most common err: sign mistake in back-sub. Check: a · inverse mod m should = 1.

Step 3: System via CRT

Solve x = a1 (mod m1), x = a2 (mod m2), ..., x = ak (mod mk).

  1. Pairwise coprime check: every pair (mi, mj), gcd(mi, mj) = 1.

    • All coprime → CRT applies directly
    • Some not coprime → check compat: each non-coprime pair, ai = aj (mod gcd(mi, mj)). Compat → reduce via lcm. Incompat → no sol.
  2. M = m1 · m2 · ... · mk (product of all).

  3. Each i, Mi = M/mi (product except mi).

  4. Each i, yi = Mi^{-1} (mod mi) via extended Euclidean (Step 2).

  5. Solution: x = sum(ai · Mi · yi for i = 1..k) mod M.

  6. State: x = [val] (mod M). Unique sol mod M.

Common totients ref:

nphi(n)nphi(n)nphi(n)
21104208
321110248
421242520
541312308
621463612
761584816
841686016
9618610040

Got: Unique sol mod M, or proof of incompat.

If err: CRT result fails verification → check Step 4 inverse calcs. Common: computing Mi^{-1} mod M instead of mod mi. Each inverse is mod individual mi, not product.

Step 4: Apply Euler / Fermat

Eval mod exponentiations | simplify w/ Euler.

  1. Euler: gcd(a, m) = 1 → a^{phi(m)} = 1 (mod m).

    • phi(m): if m = p1^e1 · p2^e2 · ... · pk^ek, phi(m) = m · product((1 - 1/pi) per prime pi | m).
  2. Fermat little (special): p prime, gcd(a, p) = 1 → a^{p-1} = 1 (mod p).

  3. Reduce exp: compute a^k (mod m):

    • r = k mod phi(m)
    • a^k = a^r (mod m)
  4. a^r (mod m) via repeated squaring (binary exp):

    • r in binary: r = b_n · 2^n + ... + b_1 · 2 + b_0
    • result = 1
    • Each bit MSB → LSB: result = result² mod m; bit = 1 → result = result · a mod m
  5. gcd(a, m) > 1: Euler doesn't apply. Factor m + CRT to combine results from prime power moduli, lifting exponent | direct compute.

Got: a^k (mod m) value via exp reduction + repeated squaring.

If err: gcd(a, m) > 1 + result wrong → don't apply Euler. Compute direct | factor m into coprime parts where some coprime to a, solve mod each, recombine via CRT.

Step 5: Verify by Substitution

Plug solution back.

  1. Single congruence: a · x mod m = b?

  2. CRT systems: each x = ai (mod mi), x mod mi = ai?

  3. Mod exponentiation: verify w/ 2nd method (direct for small | independent repeated squaring impl).

  4. Document explicit:

Solution: x = 23
Check 1: 23 mod 3 = 2 = a1. Correct.
Check 2: 23 mod 5 = 3 = a2. Correct.
Check 3: 23 mod 7 = 2 = a3. Correct.
All congruences satisfied.

Got: All eqns verified w/ explicit computation.

If err: Verify fails → trace back. Common: arith mistakes in extended Euclidean, wrong sign in back-sub, forget reduce mod M in final CRT step.

Check

  • Type ID'd (single, system, exp, inverse)
  • All coefs reduced mod respective moduli
  • ax = b (mod m): gcd(a, m) | b checked before solving
  • Extended Euclidean back-sub verified: a · inverse mod m = 1
  • CRT: pairwise coprime verified before
  • CRT non-coprime: compat checked
  • Euler applied only when gcd(a, m) = 1
  • phi(m) from prime factorization, not guessed
  • Repeated squaring uses mod reduction every step (no overflow)
  • Every sol verified by substitution

Traps

  • CRT w/o coprime check: Standard formula needs pairwise coprime. Non-coprime → wrong answer not error. Check gcd(mi, mj) = 1 first.
  • Wrong inverse: Mi^{-1} mod mi (individual), NOT mod M (product). Most common CRT impl err.
  • Euler when gcd(a, m) > 1: a^{phi(m)} = 1 (mod m) needs gcd = 1. Else theorem doesn't apply, result wrong.
  • Sign errs in extended Euclidean back-sub: Track signs each step. Final inverse may be neg → reduce mod m for positive rep.
  • Overflow in mod exp: Even repeated squaring → intermediate products overflow. Reduce mod m after every mult, not just end.
  • Forget multiple solutions: ax = b (mod m), g = gcd(a, m) > 1, g | b → exactly g incongruent solutions mod m, not just one.

  • analyze-prime-numbers — prime fact needed for phi(m) + verify coprime
  • explore-diophantine-equations — linear Diophantine ax + by = c equiv to linear congruence ax = c (mod b)
  • prove-geometric-theorem — mod arith in constructibility proofs (which regular n-gons are constructible)

Repositorio GitHub

pjt222/agent-almanac
Ruta: i18n/caveman-ultra/skills/solve-modular-arithmetic
0
agentsagentskillsai-assisted-developmentclaude-codeskillsteams
FAQ

Frequently asked questions

What is the solve-modular-arithmetic skill?

solve-modular-arithmetic is a Claude Skill by pjt222. Skills package instructions and resources that Claude loads on demand, so Claude can perform solve-modular-arithmetic-related tasks without extra prompting.

How do I install solve-modular-arithmetic?

Use the install commands on this page: add solve-modular-arithmetic to Claude Code as a plugin, or clone its repository into your skills directory, then restart Claude so it picks up the skill.

What category does solve-modular-arithmetic belong to?

solve-modular-arithmetic is in the Other category, tagged general.

Is solve-modular-arithmetic free to use?

Yes. solve-modular-arithmetic is listed on AIMCP and free to install. It runs inside Claude, so no separate service account is required to use the skill itself.

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