Remainder of $35! $ when divided by $41$

Question

I use $p=41 $ is a prime, so

$\begin{align} (41-1)! &\equiv -1 \mod 41\\ (40)! &\equiv -1 \mod 41\\ 40\times39\times38\times37\times36\times(35)! &\equiv -1 \mod 41\\ (-1)\times(-2)\times(-3)\times(-4)\times(-5)\times(35)! &\equiv -1 \mod 41\\ 120(35)! &\equiv +1 \mod 41\\ (-3)(35)! &\equiv +1 \mod 41\\ (3)(35)! &\equiv -1 \mod 41 \\ (3)(35)! &\equiv -42 \mod 41 \ \to \div 3 \\(35)! &\equiv -14\equiv 27 \mod 41 \\ \end{align}$

IS my work true ? can anyone says another idea ?

Yes, it is correct. You used a special case of the Wilson Reflection Formula — Bill Dubuque, Sep 05 '17 at 20:12
@miracle173 That CAS does not tell him whether the passages are correct or not. — , Sep 05 '17 at 20:28
you might want to explain how you computed the inverse of $3$ in $\mathbb Z/41 \mathbb Z$. — Gabriel Romon, Sep 05 '17 at 20:30
@LeG He used $,3x \equiv 3(-14),\Rightarrow, x\equiv -14,,$ which is true for any modulus coprime to $3$ (since then, by Bezout, $3$ is invertible so cancellable). $\ \ $ — Bill Dubuque, Sep 05 '17 at 20:34
@G.Sassatelli Doesn't it? I think each line can be verified by the CAS. What "passage" can't be checked by the CAS? — miracle173, Sep 05 '17 at 20:37
Do you realize that putting {curly braces} around {\equiv} is why you didn't have proper spacing? And using \text{mod } rather than \mod or \bmod is why you had to add spacing manually? — Michael Hardy, Sep 05 '17 at 21:23
Personally I prefer \pmod{41} for formulas like this, but perhaps not everyone expects to see parentheses there. — David K, Sep 05 '17 at 21:28
Updated link to the Wilson Reflection Formula (prior thread was deleted) — Bill Dubuque, Dec 19 '19 at 17:51

score -1 · Answer 1 · answered Sep 06 '17 at 04:45

-1

with a simple division, $$35!\equiv 27(mod41)$$

answered Sep 06 '17 at 04:45

ProblemSolver1

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Remainder of $35! $ when divided by $41$

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