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I am trying to verify directly using modular arithmetic that $|M|/(40*41) \equiv 1$ mod $41$, where $M$ is the monster group. I am not really sure how to approach this type of problem, i.e. one involving modular arithmetic in this way. My only idea so far is to use some divisibility rules for 41, but I think that there is a better way to do this, or at least one that uses modular arithmetic more explicitly. I would greatly appreciate help with this. Thank you.

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  • $\begingroup$ Well I don't know how to do this either, but I do see that you can take the $41$ out of the denominator in your congruence because $41\equiv 1\text{ mod }40$ so $41^{-1}\equiv 1\text{ mod }40$ as well. $\endgroup$ – rayradjr Nov 12 '12 at 4:42
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If you want to do this with a bit of pencil work, then a trick that may come in handy is the following. It is easy to check that $10^5\equiv 1\pmod{41}$. Therefore for all integers $a,b$ we have the reduction rule $$ 10^5a+b\equiv a+b\pmod{41}. $$

Iterating this gives you for example that $$ \begin{aligned} &808017424794512875886459904961710757005754368000000000\\ \equiv&8080+17424+79451+28758+86459+90496+17107+57005+75436+80000+0\pmod{41}. \end{aligned} $$ You can use this, if you know the decimal expansion of the order of the monster group divided by $40\cdot41$.

Alternatively (assuming you only have the factorization of the order of the monster) you can try to find partial products that give you $\equiv\pm1\pmod{41}$. For example $31\equiv-10$, so $2^2\cdot31\equiv-40\equiv1$, $2\cdot3\cdot7=42\equiv1$, $47\equiv6\implies 47\cdot7\equiv42\equiv1$ et cetera.

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what's the problem just compute it

? Mod(808017424794512875886459904961710757005754368000000000/(40*41),41)
% = Mod(1, 41)
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  • $\begingroup$ If you're going to say just compute it, perhaps you should mention which program you used. $\endgroup$ – Graphth Nov 12 '12 at 17:39
  • $\begingroup$ @Graphth, PARI/GP pari.math.u-bordeaux.fr it's good free software $\endgroup$ – sperners lemma Nov 12 '12 at 17:41

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