How do I solve $(68!+68^{86} ) \mathrm{ \ mod \ }75$?

Question

I feel like it's really easy and i'm just misunderstanding something. It's obvious that I need to use Wilson' and Fermat's theorems:

$(p-1)!≡-1\mathrm{ \ mod \ }p$
$a^{p-1}≡1\mathrm{ \ mod \ }p$ if $p$ does not divide $a$.

i figured i might divide $75$ (since it's not prime) into $5^2$ and $3$ and get the system:

$68!\mathrm{ \ mod \ }3+68^{86}\mathrm{ \ mod \ }3$

$68!\mathrm{ \ mod \ }5+68^{86}\mathrm{ \ mod \ }5$

Am i going in the right direction?

You can do it separately. First see that $68^{86}\equiv 49\bmod 75$, see here. — Dietrich Burde, Jun 21 '22 at 15:12
First off, do you see that $68!$ is just an obfuscation. $75 = 3\cdot 25$ and $3< 68$ and $25 < 68$ so $3\cdot 25\mid 68!$ so $68! \equiv 0 \pmod{75}$ and so all you need to do is solve $68^{86}\pmod {75}$. .... And as $68$ and $75$ are relatively prime you can use Eulers theorem. Tossing in a bit of Chinese Remainder theorem should finish it. — fleablood, Jun 21 '22 at 15:13
Thank you a lot!!! I forgot about Eulers... I appreciate it ! — mnogoetashka, Jun 21 '22 at 15:22
There are no unknowns, and therefor nothing to solve, you need to calculate something. — Henrik supports the community, Jun 21 '22 at 15:22
$68! = 1 ... 3 ... 25 ... 68 = 75k$, $68^{86}=(75-7)^{86}=75m+7^{86}$, $7^4=2401=75\cdot 32+1$. $7^{86}=(7^4)^{22}\cdot 7^2$. — Ivan Kaznacheyeu, Jun 21 '22 at 15:56

score 1 · Answer 1 · answered Jun 21 '22 at 17:02

There are several approaches and your approach seems essentially fine, except that knowing $68^{86} \bmod 3$ and $68^{86} \bmod 5$ is not sufficient to obtain $68^{86} \bmod 75$. Instead, you need $68^{86} \bmod 3$ and $68^{86} \bmod 25$.

Fermat's theorem says $68^{4}\equiv 1 \pmod 5$ , but you need more precise evaluation. You need to calculate $68^{4} \equiv 5a+1 \pmod {25}$ to find $68^{4k} \equiv 5ak+1 \pmod {25}$ using binomial expansion. In this case $68^{4} \equiv 1 \pmod {25}$, so it is rather easy.

score 1 · Answer 2 · answered Jun 21 '22 at 17:12

Note that $3\mid 68! $ and $ 25\mid 68! $. So $68!\equiv 0$(mod 75).Now for the second part Fermat's little theorem is of no use as 75 is not a prime,here you need to use the Euler's totient function.You can find it here .

We have $\phi (75)=(25-5)(3-2)=40$

Now by Euler's formula $68^{\phi (75)}=68^{40}\equiv 1 $(mod 75) $\Rightarrow 68^{80}\equiv 1$(mod 75).

Also $$68\equiv -7(mod 75)$$ $$\Rightarrow 68^6\equiv 7^6(mod75)$$ $$\Rightarrow 68^6\equiv 343^2\equiv 43^2\equiv 1849\equiv 49(mod 75)$$ $$\Rightarrow 68^{86}\equiv 68^{80}\times 68^6\equiv 49(mod 75)$$ So the whole thing is equivalent to 49 mod 75

How do I solve $(68!+68^{86} ) \mathrm{ \ mod \ }75$?

2 Answers2