multiply different residue classes

Question

I am trying to solve a problem. Proof that $300^{3000} \equiv 1 \bmod 1001$ So after a bit of puzzeling I found that $1001 = 7*11*13$ and I have prooved that:

• $300^{3000} \equiv 1 \;(\bmod 7\;)$
• $300^{3000} \equiv 1 \;(\bmod 11\;)$
• $300^{3000} \equiv 1 \;(\bmod 13\;)$

Now I wish to conclude that therefore $300^{3000} \equiv 1 \bmod 1001$

But that hinges on the assumption that I can multiply residue classes somehow like:

$[1]_7 * [1]_{11} * [1]_{13} = [1]_{1001}$ or more general maybe: $[a]_p * [b]_q * [c]_{13} = [abc]_{pqr}$

Now we have some things that may help such as the factors $p,r,r$ being (co)prime in this case... but exaclty based on what can we draw such a conclusion ?

I was thiking of that somehow we can use Bezout, which guarantees that if $gcd(p,q)=1 => (\exists x,y \in \mathbb{Z})\;(px+qy=1)$

but after scribbling some papers full of almosts...I want to ask you guys for some direction. Thanks!

Answer 1

Since $1$ is a solution to all three congruences, the Chinese remainder theorem tells us that it is the unique solution $\pmod{1001}$.

Answer 2

If you have a calculator, here's a long way to do it. $$300^{3000}\equiv(-90)^{1500}\equiv92^{750}\equiv456^{375}\equiv(456^2)^{187}\cdot 456\equiv((-272)^2)^{93}\cdot(-272)\cdot456\equiv(-90)^{93}\cdot92$$$$(-90)^{93}\cdot92\equiv(90^2)^{46}\cdot90\cdot92\equiv92^{46}\cdot272\equiv(456^2)^{11}\cdot456\cdot272\equiv -272^{12}\cdot456\equiv-90^6\cdot456$$$$-90^6\cdot456\equiv-92^3\cdot456\equiv(-90)\cdot456\equiv1$$

Answer 3

You can't multiply the residue classes like that. If all three were congruent to $2$, the answer would be $2$, not $8$. As others have said, the Chinese Remainder Theorem is the thing. But you can proceed as follows. Let $x=300^{3000}$. Then $x\equiv 1 \pmod{7}$ tells us that $x=7r+1$ for some integer $r$. Put this in the second congruence to see that

$$x = 7r+1 \equiv 1 \pmod{11}.$$

Solve this for $r$ in the usual way to get $r\equiv 0 \pmod{11}$, which is to say $r=11s$ for some integer $s.$ Then we have $x=7(11s)+1$. Plug this into the last congruence to get

$$x=77s+1 \equiv 1\pmod{13}.$$

Solve this in the usual way to get $s\equiv 0 \pmod{13}$, or $s=13t$ for some integer $t$. Now you have

$$x = 77(13t)+1 =1001t+1 \equiv 1 \pmod{1001}.$$

Answer 4

The inference follows immediately from the $\rm\color{#c00}{universal}\,$ property of $\,{\rm lcm}$ = least common multiple, viz.

$\ x\equiv a\pmod{\!k,m,n}\!\iff\!$ $\smash[t]{\overbrace{k,m,n\mid x\!-\!a\!\color{#c00}\iff\! \ell := {\rm lcm}(k,m,n)\mid x\!-\!a}}$ $\!\iff\! x\equiv a\pmod{\!\ell}$

So in the OP we have $\ x\equiv 1\pmod{\!7,11,13}\iff x\equiv 1\pmod{7\cdot 11\cdot 13}$

Remark $ $ Alternatively we can use CCRT = Constant-case of CRT (Chinese Remainder Theorem), which is equivalent to the above $\,{\rm lcm}\,$ property.