Show that $1000000! \equiv 500001 \mod 1000003$

Question

I'm working in the following exercise:

Show that $1000000! \equiv 500001 \mod 1000003$

Trying to find a way to apply Wilson's theorem I'm trying the following:

\begin{align*} 1000002! &\equiv -1 \mod 1000003!\\ 1000002\cdot1000001! &\equiv -1 \mod 1000003!\\ 1000002\cdot1000001\cdot1000000! &\equiv -1 \mod 1000003!\\ (-1)\cdot1000001\cdot1000000! &\equiv -1 \mod 1000003! \end{align*}

This is as close as I've been able to be to the exercise, I don't know what path to follow to reach that $ 500001 $, any hint or help will be greatly appreciated.

All of those should be mod $1000003$ - not mod the factorial. — jmerry, Feb 04 '19 at 06:43
It should also be clear that one should mention the fact that $1000003$ is a prime. - For example, with an additional digit, we have $10000000!\equiv 0\pmod {10000003}$ instead — Hagen von Eitzen, Feb 04 '19 at 06:45
$1000002! \equiv \pmod{100003}$ is not that exciting a remark. It doesn't need an exclamation point. — fleablood, Feb 04 '19 at 06:49

Arthur · Answer 1 · 2019-02-04T06:45:06.113

2

Hint: $$500\,001\cdot 2\equiv -1$$

edited Feb 04 '19 at 06:45

answered Feb 04 '19 at 06:42

Arthur

199,419

It's 1 000 003, not 100 003. – jmerry Feb 04 '19 at 06:43
@jmerry I just recounted the zeroes and saw that. – Arthur Feb 04 '19 at 06:45

score 2 · Accepted Answer · edited Feb 04 '19 at 07:34

2

As you already noticed, $$2\cdot1000000!\equiv1000000!\cdot1000001\cdot1000002=1000002!\equiv-1\pmod{1000003}$$

Now, the inverse of 2 $\pmod{1000003}$ is $\frac{1000004}{2}=500002$. So, $$1000000!\equiv-1\cdot500002=-500002\equiv500001\pmod{1000003}$$

edited Feb 04 '19 at 07:34

abc...

4,904

answered Feb 04 '19 at 07:02

Jonas De Schouwer

535
4
11

score 1 · Answer 3 · answered Feb 04 '19 at 06:42

1

The next step? $1000001\equiv -2\mod 1000003$. Now, you just have to divide $-1$ by $(-1)\cdot (-2)=2$.

answered Feb 04 '19 at 06:42

jmerry

19,403

Show that $1000000! \equiv 500001 \mod 1000003$

3 Answers3