1. ## nonegative residu

Find the least nonnegative residue of 2(100!) modulo 103

This is what I have so far but dont know where to go form here:

2(100!)x≡1 mod 103
2(100!)x≡104 mod 103
100!x≡52 mod 103

2. Hello,
Originally Posted by rmpatel5
Find the least nonnegative residue of 2(100!) modulo 103

This is what I have so far but dont know where to go form here:

2(100!)x≡1 mod 103
2(100!)x≡104 mod 103
100!x≡52 mod 103
I don't understand what x≡ is o.O

Okay, by Wilson's theorem, you know that $(102)! \equiv -1 (\bmod 103)$

$(102)!=(100)! \cdot 101 \cdot 102=2(100)! \cdot (101 \times 51)$

$101 \times 51=(103-2) \times 51=103 \times 51 -2 \times 51 \equiv -102 (\bmod 103) \equiv 1 (\bmod 103)$

Hence $(102)!=2(100)! \cdot (101 \times 51) \begin{array}{ll} \equiv 2(100)! (\bmod 103) \\ \equiv -1 (\bmod 103) \end{array}$

-1 mod 103 = 102 mod 103. this is the least nonnegative residue.

3. Originally Posted by Moo
Hello,

I don't understand what x≡ is o.O

Okay, by Wilson's theorem, you know that $(102)! \equiv -1 (\bmod 103)$

$(102)!=(100)! \cdot 101 \cdot 102=2(100)! \cdot (101 \times 51)$

$101 \times 51=(103-2) \times 51=103 \times 51 -2 \times 51 \equiv -102 (\bmod 103) \equiv 1 (\bmod 103)$

Hence $(102)!=2(100)! \cdot (101 \times 51) \begin{array}{ll} \equiv 2(100)! (\bmod 103) \\ \equiv -1 (\bmod 103) \end{array}$

-1 mod 103 = 102 mod 103. this is the least nonnegative residue.
I understand what you did by Wilson's theorem and then breaking down the factorial of 102 to 2(100!)101*51. But after that i am lost to what your doing. I dont know why your doing anything.

4. Originally Posted by rmpatel5
I understand what you did by Wilson's theorem and then breaking down the factorial of 102 to 2(100!)101*51. But after that i am lost to what your doing. I dont know why your doing anything.
Yeah I know, it's a real mess...

Keep in mind that we want to find x in ${\color{red}2(100)!} \equiv x (\bmod 103)$

We know that $(102)! \equiv -1 (\bmod 103)$, that is to say $(102)! \equiv 102 (\bmod 103) \quad (1)$

But $(102)!={\color{red}2(100)!} \times (51 \times 101)$

So $\boxed{(102)! \equiv 2(100)! \times (51 \times 101) (\bmod 103)}$

So if I can reduce $51 \times 101 (\bmod 103)$, it would be nice !
101 is 103-2, so writing it this way will make appear 103.
$51 \times 101=51 \times (103-2)=103 \times 51 - 2 \times 51 \equiv 0-2 \times 51 (\bmod 103)$

$-2 \times 51=-102 \equiv 1 (\bmod 103)$

Hence, if we go back to the boxed expression, we'll get :

$(102)! \equiv 2(100)! \times (51 \times 101) (\bmod 103) \equiv 2(100)! \times 1 (\bmod 103)$

But we know that $(102)! \equiv 102 (\bmod 103)$ (1)

Therefore $2(100)! \equiv 102 (\bmod 103)$

$\boxed{x=102}$