Number Theory - Wilson's Thm

[Jamers328]

a. Find the remainder when 15! is divided by 17.
b. Find the remainder when 2(26!) is divided by 29.

a.
16! = -1 (mod 17)
16! = 16 (mod 17)
15! = 1 (mod 17)

Is the answer 1? :?

b.
28! = -1 (mod 29)
27! = 1 (mod 29)

I don't know where to go from here... I need to get to 26! obviously, I know that.

 

[daon]

a. Find the remainder when 15! is divided by 17.
b. Find the remainder when 2(26!) is divided by 29.

a.
16! = -1 (mod 17)
16! = 16 (mod 17)
15! = 1 (mod 17)

Is the answer 1? :?

b.
28! = -1 (mod 29)
27! = 1 (mod 29)

I don't know where to go from here... I need to get to 26! obviously, I know that.

Your answer for (a) is correct.

For (b) notice that $\displaystyle 2(27!)= (27)(2)(26!)\,\, (mod \,\, 29) = 2 \,\, (mod \,\, 29)$.
Also, $\displaystyle 2 \,\, (mod \,\, 29) = -27 \,\, (mod \,\, 29)$.

 

[Jamers328]

a. Find the remainder when 15! is divided by 17.
b. Find the remainder when 2(26!) is divided by 29.

a.
16! = -1 (mod 17)
16! = 16 (mod 17)
15! = 1 (mod 17)

Is the answer 1? :?

b.
28! = -1 (mod 29)
27! = 1 (mod 29)

I don't know where to go from here... I need to get to 26! obviously, I know that.

Your answer for (a) is correct.

For (b) notice that $\displaystyle 2(27!)= (27)(2)(26!)\,\, (mod \,\, 29) = 2 \,\, (mod \,\, 29)$.
Also, $\displaystyle 2 \,\, (mod \,\, 29) = -27 \,\, (mod \,\, 29)$.

Thanks!

:? so what about just 2(26!)?

 

[daon]

Thanks!

:? so what about just 2(26!)?

Well, if $\displaystyle (27)(2(26!))\,\, (mod \,\, 29) \,\, = \,\, -27 \,\, (mod \,\, 29)$, what can you say about $\displaystyle 2(26!) \,\, (mod \,\, 29)$?

 

[Jamers328]

Thanks!

:? so what about just 2(26!)?

Well, if $\displaystyle (27)(2(26!))\,\, (mod \,\, 29) \,\, = \,\, -27 \,\, (mod \,\, 29)$, what can you say about $\displaystyle 2(26!) \,\, (mod \,\, 29)$?

would it be 2(26!) = -1 (mod 29) by dividing by 27??

 

[daon]

Exactly.

 

[Jamers328]

Awesome, thank you so much for your help.