Number Theory - Wilson's Thm

Jamers328

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Sep 20, 2007
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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.
 
Jamers328 said:
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, \(\displaystyle 2 \,\, (mod \,\, 29) = -27 \,\, (mod \,\, 29)\).
 
daon said:
Jamers328 said:
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!)?
 
Jamers328 said:
Thanks!

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

Well, if \(\displaystyle (27)(2(26!))\,\, (mod \,\, 29) \,\, = \,\, -27 \,\, (mod \,\, 29)\), what can you say about \(\displaystyle 2(26!) \,\, (mod \,\, 29)\)?
 
daon said:
Jamers328 said:
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??
 
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