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(p-1)! = -1(mod p) | (p-1)! = -1(mod p) | ||
-Sarah | -Sarah | ||
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| + | I'm still a little unsure how to use Wilson's Theorem here, can anyone help? | ||
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| + | ---- | ||
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| + | Yep... | ||
| + | Since 101 is prime, by wilson's thm. 100! mod 101 = 100 mod 101. | ||
| + | |||
| + | (100*99) * 98! = 100<br /> | ||
| + | 98! = 1/99 = -1/2 (because -2 mod 101 = 99 mod 101)<br /> | ||
| + | 98! = 100 * 1/2(because -1 mod 101 = 100 mod 101)<br /> | ||
| + | 98! = 100 * 51 (because 51 * 2 = 1)<br /> | ||
| + | 98! = 5100 mod 101 = 50 | ||
Latest revision as of 13:23, 5 November 2008
this theorem is helpful
http://en.wikipedia.org/wiki/Wilson's_Theorem
I got 50, can anybody confirm? ________________________ Yes, use Wilson's Theorem
(p-1)! = -1(mod p) -Sarah
I'm still a little unsure how to use Wilson's Theorem here, can anyone help?
Yep... Since 101 is prime, by wilson's thm. 100! mod 101 = 100 mod 101.
(100*99) * 98! = 100
98! = 1/99 = -1/2 (because -2 mod 101 = 99 mod 101)
98! = 100 * 1/2(because -1 mod 101 = 100 mod 101)
98! = 100 * 51 (because 51 * 2 = 1)
98! = 5100 mod 101 = 50
