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secondary 4 | A Maths
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please help! (wilson’s theorem)
Date Posted: 3 years ago
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25! = 25 × 24 × 23 × 22 × ... × 3 × 2 × 1
= (23 + 2)(23 + 1)(23)(22!)
We know from Wilson's Theorem that for a natural number bigger than 1 :
If and only if the sum of the integers smaller than a number is 1 less than a multiple of that number, the number is prime.
Now 23 is prime.
So let 22! = 23n - 1, where n is some positive integer.
Then, 25! = 25 × 24 × 23 × 22 × ... × 3 × 2 × 1
= (25 × 24)(23)(22!)
= 600(23)(23n - 1)
= 23(23(600n) - 600)
Now we need to get rid of the negative sign as r is positive.
Since 600 ÷ 23 = 26 R2, 600 = 23 × 26 + 2. We can rewrite 600 as : 23 × 27 - 23 + 2 = 23 × 27 - 21
Then the expression becomes :
= 23(23(600n) - (23 × 27 - 21))
= 23(23(600n) - 23 × 27 + 21)
= 23(23(600n - 27) + 21)
Comparing this to 23ᵏ(23q + r), we deduce that:
k = 1, q = 600n - 27, r = 21
= (23 + 2)(23 + 1)(23)(22!)
We know from Wilson's Theorem that for a natural number bigger than 1 :
If and only if the sum of the integers smaller than a number is 1 less than a multiple of that number, the number is prime.
Now 23 is prime.
So let 22! = 23n - 1, where n is some positive integer.
Then, 25! = 25 × 24 × 23 × 22 × ... × 3 × 2 × 1
= (25 × 24)(23)(22!)
= 600(23)(23n - 1)
= 23(23(600n) - 600)
Now we need to get rid of the negative sign as r is positive.
Since 600 ÷ 23 = 26 R2, 600 = 23 × 26 + 2. We can rewrite 600 as : 23 × 27 - 23 + 2 = 23 × 27 - 21
Then the expression becomes :
= 23(23(600n) - (23 × 27 - 21))
= 23(23(600n) - 23 × 27 + 21)
= 23(23(600n - 27) + 21)
Comparing this to 23ᵏ(23q + r), we deduce that:
k = 1, q = 600n - 27, r = 21
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