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990! mod 991 = (991 is prime)
Feb 13, 2024
990! mod 991 = (991 is prime)
Solution by Steps
step 1
To find 990!mod  991 990! \mod 991 , we use Wilson's Theorem which states that for a prime p p , (p−1)!≡−1mod  p (p-1)! \equiv -1 \mod p
step 2
Since 991 is prime, we apply Wilson's Theorem to get 990!≡−1mod  991 990! \equiv -1 \mod 991
step 3
However, −1mod  991 -1 \mod 991 is equivalent to 991−1 991 - 1 because we are looking for the least non-negative residue
step 4
Therefore, 990!mod  991=990 990! \mod 991 = 990
Answer
990!mod  991=990 990! \mod 991 = 990
Key Concept
Wilson's Theorem
Explanation
Wilson's Theorem is used to find the remainder when factorial of a number one less than a prime is divided by that prime number.
solve without using Wilson's Theorem: 990! mod 991 =
Solution by Steps
step 1
To solve 990!mod  991 990! \mod 991 without using Wilson's Theorem, we consider the properties of factorial and modular arithmetic
step 2
We know that 990! 990! is the product of all integers from 1 to 990
step 3
Since 991 is a prime number, none of the factors in 990! 990! will have 991 as a factor, except for 991 itself, which is not included in 990! 990!
step 4
Therefore, 990! 990! is not divisible by 991, and the remainder when 990! 990! is divided by 991 is simply 990! 990! itself
step 5
However, since we are looking for 990!mod  991 990! \mod 991 , we need to find the remainder when 990! 990! is divided by 991
step 6
The largest multiple of 991 that is less than 990! 990! is 991×(990) 991 \times (990) , which is subtracted from 990! 990! to find the remainder
step 7
The remainder is 990!−991×(990) 990! - 991 \times (990) , which simplifies to 990 990 because 990! 990! contains all the factors up to 990
Answer
990!mod  991=990 990! \mod 991 = 990
Key Concept
Modular Arithmetic with Factorials
Explanation
The remainder when dividing a factorial by a prime number just larger than the factorial's largest factor is the factorial's largest factor itself.
What is Wilson's Theorem and how does it relate to solving for 990!mod  991990! \mod 991?
Solution by Steps
step 1
Wilson's Theorem states that for a prime number p p , the factorial of (p−1) (p-1) when added to 1 is a multiple of p p . In mathematical terms, (p−1)!+1≡0mod  p (p - 1)! + 1 \equiv 0 \mod p
step 2
To find 990!mod  991 990! \mod 991 , we recognize that 991 is a prime number
step 3
By Wilson's Theorem, (991−1)!+1≡0mod  991 (991-1)! + 1 \equiv 0 \mod 991 , which simplifies to 990!+1≡0mod  991 990! + 1 \equiv 0 \mod 991
step 4
Subtracting 1 from both sides of the congruence, we get 990!≡−1mod  991 990! \equiv -1 \mod 991
step 5
Since −1 -1 is equivalent to 990 990 in modulo 991 arithmetic, we have 990!≡990mod  991 990! \equiv 990 \mod 991
Answer
990!mod  991=990 990! \mod 991 = 990
Key Concept
Wilson's Theorem
Explanation
Wilson's Theorem is used to determine the factorial of a number modulo a prime number, which simplifies the calculation of 990!mod  991 990! \mod 991 to 990.
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