Find the remainder when 33! is divided by 37.
Please give full solution.
If u is a primitive element of Z_{37}, then:
u^(36/2) = -1
We thus have that:
36! = Product from k = 0 to 36 of u^k = u^(36*37/2) = (-1)^37 = -1.
So, we have:
33! = -34^(-1) 35^(-1)36^(-1) =
2^(-1) 3^(-1)
2^(-1) = 1/2 = 38/2 = 19
3^(-1) = 1/3 = 75/3 = 25
------>
33! = 19*25 = 31
Sure, let me calculate that for you.
*Calculating...*
Hey, while I was calculating, a clown walked in and told me a joke. Want to hear it?
Why don't scientists trust atoms?
Because they make up everything!
*Ahem* Okay, let's get back to the math.
According to Wilson's theorem, if p is a prime number, then (p-1)! ≡ -1 (mod p).
So, we can rewrite 33! ≡ (37-1)! ≡ -1 (mod 37).
Thus, the remainder when 33! is divided by 37 is -1.
I hope that helps! And remember, laughter is the best remainder.
To find the remainder when 33! is divided by 37, we can use Wilson's theorem.
Wilson's theorem states that if p is a prime number, then (p - 1)! ≡ -1 (mod p).
In this case, p = 37, so we have (37 - 1)! ≡ -1 (mod 37).
Simplifying, we get 36! ≡ -1 (mod 37).
Now, we can break the factorial down into smaller parts.
36! = 36 * 35 * 34 * ... * 3 * 2 * 1
Notice that 37 is a prime number, so all numbers from 1 to 36 are relatively prime to 37 (meaning they have no common factors other than 1).
Therefore, we can write the congruence as:
(36 * 35 * 34 * ... * 3 * 2 * 1) ≡ -1 (mod 37)
Now, we can take the product of the terms on the left-hand side:
36! ≡ -1 (mod 37)
(36 * 35 * 34 * ... * 3 * 2 * 1) ≡ -1 (mod 37)
Since all the numbers from 1 to 36 are relatively prime to 37, we can cancel them out:
1 ≡ -1 (mod 37)
Therefore, the remainder when 33! is divided by 37 is 1.
To find the remainder when 33! is divided by 37, we can use Wilson's Theorem and the properties of modular arithmetic.
1. Wilson's Theorem states that for any prime number p, (p-1)! ≡ -1 (mod p). Since 37 is a prime number, (37-1)! ≡ -1 (mod 37).
2. Expand (37-1)! = 36! = 36 × 35 × 34 × ... × 2 × 1.
3. Notice that 37 is one more than a multiple of 36, so we can rewrite (37-1)! as 37 × (36!) ≡ -1 (mod 37).
4. Now, let's find the remainder when (36!) is divided by 37.
a) Consider the numbers 1, 2, 3, ..., 36.
b) We can pair each number with its modular inverse modulo 37.
For example:
1 is paired with 1
2 is paired with 19 (because 2*19 ≡ 1 (mod 37)).
3 is paired with 25 (because 3*25 ≡ 1 (mod 37)).
...
36 is paired with 34 (because 36*34 ≡ 1 (mod 37)).
Notice that every number in this list has a modular inverse except for 1 and 37.
c) Each pair multiplies to 1 modulo 37, so their product is congruent to 1 modulo 37.
Thus, (36!) ≡ 1 (mod 37).
5. We can now substitute this result in step 3:
37 × (36!) ≡ -1 (mod 37).
Since (36!) ≡ 1 (mod 37), we have:
37 × 1 ≡ -1 (mod 37).
This simplifies to:
37 ≡ -1 (mod 37).
6. To find the remainder when 33! is divided by 37, we can rewrite 33! as:
33 × 32 × 31 × ... × 3 × 2 × 1.
7. Notice that 33! can be expressed as:
33! = 37 × (32! ≡ -1 (mod 37)).
This is because we have 37 as a factor, and the remaining factors (32!) will cancel out the -1.
8. Thus, the remainder when 33! is divided by 37 is -1, which is equivalent to 37 - 1 = 36.
Therefore, the remainder when 33! is divided by 37 is 36.