Position 1, can choose a book in 50 different ways.
position 2, can choose a book in 49 different ways.
position 3, can choose a book in 48 different ways.
...
position 50, can choose a book in 1 different ways.
By multiplication rule, you can arrange the books by 50! or about 3.04×10⁶⁴ ways
In how many ways can i arrange the 50 books i have
This has to do with combinatorial. (Education on counting) I'll create a new topic.
Let me start with the problem:
If p is a prime number greater than or equal to 5. Prove that "24 | p²-1".
Absolutely correct, you did it, great!
Problem 2:
For all integer x prove that (3x+2)/(4x+3) is irreducible fraction.
Let n be an odd number, there will be some k ∈ Z such that n=2k+1
Then n(n-1)(n+1) = n³-n = (2k+1)³-(2k+1) = 8k³+12k²+6k+1-2k-1 = 8k³+12k²+4k = 4k(2k²+3k+1) = 4k(k+1)(2k+1) ...[*]
Since k ∈ Z, Remainder from dividing k by 3, there are 3 ways: remainders 0, 1, or 2. So I'll divide the cases as follows.
Case 1: k=3q ; q ∈ Z -> [*] = 4k(k+1)(2k+1) = 12q(3q+1)(6q+1) ...[1]
Since q ∈ Z, then q can be an even or odd number.
case 1.1: q=2r ; r ∈ Z -> [1] = 24r(3q+1)(6q+1) ; notice r(3q+1)(6q+1) ∈ Z. So 24 | [1]
case 1.2: q=2r+1 ; r ∈ Z -> [1] = 12q(6r+3+1)(12q+1) = 24q(3r+2)(12q+1) ; notice q(3r+2)(12q+1) ∈ Z. So 24 | [1]
From the two subcase, therefore 24| [*]
Case 2: k=3q+1 ; q ∈ Z -> [*] = 4k(k+1)(2k+1) = 4(3q+1)(3q+2)(6q+3) = 12(3q+1)(3q+2)(2q+1) ...[2]
Since q ∈ Z, then q can be an even or odd number.
case 2.1: q=2r ; r ∈ Z -> [2] = 12(3q+1)(6r+2)(2q+1) = 24(3q+1)(3r+1)(2q+1) ; notice (3q+1)(3r+1)(2q+1) ∈ Z. So 24 | [2]
case 2.2: q=2r+1 ; r ∈ Z -> [2] = 12(6r+3+1)(3q+2)(2q+1) = 24(3r+2)(3q+2)(2q+1) ; notice (3r+2)(3q+2)(2q+1) ∈ Z. So 24 | [2]
From the two subcase, therefore 24| [*]
Case 3: k=3q+2 ; q ∈ Z -> [*] = 4k(k+1)(2k+1) = 4(3q+2)(3q+3)(6q+5) = 12(3q+2)(q+1)(6q+5) ...[3]
Since q ∈ Z, then q can be an even or odd number.
case 3.1: q=2r ; r ∈ Z -> [3] = 12(6r+2)(q+1)(6q+5) = 24(3r+1)(q+1)(6q+5) ; notice (3r+1)(q+1)(6q+5) ∈ Z. So 24 | [3]
case 3.2: q=2r+1 ; r ∈ Z -> [3] = 12(3q+2)(2r+1+1)(6q+5) = 24(3q+2)(r+1)(6q+5) ; notice (3q+2)(r+1)(6q+5) ∈ Z. So 24 | [3]
From the two subcase, therefore 24 | [*]
From all 3 case, it can be concluded that 24 | n(n-1)(n+1) for all n ∈ Z.
Yes, this property still true when n≠0 and divisible by 8.
No it only works for odds it doesnt work for 4 for example
I defined n=2k+1 ; k∈Z from the first line. That means n is an odd number.
No it only works for odds it doesnt work for 4 for example
I defined n=2k+1 ; k∈Z from the first line. That means n is an odd number.
Oh sorry, my conclusion in the last line was wrong. Is only for all n is an odd number.
PROVE BY INDUCTION
Let p(n) represent “1+3+5+7+...+(2n-1) = n²” where n=k.
Base case: for k=1; 1=1² so p(1) is true.
induction step: assume p(k) is true, we will prove p(k+1) is also true.
Since 1+3+5+...+(2k-1) = k². we will get 1+3+5+...+(2k-1)+(2k+1) = k²+(2k+1) = (k+1)²
That mean 1+3+5+...+(2(k+1)-1) = (k+1)². So p(k+1) is also true.
So p(n) is true for all n∈N.
Yes.
1+3+5+...+2n-1 = (1+2+3+...+2n) - (2+4+6+...+2n) = [2n(2n+1)/2] - 2[n(n+1)/2] = n(2n+1)-n(n+1) = n²
I can prove that 1+2+3+...+n = n(n+1)/2 without induction.
Case 1: n is even
1+2+3+...+n = (n+1)+(n-1+2)+(n-2+3)+... [n/2 term] = n(n+1)/2
Case 2: n is odd
1+2+...+(n+1)/2+...+n = (n+1)+(n-1+2)+... [(n-1)/2 term] + (n+1)/2 = (n-1)(n+1)/2 + (n+1)/2 = n(n+1)/2
Induction is kijd of the weapon of a lazy mathematician, yes i use it many times but it doesnt give you much insight into the problem, isnt it, is there a deductive way?
I think induct is a good method. To represent some property of all counting numbers. Its effectiveness is better in proving recurrence relation.
Problem 3:
Find the positive integer m,n that satisfy the following conditions.
M is 360 and N is 15
Almost yes, but with the condition n>15
This is known as Wilson's theorem. "p is prime if and only if, (p-1)! ≡ -1 (mod p)"
Proved to be necessery conditions(->)
p=2 ; (2-1)! = 1 ≡ -1 (mod 2)
p=3 ; (3-1)! = 2 ≡ -1 (mod 3)
for p ≥ 3 ; Since p is a prime number Therefore, all elements in S = {1,2,3,...,p-1} are relative primes to p. So for a∈S there is another b that ab ≡ 1 (mod p)* [ a≡b (mod c) means that c | a-b ]
*Prove: Since the theoretical diophantic equation ab+py = 1, where y is any number, there is always answer b in modulo p, because g.c.d.(a,p) = 1 and there is no more than 1 solution for b for each a. Otherwise, if b₁,b₂ < p such that ab₁ ≡ ab₂ ≡ 1 (mod p), then ab₁-ab₂ = a(b₁-b₂) is evenly divisible by p, which is not possible. There is a conflict, so we can match all a,b (p-1)/2 different pairs from S such that ab ≡ 1 (mod p).
However, if a=b only occurs when a=1 or p-1, we find that 1² ≡ (p-1)² ≡ p²-2p+1 ≡ 1 (mod p) where a∈S-{1,p -1} is a≠b because assuming a=b then a²≡1 (mod p), that is a²-1 = (a-1)(a+1) divisible by p, so a-1 or a+1 Divisible by p is not possible.
Then we can be grouped into pairs whose product thus making (2)(3)(4)...(p-2) ≡ 1 (mod p)
So (p-1)! = 1(p-1)×(2)(3)(4)...(p-2) ≡ (p-1)×1 ≡ p-1 ≡ -1 (mod p) that is p | (p-1)!+1
Proved to be a sufficient condition (<-)
I will first prove that n | (n-1)! when n is a composite number greater than 4.
Prove: in case n is not an absolute square, then n can be written in the form n=a×b where 1 < a≠b < n. We see that a,b ∈ {1,2,3,...,n-1} so a×b = n is a factor of (n-1)!
If n is an absolute square, then assume n=k². When k is a positive integer greater than 2, then 2k < k² = n, so k,2k ∈ {1,2,3,...,n-1} So 2k² and k² are factors of (n-1)!
Returning to the main theory, given p | (p-1)!+1, we assume that p is composite, so p | (p-1)! Therefore there is a conflict, so p must be prime only. as proved.
Find the positive integer m,n that satisfy the following conditions.
Answer: m=120, n=45
Solution: Since gcd(m,n) = 15, then there will be p,q ∈ N such that m=15p, n=15q where the greatest common factor of p and q equals 1 (otherwise gcd(m, n) > 15)
From lcm(m,n) = 360 we get lcm(15p, 15q) = 15pq = 360 so pq=24.
With the condition gcd(p,q)=1, it means that only p=8 and q=3 (if q=1 then n=15 is still inconsistent), then m=15(8)=120, n=15(3)=45.
Problem 4:
Find all the absolute squares in the sequence.
1, 11, 111, 1111, 11111, ... , (10ⁿ-1)/9, ...
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