you can actually prove this purely through the method of finding 3. notice at the end there that the first term had a coefficient of 24. this means that all you need to do is prove that (n-1)n(n+1)(n+2) is a multiple of 8. this can be done by applying the same logic that proves that the product of 3 consecutive numbers is divisible by 3, in a sequence of 4 consecutive numbers, there is always one multiple of 4 (because every 4th number is a multiple of 4). this means that this product of 4 consecutive numbers contains a multiple of three and two even numbers, one of which is also a multiple of 4. (ie. 1x2x3x4 = 24, 4x5x6x7 = 840) edit: yes, this conclusion can also be reached by the consecutive even integers logic mentioned in the proving for 8 part, but this one is more immediately apparent since it uses the logic required to reach this point.
@@dan-florinchereches4892 That's so obvious, but I never really thought this through! I just knew, that the product of (2) consecutive numbers was even, 3 would divisible by 6, and this proves 4 are divisible by 24. (1), 2, 6, 24, ... where have I heard of this sequence? Proof is easy...
That's "nice mathematics", but I must admit, that I may have known that underneath. We know, that n numbers that are consecutive, have exactly one of them divisible by n, and so is the product. But n consecutive numbers are also (n-1) consecutive numbers (n>1), so also (n-1) applies. Divisible by n, n-1, n-2, ..., 2, (1) => by n!
Honestly I enjoy seeing the steps that lead up to the "obvious" conclusion. Yes, it's clear at the end, but to me that's part of what makes the journey so delightful.
@prime I liked your proof about the divisibility of two consecutive even numbers by 8 but that is not needed at all. If you have an expression like this 24n^3(n+2) + (n-1)n(n+1)(n+2) then the first part is obviously divisible by 24 but the second part is a multiplication of 4 consecutive numbers which automatically makes it divisible by 24. That is true because one of the consecutive numbers is divisible by 2, one by 3 and one by 4. Therefore the entire expression is divisible by 24.
Another way to show that the product n(n+2)(5n-1)(5n+1) is divisible by 3: Since n is an integer, it can be in exactly one of the following forms (k is also an integer): Either n = 3k, OR n = 3k+1, and thus n+2 = 3k+3 = 3(k+1), OR n = 3k+2, and thus 5n-1 = 15k + 10 - 1 = 15k + 9 = 3(5k+3). In any case, the product above has a factor that is divisible by 3, and thus the product itself is always divisible by 3. Nice video!
In words, the thesis derives from these two general properties: - if two even numbers differ by 2, either one or the other is divisible by 4. - if I have three consecutive numbers, one of them is divisible by 3.
Classic approach by induction: 1) n=1: Exp(1)=1(1+3)(5(1)-1)(5(1)+1)=1(3)(4)(6)=(3)(24) √ Ok - divisible by 24 2) n=k: Exp(k)=k(k+2)(5k-1)(5k+1) = 25k⁴+50k³-k²-2k = 24m - assumption that Exp(k) is divisible by 24 3) n=k+1: Exp(k+1)=(k+1)(k+3)(5k+4)(5k+6) = 25k⁴+150k³+299k²+246k+72 = (25k⁴+50k³-k²-2k)+100k³+300k²+248k+72 = 24m+4(25k+75k+62k+18) The expression A=25k+75k+62k+18 can be written as 25(k³+3k+2k)+12k+18 i.e. 25k(k+1)(k+2)+6(2k+3). k(k+1)(k+2) is divisible by 3 and by 2 (i.e. by 6) as a product of three consecutive numbers hence A is divisible by 6 i.e. A=6n Exp(k+1)=24m+4A = 24m+4(6)n=24(m+n) is also divisible by 24 - what had to be proven
I did it slightly differently. Noting that (5n-1)(5n+1)=25n^2-1=24n^2+(n-1)(n+1). Expanding the expression gives 24n^3(n+2) + (n-1)(n)(n+1)(n+2). First term is divisible by 24 and second term is divisible by 8 and 3.
n(n+2)(5n-1)(5n+1) n(n+2)(5n-1)(5n+1) * 1 n(n+2)(5n-1)(5n+1) * (5/5) (n+2)(5n-1)(5n)(5n+1) / 5 Since 5 is not divisible by 3, we can conclude that any three consecutive numbers of the form (5n-1)(5n)(5n+1) are divisible by 3. In fact (5n-1)(5n)(5n+1) is divisible by both 3 and 5. (an-1)n(an+1) is divisible by 3 as long as (a) is NOT divisible by 3.
We can see the term (25n - 1)(25n+1) can be written as [24n^2 - (n - 1)(n + 1)], so that the whole expression can be written as to 24n^2*n*(n+2) + (n-1)n(n+1)(n+2), both these terms are divisible by 24 = 4!. It is that the product of N consecutive numbers is divisible by N!| Good video like always
Prove that n(n+2)(5n-1)(5n-1) is divisible by 24. N is a natural number. The product of 2 consecutive even numbers is divisible by 8 let n,n+2 be consecutive even numbers. (2k)(2k+2)=4k^2+4k 4*2m=8m k,m is an integer n(n+2) is divisible by 8 if It’s even.
For the ‘divisible by 3’ part, you could simply say that precisely one of the consecutive numbers (5n-1), (5n) and (5n+1) is divisible by 3. For 5n to be divisible by 3, n must be divisible by 3 (since 5 is not divisible by three). Since n, 5n-1 and 5n+1 are all factors, then the expression is divisible by 3
Solution: (1) modulo 3: n(n + 2)(5n - 1)(5n + 1) = n(n + 2)(2n + 2)(2n + 1) = n(n + 2)*2(n + 1)*2(n + 2) = n(n + 1)(n + 2)^2 = 0 (mod 3), because the last product contains a product of 3 consecutive integers which is always divisible by 3. Therefore, 3 divides the given product. (2) modulo 8: n(n + 2)(5n - 1)(5n + 1) = n(n+2)*5(n - 5)*5(n + 5) = n(n + 2)(n + 3)(n + 5) = 0 (mod 8), because we always have 4 dividing one of the four factors of the last product and 2 dividing *another* one of these factors. Therefore, the last product is divisible by 8, and so is the given product (due to the congruence modulo 8). (3) 3 and 8 are coprime: If we put the results of (1) and (2) together, we arrive at the proposition, q.e.d.
n(n+2)(5n-1)(5n+1) = 24m | m is an integer n(n+2)(5n-1)(5n+1) = 4m without loss of generation (WLOG), since n = 0; 1; 2; 3 it is always divisible by 6, so n + t = 6k, which t and k are integers. By mathematical induction, let's say that this is also true for n + 1: (n+1)(n+3)(5n+4)(5n+6) = 4p 25n⁴+150n³+299n²+246n+72 = 4p n(n+2)(5n-1)(5n+1) = 4m 25n⁴+50n³-n²-2n = 4m (100n³+300n²+248n+72)+(25n⁴+50n³-n²-2n) = 4p 4m + 100n³+300n²+248n+72 = 4p m + 25n³+75n²+62n+18 = p That is the way I did 👍
Solution: n(n + 2)(5n - 1)(5n + 1) n(n + 2)(25n² - 1²) n(n + 2)(24n² + n² - 1²) 24n² * n(n + 2) + (n² - 1²)n(n + 2) 24n² * n(n + 2) is obviously a multiple of 24 (n² - 1²)n(n + 2) can be rewritten as (n - 1)n(n + 1)(n + 2) which is the product of 4 consecutive numbers. Any 4 consecutive numbers contain multiples of 2, 3 and 4. Therefore the product of those numbers can be divided by 2 * 3 * 4 = 24. As such, (n² - 1²)n(n + 2) is a multiple of 24 This proves, that the original term in total is a multiple of 24.
Dear sir, You could have done this with starting @ 12:34 .! Already YOU HAVE PRODUCT OF FOUR consecutive number which is always divisible by 24 . and the other part has a 24 (n^3+2) . no need to prove divisibility for 8 separately. QED.Thanks☺
(5n-1)(5n+1) can be expanded to 24n^2 + (n-1)(n+1). The whole statement then can be expanded to: (n-1)n(n+1)(n+2)] + n.(n+2).(24n^2). First part, 4 consecutive numbers, always divisible by 24, the right part also a multiple of 24.
24=3*8=2^3*3 n*(n+2)*(5n-1)*(5n+1) If either n or n+2 is divisible by 3 that would take care of that, if not n equiv 2 mod 3 so 5n-1 equiv 9 equiv 0 mod 3. Next to 8: if n is even then either n=4m or n=4m+2 so either n or n+2 is a multiple of 4 and we have 2*4=8, if n is odd then n=2x+1 and (5n-1)*(5n+1)=(5(2x+1)-1)*(5(2x+1)+1)=(10x+4)*(10x+6)=100x^2+100x+24=4*(25x^2+25x+6)=4*(25*(x(x+1))+6) now x*(x+1) is even so 25x^2+25x+6 and we have 4*2=8. Proofed.
Suppose 3|(n+1) then 3|(5n+5) subtracting 6, we see that 3|(5n-1) so n(n+2)(5n-1) is divisible by 3 for all integers n let n=2k+1 n+2=2k+2 5n-1=10k+4=2(5k+2) 5n+1=10k+6=2(5k+3) if k is even, 4|(5n-1) and 2|(5n+1) if k is odd, 4|(5n+1) and 2|(5n-1) in both cases, 8|(5n-1)(5n+1) if instead n=2k, 2k(2k+2)=4k(k+1) either k or k+1 will be even, so in the case of n being even, 8|n(n+2) thus, no matter the parity of n, 8|n(n+2)(5n-1)(5n+1) since 3|P and 8|P, and since 3 and 8 are coprime, 24|P.
i proved the divisibility by 8 like you in the video. But i think, the divisibility by 3 can shown a bit easier: one of 3 consecutive numbers n, n + 1, n + 2 is always divisible by 3. If n or n + 2 is divisible by 3, then the products n * (n + 2) and n * (n + 2) * (5n - 1) * (5n + 1) are obiously also divisible by 3. I neither n nor n + 2 is divisible by 3, then n + 1 must be divisible by 3. Also 5n - 1 = 5 * (n+1 - 1) - 1 = 5 * (n+1) - 5 - 1 = 5 * (n+1) - 6. Now (n + 1) and 6 are divisible by 3, so also 5 * (n+1) and 5 * (n+1) - 6 = 5n - 1 Finaly the product n * (n + 2) * (5n - 1) * (5n + 1), if n + 1 is divisible by 3.
You can prove divisibility by 24 just using the 2nd result: (1) first part ( 24 n3 (n+2) ) has a 24 factor (2) second part ( (n-1)n(n+1)(n+2) ) is the product of 4 consecutive numbers, so at least one of them is divisible by 4, at least one is divisible by 3 and at least one is divisible by 2 (but not by 4): 2*3*4=24 and that's all
P = n(n+2)(5n-1)(5n + 1) Is P divisible by 24? = n(n+2)(25n^2 - 1^2) = n(n+2)(24n^2 + n^2 - 1) = [n(n+2)]*[(24n^2) + (n-1)(n+1)] P = 24(n^2)(n)(n+2) + (n-1)(n)(n+1)(n+2). Let Q=24(n^2)(n)(n+2) and R=(n-1)(n)(n+1)(n+2). First, Q=24(n^2)(n)(n+2) == 0 mod 24 by inspection. Also, R=(n-1)(n)(n+1)(n+2) is a product of four consecutive integers. In a set of four consecutive integers, can be found: - one multiple of 4, - another even integer, - and at least multiple of 3: Giving R=(n-1)(n)(n+1)(n+2) == 4*2*3*k = 24k == 0 mod 24. {k: Z} Since Q == 0 mod 24 and R == 0 mod 24, therefore P = Q + R == 0 mod 24. Therefore, P = n(n+2)(5n-1)(5n + 1) is divisible by 24.
If n is even, n or n+2 is divisible by 4. Otherwise 5n-1 or 5n+1 is divisible by 4. So the full product is divisible by 8. n or 5n-1 or 5n+1 is divisible by 3. So the full product is divisible by 24 also.
n(n+2)(25n-1)(25n+1)=n(n+2)(n²-1)Modelo 24= (n-1)((n)(n+1)(n+2)Modelo 24 is the product of 4 consecutive numbers and from it it is read as division by 4! =24
Actually it’s easy to proof that for any n: n*(n+1)*(n+2)*(n+3) is divisible by 2, 3 and 4, so by 24 as well. And obviously 24*n^2 is also divisible by 24.
I stared at this, thinking of applying induction, then realised: If n is even, there is the product of two consecutive even numbers, so the number is divisible by 8. Additionally, either (5n-1) or (5n+1) is divisible by 3, so the expression is divisible by 24. If n is odd, there is the product of two consecutive even numbers thanks to (5n-1) and (5n+1), so the number is divisible by 8. Additionally, either n or (n+2) is divisible by 3, so the expression is divisible by 24.
n(n+2)(5n-1)(5n+1)=n(n+2)(25n^2-1) =n(n+2)(24n^2+n^2-1) =n(n+2)24n^2 + n(n+2)(n^2-1) =24n^3(n+2) + (n-1)n(n+1)(n+2) 24n^3(n+2) is a multiple of 24 and (n-1)n(n+1)(n+2) should have a multiple of 3, a multiple of 4 and an even number (multiple of 2). And multiplying multiple of 2 with multiple of 3 with multiple of 4 will be a multiple of 24 and the some of two multiples of 24 will be a multiple of 24🎉 🎉
Note that in your 24 times blap + (n-1)n(n+1)(n+2) you have 8 divides this mess. First 8|24 and if n is even then you have n(n+2)times rest if n is odd you have (n-1)(n+1) times rest.
Want to show that n(n+2)(5n-1)(5n+1) is divisible by 3? n(n+2)(5n-1)(5n+1)=n(n+2)(25n^2-1)=n(n+2)(24n^2+n^2-1)=n(n+2)(24n^2+(n-1)(n+1)) n(n+2)(5n-1)(5n+1)=25n^4+50n^3-n^2-2n=24n^3(n+2)+(n-1)n(n+1)(n+2) 24n^3(n+2) is divisible by 3 (n-1)n(n+1)(n+2) is divisible by 3. If a/b and a/c then a/b±c if 3 and 8 are factors then 24 are also factors.
Let's call Z 5n+1 times 5n-1 Z = 25n^2 - 1 Let's call U 2n+2 times n U= n^2 + 2n 25n^2-1 can be written as 24n^2 + n^2 - 1, mod 24 = n^2-1. So does ( n^2-1) * (n^2+2n) mod 24= 0 n=1, 0 n=2, 3*8,0 n=3 8*15=8*3=0 n=4 15*24=0 n=5 24*35=0 n=6 35*48,24...=0 n=7, 48*63 n=8 65*80=3*8=24=0 ... The pattern appears to be when n is a power of 2, the left will be divisible by 3, the right 8 Okay so if n is a power of 2, then n^2 = 2^(y*2) 2^(y*2) - 1 is always divisible by 3, because 2^2-1 is 3. (From my exponent divisibility formula) 2^(y*2) + 2y, since y is a power of 2, will always be div 8 past n=8, since y will be divisible by 8 past that point. Check the list above for the manual check of less than n=8. So powers of 2 are covered. Second pattern i notice is that n^2-1, when n is a prime past 3, always divisible by 24. primes past 3 are all p mod 6 =5, or p mod 6 = 1 5^2-1 = 24 mod 6 times both by 4, 24 times 4 mod 24 =0 1^2 - 1 = 0 mod 6, times both by 4, 0 times mod 24 =0 So yes all prime values of n are divisible by 24(Edit, maybe??? is this logic even correct??) So we're left with prime factorizations that are either powers of primes (-2), or mixed prime factorizations. I'm at my limit! Finished your video. Wonderful as always. What a treat!
0:58 ach your intro i also would like to have a such nice video but i am just to lost for it :D anyway sounds interesting! 1:15 that was my second thought after i thought there is an binobic formular at the end 1:50 it is easyer yes but how can you see it now? 1:54 konsekutiv i had to research for the meaning :D but yes it makes sense 2:40 let me try it by my self (n) (n+2) / 8 = a; a, n € N| if n%2 = 0 therfore: n² * 2n = a and n² is an ord number as n is one as well as 2 is one and n it self; so waht we have is the product of 3 ord numbers that means we have the form 2(x) * 2(y) * 2(z) but than we knowe we can move the factorials and we came to 2*2*2 * (xyz) = 2*4*xyz = 8*xyz and therfore we knowe it has to be devideable by 8! right? 3:58 also an option to prove it 4:26 yes i also vorget to tell that x,y,z is an integer as well but i think it is clear :D 5:30 ok now i got your idear :D 7:53 if you don´t like sth. in math just rewrite it :D i like that 8:28 i see 9:05 ok nice idear 9:48 ok yes that is clear and what about the other summant 10:27 yes we knowe: (a + b) % n = 0 if a,b % n = 0; it is logic because lets turn it 3(a+b) % 3 = 0 | (we talk about integer!) --> 3a + 3b % 3 = 0 and now call 3a x and 3b y; x % 3 = 0; y % 3 = 0 ==> x + y % 3 = 0 12:10 :D i understand! 12:24 i have quote this sentencis now about 5 times :D LG K.Furry
n(n+2)(5n-1)(5n+1)=n(n+2)(25n^2-1)=24n^3(n+2)+(n-1)n(n+1)(n+2) obviously， 24n^3(n+2)is divisible by 24 and (n-1)n(n+1)(n+2) is divisible by2 by3 by4 by6 by8 …… and 3 is coprime with 8 so it is divisible by24
@@rainerzufall42 it does make sense since (7n-1)(7n+1)=(49n^2-1) and (49n^2-1) (mod 24) = (n^2-1)=(n-1)(n+1) since 48 is divisible by 24 and rest of the steps are same. Thanks for elegant proof.
@@LakshaySura Next step: n (n + 2) (k n - 1) (k n + 1) = 0 (mod 24) for k = 6 m +/- 1, m € IZ, that's k = -1, 1, 5, 7, 11, 13, 17, 19, ..., because (6 m +/- 1)^2 = 1 (mod 24), nice!
Why (6 m +/- 1)^2 = 36 m^2 +/- 12 m + 1 = 1 (mod 24) ? Well, that's 12 m^2 +/- 12 m + 1 = 12 m (m +/- 1) + 1 (mod 24) and either m or (m +/- 1) is even! Thus it is 1 (mod 24)...
I love your videoes. It helps has to look into deep. Can we also prove that always the product of 4 consecutive numbers are devide by 4, and the product of 5consecutive numbers are devided by 5....... and so on . I checked few steps and it happened right, but no proof
It's actually pretty obvious: If you have any k consecutive integers, don't you think one of them must be divisible by k? Otherwise you would have a gap of size greater than k without a multiple of k, that's impossible! Alternatively, think about congruence classes mod k and the pigeon hole principle. However you do it, if one of the factors is already divisible by k, that applies to the product as well. But it goes much deeper! The product of 4 consecutive integers is always divisible by 24, the product of 5 consecutive integers is always divisible by 120, and so on. In general, product of k consecutive integers is always divisible by k! (k factorial). That's much much harder to prove, though. You can do it by looking at prime factorizations and counting prime factors, by leveraging the properties of binomial coefficients, or by mathematical induction (from first principles, but not simple). To give some hints. (But if you didn't see the first statement, it's probably too hard for you.)
@@Abby-hi4sf Hi Abby, in my previous answer I gave you all the needed ingredients for a complete, rigorous mathematical proof. You just have to formalize it, just a little bit. There wasn't anything missing or handwavy, actually. To help you out, I will show you an example of how formalization could look like. I give you a proof by contradiction and also using the so-called "pigeon-hole principle", two very important proof techniques you should know 😃: Look at an arbitrary sequence of k >= 1 consecutive integers starting at n: n, n+1, ..., n+k-1. Consider their remainders upon division by k: r_0, r_1, ..., r_(k-1). These r_i are integers and all satisfy 0
@@Abby-hi4sf Hi Abby, in my previous answer I gave you all the needed ingredients for a complete, rigorous mathematical proof. You just have to formalize it, just a little bit. There wasn't anything missing or handwavy, actually. To help you out, I will show you an example of how formalization could look like. I give you a proof by contradiction and also using the so-called "pigeon-hole principle", two very important proof techniques you should know 😃: Look at an arbitrary sequence of k >= 1 consecutive integers starting at n: n, n+1, ..., n+k-1. Consider their remainders upon division by k: r_0, r_1, ..., r_(k-1). These r_i are integers and all satisfy 0
Your conclusion is wrong, or at least incomplete. You have to add that if a and b are factors and a is not divisible by b or viceversa then (a*b) Will be a factor. In example. 8 is divisible by 8, and 8 is divisible by 2, but 8 is not divisible by 16.
That's not sufficient, as you can see with a=4 and b=6 both dividing n=12, and they don't divide each other. Their product still does not divide n. The correct sufficient condition is that the gcd(a,b) is 1.
You left out three words and put in an incorrect word. You need to prove that *the product of* three consecutive *integers* is divisible by three. Let n, k, and M be integers. Base case where n = 1: 1(2)(3) = 6, which is divisible by 3. Inductive step Assume it is true for n = k: k(k + 1)(k + 2) = 3M Show it is true for n = k + 1: (k + 1)(k + 2)(k + 3) must also be a multiple of 3. k(k + 1)(k + 2) + *3(k + 1)(k + 2)* = 3M + *3(k + 1)(k + 2)* (k + 1)(k + 2)(k + 3) = 3[M + (k + 1)(k + 2)] This shows that (k + 1)(k + 2)(k + 3) is a multiple of 3. Thus, by the Principle of Mathematical Induction, I have proven that the product of three consecutive integers is divisible by three.
@@robertveith6383 Excellent.👍 It is simpler to argue with congruence classes and congruence relations modulo 3, but OP asked for mathematical induction, and you provided it. What I like is the rigor and level of detail of the proof, it should be very instructive for students. 😃
@@xyz9250 its 4 consecutive numbers, one of them will divide 2 and other exclusively will divide 4. If n-1 divides 2, n+1 will divide 4, similarly for other cases.
I figured this out without having to write something down for once. Divisibility by 8 was the easiest part. If n is even, n and n+2 are two consecutive even integers. That means one is divisible by 4. 4*2 is 8. If n is odd, 5n is odd which means 5n + 1 and 5n-1 are consecutive even integers. I almost over thought divisibility by 3 but it was staring right in front of me. 5n mod 3 is congruent to n mod 3 which means 5n + 1 mod 3 is congruent to n + 1 mod 3. So we have n, n+1, and n+2, which are 3 consecutive integers, meaning one of them is divisible by 3.
The last paragraph doesn’t look right. 5n mod 3 is not congruent to n mod 3 because 5 mod 3 is not 0. Trivially: 1 mod 3 = 1 and 5 mod 3 = 2. (5n+1)mod 3 is congruent to (n+2)mod 3 and (5n-1) mod 3 is congruent to n mod 3, so for example when n =1, the expression is (1)(3)(4)(6) where 1 and 4 are congruent (to 1), 3 and 6 are congruent to 0.
@@GreenMeansGOF It's even better: n consecutive numbers contain a multiple of n! for all natural numbers n>0. Because you are guaranteed to have a multiple of 2, of 3, of 4, ..., a multiple of n in these n factors, the product is 0 (mod n!).
you can actually prove this purely through the method of finding 3. notice at the end there that the first term had a coefficient of 24. this means that all you need to do is prove that (n-1)n(n+1)(n+2) is a multiple of 8. this can be done by applying the same logic that proves that the product of 3 consecutive numbers is divisible by 3, in a sequence of 4 consecutive numbers, there is always one multiple of 4 (because every 4th number is a multiple of 4). this means that this product of 4 consecutive numbers contains a multiple of three and two even numbers, one of which is also a multiple of 4. (ie. 1x2x3x4 = 24, 4x5x6x7 = 840)
edit: yes, this conclusion can also be reached by the consecutive even integers logic mentioned in the proving for 8 part, but this one is more immediately apparent since it uses the logic required to reach this point.
I think it is pretty self explanatory that a product of k consecutive numbers is divisible by k! simply because choose(n,k) is an integer when n>=k
Yeah I noticed that at the end too.
@@dan-florinchereches4892 That's so obvious, but I never really thought this through! I just knew, that the product of (2) consecutive numbers was even, 3 would divisible by 6, and this proves 4 are divisible by 24. (1), 2, 6, 24, ... where have I heard of this sequence? Proof is easy...
That's "nice mathematics", but I must admit, that I may have known that underneath. We know, that n numbers that are consecutive, have exactly one of them divisible by n, and so is the product. But n consecutive numbers are also (n-1) consecutive numbers (n>1), so also (n-1) applies. Divisible by n, n-1, n-2, ..., 2, (1) => by n!
Honestly I enjoy seeing the steps that lead up to the "obvious" conclusion. Yes, it's clear at the end, but to me that's part of what makes the journey so delightful.
@prime I liked your proof about the divisibility of two consecutive even numbers by 8 but that is not needed at all. If you have an expression like this 24n^3(n+2) + (n-1)n(n+1)(n+2) then the first part is obviously divisible by 24 but the second part is a multiplication of 4 consecutive numbers which automatically makes it divisible by 24. That is true because one of the consecutive numbers is divisible by 2, one by 3 and one by 4. Therefore the entire expression is divisible by 24.
Such an awesome insight!
in general: n consecutive numbers is divisible by n!
Another way to show that the product n(n+2)(5n-1)(5n+1) is divisible by 3:
Since n is an integer, it can be in exactly one of the following forms (k is also an integer):
Either n = 3k,
OR n = 3k+1, and thus n+2 = 3k+3 = 3(k+1),
OR n = 3k+2, and thus 5n-1 = 15k + 10 - 1 = 15k + 9 = 3(5k+3).
In any case, the product above has a factor that is divisible by 3, and thus the product itself is always divisible by 3.
Nice video!
In words, the thesis derives from these two general properties:
- if two even numbers differ by 2, either one or the other is divisible by 4.
- if I have three consecutive numbers, one of them is divisible by 3.
Great work, as usual. I particulary like problems in the Number Theory field. Thanks
I love your videos.
Classic approach by induction:
1) n=1: Exp(1)=1(1+3)(5(1)-1)(5(1)+1)=1(3)(4)(6)=(3)(24) √ Ok - divisible by 24
2) n=k: Exp(k)=k(k+2)(5k-1)(5k+1) = 25k⁴+50k³-k²-2k = 24m - assumption that Exp(k) is divisible by 24
3) n=k+1: Exp(k+1)=(k+1)(k+3)(5k+4)(5k+6) = 25k⁴+150k³+299k²+246k+72 = (25k⁴+50k³-k²-2k)+100k³+300k²+248k+72 = 24m+4(25k+75k+62k+18)
The expression A=25k+75k+62k+18 can be written as 25(k³+3k+2k)+12k+18 i.e. 25k(k+1)(k+2)+6(2k+3).
k(k+1)(k+2) is divisible by 3 and by 2 (i.e. by 6) as a product of three consecutive numbers hence A is divisible by 6 i.e. A=6n Exp(k+1)=24m+4A = 24m+4(6)n=24(m+n) is also divisible by 24 - what had to be proven
At 3:20 , why did we "let" it if it's true for all positive consecutive even integers?
Very nicely done!
10:19 All parts are divisible by 24, not only by 3
Pour la divisibilité par 3, j’ai employé la congruence qui prend moins de temps. Nice video !
Et moi, bêtement, la congruence [8]. C'est systématique et hyper propre.
Highly enjoyed ur lesson.
Next: M=E-esinE. What are you’re thoughts on analytical solutions?
I did it slightly differently. Noting that (5n-1)(5n+1)=25n^2-1=24n^2+(n-1)(n+1). Expanding the expression gives 24n^3(n+2) + (n-1)(n)(n+1)(n+2). First term is divisible by 24 and second term is divisible by 8 and 3.
I laughed a lot for "mmmm" "mmmmmmmmm" I already see the answer... you are funny
Jó kis régi magyar matematika feladvány!
n(n+2)(5n-1)(5n+1)
n(n+2)(5n-1)(5n+1) * 1
n(n+2)(5n-1)(5n+1) * (5/5)
(n+2)(5n-1)(5n)(5n+1) / 5
Since 5 is not divisible by 3, we can conclude that any three consecutive numbers of the form (5n-1)(5n)(5n+1) are divisible by 3. In fact (5n-1)(5n)(5n+1) is divisible by both 3 and 5.
(an-1)n(an+1) is divisible by 3 as long as (a) is NOT divisible by 3.
We can see the term (25n - 1)(25n+1) can be written as [24n^2 - (n - 1)(n + 1)], so that the whole expression can be written as to 24n^2*n*(n+2) + (n-1)n(n+1)(n+2), both these terms are divisible by 24 = 4!. It is that the product of N consecutive numbers is divisible by N!|
Good video like always
There is a similar video from Numberphile with Matt Parker, showing that (p^2-1) is divisible by 24, where p is a prime number greater than 3.
Difference of two squares, two consecutive even integers, three consecutive integers but since p is prime greater than three, it's not divisible by 3.
@robertlunderwood Yes, that was the second proof. The first one was rather clumsy.
@@robertlunderwoodVery cool! Thanks.
Prove that n(n+2)(5n-1)(5n-1) is divisible by 24. N is a natural number. The product of 2 consecutive even numbers is divisible by 8 let n,n+2 be consecutive even numbers. (2k)(2k+2)=4k^2+4k 4*2m=8m k,m is an integer n(n+2) is divisible by 8 if It’s even.
not covinced... But 4k^2 + 4k = 4k(k+1) and as either k or k+1 is even then... it follows that n(n+2) is divisible by 8
For the ‘divisible by 3’ part, you could simply say that precisely one of the consecutive numbers (5n-1), (5n) and (5n+1) is divisible by 3. For 5n to be divisible by 3, n must be divisible by 3 (since 5 is not divisible by three).
Since n, 5n-1 and 5n+1 are all factors, then the expression is divisible by 3
Solution:
(1) modulo 3: n(n + 2)(5n - 1)(5n + 1) = n(n + 2)(2n + 2)(2n + 1) = n(n + 2)*2(n + 1)*2(n + 2) = n(n + 1)(n + 2)^2 = 0 (mod 3), because the last product contains a product of 3 consecutive integers which is always divisible by 3. Therefore, 3 divides the given product.
(2) modulo 8: n(n + 2)(5n - 1)(5n + 1) = n(n+2)*5(n - 5)*5(n + 5) = n(n + 2)(n + 3)(n + 5) = 0 (mod 8), because we always have 4 dividing one of the four factors of the last product and 2 dividing *another* one of these factors. Therefore, the last product is divisible by 8, and so is the given product (due to the congruence modulo 8).
(3) 3 and 8 are coprime: If we put the results of (1) and (2) together, we arrive at the proposition, q.e.d.
n(n+2)(5n-1)(5n+1) = 24m | m is an integer
n(n+2)(5n-1)(5n+1) = 4m without loss of generation (WLOG), since n = 0; 1; 2; 3 it is always divisible by 6, so n + t = 6k, which t and k are integers.
By mathematical induction, let's say that this is also true for n + 1:
(n+1)(n+3)(5n+4)(5n+6) = 4p
25n⁴+150n³+299n²+246n+72 = 4p
n(n+2)(5n-1)(5n+1) = 4m
25n⁴+50n³-n²-2n = 4m
(100n³+300n²+248n+72)+(25n⁴+50n³-n²-2n) = 4p
4m + 100n³+300n²+248n+72 = 4p
m + 25n³+75n²+62n+18 = p
That is the way I did 👍
Solution:
n(n + 2)(5n - 1)(5n + 1)
n(n + 2)(25n² - 1²)
n(n + 2)(24n² + n² - 1²)
24n² * n(n + 2) + (n² - 1²)n(n + 2)
24n² * n(n + 2) is obviously a multiple of 24
(n² - 1²)n(n + 2)
can be rewritten as
(n - 1)n(n + 1)(n + 2)
which is the product of 4 consecutive numbers.
Any 4 consecutive numbers contain multiples of 2, 3 and 4. Therefore the product of those numbers can be divided by 2 * 3 * 4 = 24.
As such, (n² - 1²)n(n + 2) is a multiple of 24
This proves, that the original term in total is a multiple of 24.
As a captive of the mighty, i can confirm i got taken away
n(n+2)=8m (n is even) or (5n-1)(5n+1)=8m (n is odd) n(n+2)(5n-1)(5n+1) is divisible by 8.
Dear sir, You could have done this with starting @ 12:34 .! Already YOU HAVE PRODUCT OF FOUR consecutive number which is always divisible by 24 . and the other part has a 24 (n^3+2) . no need to prove divisibility for 8 separately. QED.Thanks☺
(5n-1)(5n+1) can be expanded to 24n^2 + (n-1)(n+1). The whole statement then can be expanded to: (n-1)n(n+1)(n+2)] + n.(n+2).(24n^2). First part, 4 consecutive numbers, always divisible by 24, the right part also a multiple of 24.
Exactly, product of N consecutive numbers is divisible by N factorial.
I think it's safe to take a base case to ensure that the given expression is divisible by "8 & 3" and not "either 8 or 3".
24=3*8=2^3*3
n*(n+2)*(5n-1)*(5n+1)
If either n or n+2 is divisible by 3 that would take care of that, if not n equiv 2 mod 3 so 5n-1 equiv 9 equiv 0 mod 3.
Next to 8: if n is even then either n=4m or n=4m+2 so either n or n+2 is a multiple of 4 and we have 2*4=8, if n is odd then n=2x+1 and (5n-1)*(5n+1)=(5(2x+1)-1)*(5(2x+1)+1)=(10x+4)*(10x+6)=100x^2+100x+24=4*(25x^2+25x+6)=4*(25*(x(x+1))+6) now x*(x+1) is even so 25x^2+25x+6 and we have 4*2=8. Proofed.
Suppose 3|(n+1)
then 3|(5n+5)
subtracting 6, we see that 3|(5n-1)
so n(n+2)(5n-1) is divisible by 3 for all integers n
let n=2k+1
n+2=2k+2
5n-1=10k+4=2(5k+2)
5n+1=10k+6=2(5k+3)
if k is even, 4|(5n-1) and 2|(5n+1)
if k is odd, 4|(5n+1) and 2|(5n-1)
in both cases, 8|(5n-1)(5n+1)
if instead n=2k, 2k(2k+2)=4k(k+1)
either k or k+1 will be even, so in the case of n being even, 8|n(n+2)
thus, no matter the parity of n, 8|n(n+2)(5n-1)(5n+1)
since 3|P and 8|P, and since 3 and 8 are coprime, 24|P.
i proved the divisibility by 8 like you in the video. But i think, the divisibility by 3 can shown a bit easier: one of 3 consecutive numbers n, n + 1, n + 2 is always divisible by 3. If n or n + 2 is divisible by 3, then the products n * (n + 2) and n * (n + 2) * (5n - 1) * (5n + 1) are obiously also divisible by 3. I neither n nor n + 2 is divisible by 3, then n + 1 must be divisible by 3. Also
5n - 1 = 5 * (n+1 - 1) - 1 = 5 * (n+1) - 5 - 1 = 5 * (n+1) - 6. Now (n + 1) and 6 are divisible by 3, so also 5 * (n+1) and 5 * (n+1) - 6 = 5n - 1 Finaly the product n * (n + 2) * (5n - 1) * (5n + 1), if n + 1 is divisible by 3.
You can prove divisibility by 24 just using the 2nd result: (1) first part ( 24 n3 (n+2) ) has a 24 factor
(2) second part ( (n-1)n(n+1)(n+2) ) is the product of 4 consecutive numbers, so at least one of them is divisible by 4, at least one is divisible by 3 and at least one is divisible by 2 (but not by 4): 2*3*4=24 and that's all
That was beautiful
P = n(n+2)(5n-1)(5n + 1) Is P divisible by 24?
= n(n+2)(25n^2 - 1^2)
= n(n+2)(24n^2 + n^2 - 1)
= [n(n+2)]*[(24n^2) + (n-1)(n+1)]
P = 24(n^2)(n)(n+2) + (n-1)(n)(n+1)(n+2).
Let Q=24(n^2)(n)(n+2) and R=(n-1)(n)(n+1)(n+2).
First, Q=24(n^2)(n)(n+2) == 0 mod 24 by inspection.
Also, R=(n-1)(n)(n+1)(n+2) is a product of four consecutive integers.
In a set of four consecutive integers, can be found:
- one multiple of 4,
- another even integer,
- and at least multiple of 3:
Giving R=(n-1)(n)(n+1)(n+2) == 4*2*3*k = 24k == 0 mod 24. {k: Z}
Since Q == 0 mod 24 and R == 0 mod 24, therefore P = Q + R == 0 mod 24.
Therefore, P = n(n+2)(5n-1)(5n + 1) is divisible by 24.
The LastPass part of the development si enough to démonstrate thé 24 ils a commun factor.
In the end, "just expand it"
If n is even, n or n+2 is divisible by 4. Otherwise 5n-1 or 5n+1 is divisible by 4. So the full product is divisible by 8.
n or 5n-1 or 5n+1 is divisible by 3.
So the full product is divisible by 24 also.
Solutie frumoasa!
Also by mathenatical induction
Actually for the part (n-1)n(n+1)(n+2) it’s easy to prove it’s divisible by both3 and 8, hence 24. Your earlier effort on 8 m wasn’t that necessary.
n(n+2)(25n-1)(25n+1)=n(n+2)(n²-1)Modelo 24= (n-1)((n)(n+1)(n+2)Modelo 24 is the product of 4 consecutive numbers and from it it is read as division by 4! =24
Honestly loved it
Simply put n(n+2)(5n-1)(5n+1) == 24n^3(n+2) + (n-1)n(n+1)(n+2) ...
And now
24n^3(n+2) and (n-1)n(n+1)(n+2)
are multiple of 24 :)
Actually it’s easy to proof that for any n: n*(n+1)*(n+2)*(n+3) is divisible by 2, 3 and 4, so by 24 as well. And obviously 24*n^2 is also divisible by 24.
I stared at this, thinking of applying induction, then realised:
If n is even, there is the product of two consecutive even numbers, so the number is divisible by 8. Additionally, either (5n-1) or (5n+1) is divisible by 3, so the expression is divisible by 24.
If n is odd, there is the product of two consecutive even numbers thanks to (5n-1) and (5n+1), so the number is divisible by 8.
Additionally, either n or (n+2) is divisible by 3, so the expression is divisible by 24.
Day 2 of asking you to start making videos on integrals
greetings from Hungary (this question is from Hungary 😀 )
n(n+2)(5n-1)(5n+1)=n(n+2)(25n^2-1)
=n(n+2)(24n^2+n^2-1)
=n(n+2)24n^2 + n(n+2)(n^2-1)
=24n^3(n+2) + (n-1)n(n+1)(n+2)
24n^3(n+2) is a multiple of 24
and (n-1)n(n+1)(n+2) should have a multiple of 3, a multiple of 4 and an even number (multiple of 2). And multiplying multiple of 2 with multiple of 3 with multiple of 4 will be a multiple of 24
and the some of two multiples of 24 will be a multiple of 24🎉 🎉
Good one sir!
Nice one sir!
Note that in your 24 times blap + (n-1)n(n+1)(n+2) you have 8 divides this mess. First 8|24 and if n is even then you have n(n+2)times rest if n is odd you have (n-1)(n+1) times rest.
Want to show that n(n+2)(5n-1)(5n+1) is divisible by 3? n(n+2)(5n-1)(5n+1)=n(n+2)(25n^2-1)=n(n+2)(24n^2+n^2-1)=n(n+2)(24n^2+(n-1)(n+1)) n(n+2)(5n-1)(5n+1)=25n^4+50n^3-n^2-2n=24n^3(n+2)+(n-1)n(n+1)(n+2) 24n^3(n+2) is divisible by 3 (n-1)n(n+1)(n+2) is divisible by 3. If a/b and a/c then a/b±c if 3 and 8 are factors then 24 are also factors.
Let's call Z 5n+1 times 5n-1
Z = 25n^2 - 1
Let's call U 2n+2 times n
U= n^2 + 2n
25n^2-1 can be written as 24n^2 + n^2 - 1, mod 24 = n^2-1.
So does ( n^2-1) * (n^2+2n) mod 24= 0
n=1, 0
n=2, 3*8,0
n=3 8*15=8*3=0
n=4 15*24=0
n=5 24*35=0
n=6 35*48,24...=0
n=7, 48*63
n=8 65*80=3*8=24=0
...
The pattern appears to be when n is a power of 2, the left will be divisible by 3, the right 8
Okay so if n is a power of 2, then n^2 = 2^(y*2)
2^(y*2) - 1 is always divisible by 3, because 2^2-1 is 3. (From my exponent divisibility formula)
2^(y*2) + 2y, since y is a power of 2, will always be div 8 past n=8, since y will be divisible by 8 past that point. Check the list above for the manual check of less than n=8.
So powers of 2 are covered.
Second pattern i notice is that n^2-1, when n is a prime past 3, always divisible by 24.
primes past 3 are all p mod 6 =5, or p mod 6 = 1
5^2-1 = 24 mod 6 times both by 4, 24 times 4 mod 24 =0
1^2 - 1 = 0 mod 6, times both by 4, 0 times mod 24 =0
So yes all prime values of n are divisible by 24(Edit, maybe??? is this logic even correct??)
So we're left with prime factorizations that are either powers of primes (-2), or mixed prime factorizations.
I'm at my limit!
Finished your video. Wonderful as always. What a treat!
Nice explanation
(n-1)n(n+1)(n+2) spans 4 consecutive integers, so it must be divisible by 2, 3, and 4. Thus it is divisible by 2*3*4=24.
I see it, too. Using just the second part of the demonstration works fine 🤩
0:58 ach your intro i also would like to have a such nice video but i am just to lost for it :D anyway sounds interesting!
1:15 that was my second thought after i thought there is an binobic formular at the end
1:50 it is easyer yes but how can you see it now?
1:54 konsekutiv i had to research for the meaning :D but yes it makes sense
2:40 let me try it by my self (n) (n+2) / 8 = a; a, n € N| if n%2 = 0
therfore: n² * 2n = a
and n² is an ord number as n is one as well as 2 is one and n it self; so waht we have is the product of 3 ord numbers that means we have the form
2(x) * 2(y) * 2(z) but than we knowe we can move the factorials and we came to
2*2*2 * (xyz) = 2*4*xyz = 8*xyz and therfore we knowe it has to be devideable by 8! right?
3:58 also an option to prove it
4:26 yes i also vorget to tell that x,y,z is an integer as well but i think it is clear :D
5:30 ok now i got your idear :D
7:53 if you don´t like sth. in math just rewrite it :D i like that
8:28 i see
9:05 ok nice idear
9:48 ok yes that is clear and what about the other summant
10:27 yes we knowe: (a + b) % n = 0 if a,b % n = 0; it is logic because lets turn it
3(a+b) % 3 = 0 | (we talk about integer!)
--> 3a + 3b % 3 = 0
and now call 3a x and 3b y;
x % 3 = 0; y % 3 = 0 ==> x + y % 3 = 0
12:10 :D i understand!
12:24 i have quote this sentencis now about 5 times :D
LG K.Furry
i had a mistake i wrote * instead of + sorry
The product of two adjacent even numbers is always divisible by 8
n(n+2)(5n-1)(5n+1)=n(n+2)(25n^2-1)=24n^3(n+2)+(n-1)n(n+1)(n+2)
obviously，
24n^3(n+2)is divisible by 24
and (n-1)n(n+1)(n+2) is divisible by2 by3 by4 by6 by8 …… and 3 is coprime with 8 so it is divisible by24
Thank you for tour great video and job
Tour conclusion may Be wrong Because 8 is not a prime Numbers
according to Gauss theorem
Regards
I thought all u have to do it make n 5, so one of the factors of this number is 24, and if one of it’s factors is 24, the number is divisible by 24🤯
You could have factorized the polynomial 25n^4 + 50n^3 - n^2 - 2n for starters. ;-)
Oh, I just get, that there's a fast short track! Claim: n (n + 2) (5n - 1) (5n + 1) = 0 (mod 24)
Well, (5n - 1) (5n + 1) = (25n² - 1) = (1n² - 1) = (n - 1) (n + 1) (mod 24), so new claim: n (n + 2) (n - 1) (n + 1) = 0 (mod 24)
But that's just the product (n - 1) n (n + 1) (n + 2) of 4 consecutive whole numbers, thus divisible by 4! = 24. QED
Without thinking, I'd say, that n (n + 2) (7n - 1) (7n + 1) is also divisible by 24! Samples:
1: 1*3*6*8 = 144 = 6 * 24
2: 2*4*13*15 = 1560 = 65 * 24
3: 3*5*20*22 = 6600 = 275 * 24
@@rainerzufall42 it does make sense since (7n-1)(7n+1)=(49n^2-1) and (49n^2-1) (mod 24) = (n^2-1)=(n-1)(n+1) since 48 is divisible by 24 and rest of the steps are same. Thanks for elegant proof.
@@LakshaySura Next step: n (n + 2) (k n - 1) (k n + 1) = 0 (mod 24)
for k = 6 m +/- 1, m € IZ, that's k = -1, 1, 5, 7, 11, 13, 17, 19, ...,
because (6 m +/- 1)^2 = 1 (mod 24), nice!
Why (6 m +/- 1)^2 = 36 m^2 +/- 12 m + 1 = 1 (mod 24) ?
Well, that's 12 m^2 +/- 12 m + 1 = 12 m (m +/- 1) + 1 (mod 24)
and either m or (m +/- 1) is even!
Thus it is 1 (mod 24)...
4 factors, first factor is n, if n=24 will be divisible by 24, did I miss something …..
I love your videoes. It helps has to look into deep.
Can we also prove that always the product of 4 consecutive numbers are devide by 4, and the product of 5consecutive numbers are devided by 5....... and so on . I checked few steps and it happened right, but no proof
It's actually pretty obvious: If you have any k consecutive integers, don't you think one of them must be divisible by k? Otherwise you would have a gap of size greater than k without a multiple of k, that's impossible! Alternatively, think about congruence classes mod k and the pigeon hole principle. However you do it, if one of the factors is already divisible by k, that applies to the product as well.
But it goes much deeper! The product of 4 consecutive integers is always divisible by 24, the product of 5 consecutive integers is always divisible by 120, and so on. In general, product of k consecutive integers is always divisible by k! (k factorial). That's much much harder to prove, though. You can do it by looking at prime factorizations and counting prime factors, by leveraging the properties of binomial coefficients, or by mathematical induction (from first principles, but not simple). To give some hints. (But if you didn't see the first statement, it's probably too hard for you.)
@@Grecks75 Obvious by inspection is true. I was looking for mathmatical proof theory
Thanks
@@Abby-hi4sf Hi Abby, in my previous answer I gave you all the needed ingredients for a complete, rigorous mathematical proof. You just have to formalize it, just a little bit. There wasn't anything missing or handwavy, actually.
To help you out, I will show you an example of how formalization could look like. I give you a proof by contradiction and also using the so-called "pigeon-hole principle", two very important proof techniques you should know 😃:
Look at an arbitrary sequence of k >= 1 consecutive integers starting at n: n, n+1, ..., n+k-1. Consider their remainders upon division by k: r_0, r_1, ..., r_(k-1). These r_i are integers and all satisfy 0
@@Abby-hi4sf Hi Abby, in my previous answer I gave you all the needed ingredients for a complete, rigorous mathematical proof. You just have to formalize it, just a little bit. There wasn't anything missing or handwavy, actually.
To help you out, I will show you an example of how formalization could look like. I give you a proof by contradiction and also using the so-called "pigeon-hole principle", two very important proof techniques you should know 😃:
Look at an arbitrary sequence of k >= 1 consecutive integers starting at n: n, n+1, ..., n+k-1. Consider their remainders upon division by k: r_0, r_1, ..., r_(k-1). These r_i are integers and all satisfy 0
Nice!
If we let n=1 then we get 3x4x6 = 3x24.
I like the old chalk board better than the white boards.
So do I
The chalk board is more relaxing on the eyes if I am not mistaken.
The sound of the chalk is also easier on the ears than the squeak of markers 😄
A sequence of n consecutive numbers is always divisible by n!
Your conclusion is wrong, or at least incomplete. You have to add that if a and b are factors and a is not divisible by b or viceversa then (a*b) Will be a factor.
In example. 8 is divisible by 8, and 8 is divisible by 2, but 8 is not divisible by 16.
Forgot that part
That's not sufficient, as you can see with a=4 and b=6 both dividing n=12, and they don't divide each other. Their product still does not divide n. The correct sufficient condition is that the gcd(a,b) is 1.
-1
use the fact that n! divides n consecutive integers' product
So... my question, as a math enjoyer, how do you prove that 3 consecutive numbers is divisible by 3 with induction.
You left out three words and put in an incorrect word. You need to prove that *the product of*
three consecutive *integers* is divisible by three.
Let n, k, and M be integers.
Base case where n = 1: 1(2)(3) = 6, which is divisible by 3.
Inductive step
Assume it is true for n = k:
k(k + 1)(k + 2) = 3M
Show it is true for n = k + 1:
(k + 1)(k + 2)(k + 3) must also be a multiple of 3.
k(k + 1)(k + 2) + *3(k + 1)(k + 2)* = 3M + *3(k + 1)(k + 2)*
(k + 1)(k + 2)(k + 3) = 3[M + (k + 1)(k + 2)]
This shows that (k + 1)(k + 2)(k + 3) is a multiple of 3.
Thus, by the Principle of Mathematical Induction, I have proven that
the product of three consecutive integers is divisible by three.
@@robertveith6383 Excellent.👍 It is simpler to argue with congruence classes and congruence relations modulo 3, but OP asked for mathematical induction, and you provided it.
What I like is the rigor and level of detail of the proof, it should be very instructive for students. 😃
@@robertveith6383 Thank you
Divisible by 8 = 0 mod 8.
Ah Hungarian paper.
@Prime:
From (n-1) * n * (n+1) * (n+2): Can we say that since its 4 consective number, it divides 2,3,4 so its divides 2*3*4 = 24 ??
Because 4 contains 2 you have to show in addition to a number is divisible by 4 a diff number is divisible by 2, which is true in this case.
@@xyz9250 its 4 consecutive numbers, one of them will divide 2 and other exclusively will divide 4. If n-1 divides 2, n+1 will divide 4, similarly for other cases.
I saw this after doing my proof for divisibility by 3 which was similar to this.
24 sq plus I sq is not equal to 25 sq .
lies...
Hungary mentioned... WTF is an easy math problem
I figured this out without having to write something down for once. Divisibility by 8 was the easiest part. If n is even, n and n+2 are two consecutive even integers. That means one is divisible by 4. 4*2 is 8. If n is odd, 5n is odd which means 5n + 1 and 5n-1 are consecutive even integers.
I almost over thought divisibility by 3 but it was staring right in front of me. 5n mod 3 is congruent to n mod 3 which means 5n + 1 mod 3 is congruent to n + 1 mod 3. So we have n, n+1, and n+2, which are 3 consecutive integers, meaning one of them is divisible by 3.
The last paragraph doesn’t look right. 5n mod 3 is not congruent to n mod 3 because 5 mod 3 is not 0. Trivially: 1 mod 3 = 1 and 5 mod 3 = 2. (5n+1)mod 3 is congruent to (n+2)mod 3 and (5n-1) mod 3 is congruent to n mod 3, so for example when n =1, the expression is (1)(3)(4)(6) where 1 and 4 are congruent (to 1), 3 and 6 are congruent to 0.
Does someone know the name of the subject ? I want to try some exercise but I don't find it.
Number theory
What, you don't understand Hungarian? Never stop learning... 😉
You didn't need the first part, as it's clear, that the term you had with "divisibility by 3" is clearly divisible by 24!
Yeah. We could say that four consecutive numbers contains a multiple of 4 and another even number. Done.
@@GreenMeansGOF It's even better: n consecutive numbers contain a multiple of n! for all natural numbers n>0. Because you are guaranteed to have a multiple of 2, of 3, of 4, ..., a multiple of n in these n factors, the product is 0 (mod n!).
Corollary: n * (n² - 1) * (n² - 4) is divisible by 120:
3 * 8 * 5 = 3 * 40 = 120 * 1
4 * 15 * 12 = 60 * 120 = 720 = 120 * 6
5 * 24 * 21 = 120 * 21
That's: 120 | (n^5 - 5 n^3 + 4 n) for all n € IN
It's even easy to calculate: n^5 - 5 n^3 + 4 n = 120 * (n+2 \over 5)
like @rainerzufall42 said