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-florinchereches48927 күн бұрын
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
@blackovich7 күн бұрын
Yeah I noticed that at the end too.
@rainerzufall427 күн бұрын
@@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...
@rainerzufall427 күн бұрын
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!
@lornacy6 күн бұрын
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.
@absolutemystery7 күн бұрын
@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.
@tygrataps5 күн бұрын
Such an awesome insight!
@larswilms82756 сағат бұрын
in general: n consecutive numbers is divisible by n!
@annacerbara42577 күн бұрын
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.
@quzpolkas7 күн бұрын
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!
@StaR-uw3dc7 күн бұрын
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
@rainerzufall427 күн бұрын
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
@rainerzufall427 күн бұрын
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
@LakshaySura5 күн бұрын
@@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.
@rainerzufall425 күн бұрын
@@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!
@rainerzufall425 күн бұрын
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)...
@jamesmarshall77566 күн бұрын
Pour la divisibilité par 3, j’ai employé la congruence qui prend moins de temps. Nice video !
@simplicity530Күн бұрын
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.
@davidplanet39196 күн бұрын
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.
@Grecks756 күн бұрын
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.
@Christian_Martel7 күн бұрын
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.
@maxvangulik19886 күн бұрын
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.
@PrithwirajSen-nj6qqКүн бұрын
Highly enjoyed ur lesson.
@RyanLewis-Johnson-wq6xs7 күн бұрын
I love your videos.
@nothingbutmathproofs71506 күн бұрын
Very nicely done!
@klausao6 күн бұрын
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
@ahmedrafea85425 күн бұрын
Great work, as usual. I particulary like problems in the Number Theory field. Thanks
@RyanLewis-Johnson-wq6xs7 күн бұрын
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.
@RyanLewis-Johnson-wq6xs7 күн бұрын
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.
@wannabeactuary016 күн бұрын
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
@martinfenner32223 күн бұрын
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.
@Tentin.Quarantino7 күн бұрын
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
@GiovanfidoPonzio7 күн бұрын
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.
@rohangt17 күн бұрын
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".
@MikeGz927 күн бұрын
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
@JossWainwright7 күн бұрын
Haven't done it yet, and am looking forward to seeing your approach after I'm done. I'm going to do the long slog version: looking at each factor of the expression in terms of mod_2 and mod_3. It should show that in all of the handful of separate cases, the expression generates at least three twos in the factors and at least one factor divisible by three. Edit: Watching your solution, indeed, something juicier did show up. Nice.
@LajosRodé7 күн бұрын
Jó kis régi magyar matematika feladvány!
@ناصريناصر-س4ب7 күн бұрын
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
@vafamoshtagh51307 күн бұрын
(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.
@klausao6 күн бұрын
Exactly, product of N consecutive numbers is divisible by N factorial.
@oida100007 күн бұрын
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.
@ionelpatriche68663 күн бұрын
Solutie frumoasa!
@MrAlvaro4126 күн бұрын
I laughed a lot for "mmmm" "mmmmmmmmm" I already see the answer... you are funny
@xyz92507 күн бұрын
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.
@ashokchaudhary24774 күн бұрын
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 :)
@dean5326 күн бұрын
Next: M=E-esinE. What are you’re thoughts on analytical solutions?
@RyanLewis-Johnson-wq6xs7 күн бұрын
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.
@igorgorkoff7 күн бұрын
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.
@kereric_c6 күн бұрын
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
@rainerzufall427 күн бұрын
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!
@GreenMeansGOF7 күн бұрын
Yeah. We could say that four consecutive numbers contains a multiple of 4 and another even number. Done.
@rainerzufall427 күн бұрын
@@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!).
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)
@robertpearce83947 күн бұрын
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.
@robertlunderwood7 күн бұрын
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.
@robertpearce83947 күн бұрын
@robertlunderwood Yes, that was the second proof. The first one was rather clumsy.
@Grecks756 күн бұрын
@@robertlunderwoodVery cool! Thanks.
@elunedssong89097 күн бұрын
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!
@vishwaspawaskar36196 күн бұрын
Nice explanation
@Abby-hi4sf7 күн бұрын
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
@Grecks756 күн бұрын
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-hi4sf5 күн бұрын
@@Grecks75 Obvious by inspection is true. I was looking for mathmatical proof theory Thanks
@Grecks754 күн бұрын
@@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
@Grecks754 күн бұрын
@@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
@adampiechuta57747 күн бұрын
10:19 All parts are divisible by 24, not only by 3
@pschymit6 күн бұрын
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.
@backgammonmaster6 күн бұрын
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☺
@tophatjones6241Күн бұрын
That was beautiful
@Satyam-b6f4 күн бұрын
At 3:20 , why did we "let" it if it's true for all positive consecutive even integers?
@spacer9997 күн бұрын
(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.
@MikeGz927 күн бұрын
I see it, too. Using just the second part of the demonstration works fine 🤩
@zyxpip83635 күн бұрын
In the end, "just expand it"
@ibrahimali31927 күн бұрын
As a captive of the mighty, i can confirm i got taken away
@bchap51975 күн бұрын
The LastPass part of the development si enough to démonstrate thé 24 ils a commun factor.
@MomoBetch-b9g7 күн бұрын
Honestly loved it
@venkatk19687 күн бұрын
Good one sir!
@johnroberts75297 күн бұрын
Nice one sir!
@RJ-cx1gt7 күн бұрын
4 factors, first factor is n, if n=24 will be divisible by 24, did I miss something …..
@terryendicott29397 күн бұрын
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.
@Dominus_Potatus6 күн бұрын
So... my question, as a math enjoyer, how do you prove that 3 consecutive numbers is divisible by 3 with induction.
@robertveith63836 күн бұрын
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.
@Grecks756 күн бұрын
@@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. 😃
@Dominus_Potatus6 күн бұрын
@@robertveith6383 Thank you
@markfarsang7 күн бұрын
greetings from Hungary (this question is from Hungary 😀 )
@younesalouch116 күн бұрын
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🎉 🎉
@xavierpech74497 күн бұрын
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
@9adam47 күн бұрын
The product of two adjacent even numbers is always divisible by 8
@KPunktFurry2 күн бұрын
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
@KPunktFurry2 күн бұрын
i had a mistake i wrote * instead of + sorry
@barygol7 күн бұрын
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.
@PrimeNewtons7 күн бұрын
Forgot that part
@Grecks755 күн бұрын
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.
@pogbacr77495 күн бұрын
Does someone know the name of the subject ? I want to try some exercise but I don't find it.
@ilikemath14243-masterfail4 күн бұрын
Number theory
@AubreyForever7 күн бұрын
I like the old chalk board better than the white boards.
@nothingbutmathproofs71506 күн бұрын
So do I
@AubreyForever6 күн бұрын
The chalk board is more relaxing on the eyes if I am not mistaken.
@lornacy6 күн бұрын
The sound of the chalk is also easier on the ears than the squeak of markers 😄
@aabhasmajumdar7 күн бұрын
@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 ??
@xyz92507 күн бұрын
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.
@aabhasmajumdar7 күн бұрын
@@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.
@robertlunderwood7 күн бұрын
I saw this after doing my proof for divisibility by 3 which was similar to this.
@Metaverse-d9f7 күн бұрын
use the fact that n! divides n consecutive integers' product
@srisaishravan55124 күн бұрын
Day 2 of asking you to start making videos on integrals
@decidueye987 күн бұрын
A sequence of n consecutive numbers is always divisible by n!
@BartBuzz7 күн бұрын
Nice!
@nedmerrill57057 күн бұрын
Divisible by 8 = 0 mod 8.
@AmY-l8i5o6 күн бұрын
-1
@arcangyal22697 күн бұрын
Hungary mentioned... WTF is an easy math problem
@298chandra6 күн бұрын
24 sq plus I sq is not equal to 25 sq .
@EHMM5 күн бұрын
lies...
@harrys23314 күн бұрын
Ah Hungarian paper.
@andrasferencz794822 сағат бұрын
What, you don't understand Hungarian? Never stop learning... 😉
@robertlunderwood7 күн бұрын
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.
@mikecaptain19677 күн бұрын
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.