3003 is the first number other than 1 to appear in two consecutive rows of Pascal’s triangle. It is in the 14th and 15th row. I knew that, and I knew it was because 1001,2002,3003 appeared in the 14th row, so I had the answer right when you stated the problem. Man, that has never come in handy before!
@@jaehyunbaig1885 It is a mistake, he's used both k and r when he should have used one or the other. If the numerator is (n-k+1) then the denominator should have been k!. Likewise if the numerator is (n-r+1) the denominator is r!
@@jaehyunbaig1885 no like Ronald said below it has to be both k or both r..otherwise what the heck isbthe relation between k and r..if you stop at k minus 1 in the numerator, then the denominator has to be k factorial NOT r or anything else factorial..unless k equals r of course..
This guy climbs at an elite level of rock climbing. In his video celebrating 50,000 subscribers, he mentioned he had climbed a few 5.14s which only a tiny handful of the climbers can do. So it makes sense that you admire his biceps since he can probably do one- fingered one-armed pullups.
Why would anyone think of defining Pascal triangle in terms of binomials or combinatorics..it's random and not smart..isn't there are a more intuitive logical way..like plugging in multiples of 3 seeing which ones can be made from some multipken of 2 since it has to be an even number and etc...
One fact worth knowing for this question, is that, in row n of Pascal's triangle, C(n,k), successive terms are obtainable from their immediate predecessors by multiplying by factors n/1, (n-1)/2, (n-2)/3, ... because C(n,1) = (n/1) C(n,0) , C(n,2) = ½(n-1) C(n,1) , C(n,3) = ⅓(n-2) C(n,2) , ... , C(n,k+1) = [(n-k)/(k+1)] C(n,k), etc. This is quickly verified by using: C(n,k) = n!/[k!(n-k)!] So in order to find consecutive terms that stand in the ratio 1:2:3, requires finding positive integers n and k for which (n-k+1)/k = 2, and (n-k)/(k+1) = 3/2 This is a 2E2U problem (2 equations in 2 unknowns). And the equations are even linear! n - k + 1 = 2k; n - 3k = -1 2(n - k) = 3(k+1); 2n - 5k = 3 ===> (n, k) = (14, 5) . . . Check: (n-k+1)/k = (14-5+1)/5 = 10/5 = 2 (n-k)/(k+1) = (14-5)/(5+1) = 9/6 = 3/2 √ [C(n,k-1), C(n,k), C(n,k+1)] = [C(14, 4), C(14, 5), C(14, 6)] = [14·13·12·11/4!, 14·13·12·11·10/5!, 14·13·12·11·10·9/6!] = [1001, 2002, 3003] It works!! And according to this method, it's the only solution. Fred
I have a problem: Let a,b,c,d positive integers (0 isn’t include) such that a*b=c*d, show that: a=gcd(a,c)*gcd(a,d)/gcd(a,b,c,d) If you had time please I need help with this exercise. Thanks, I admire you a lot.
If you want the short answer instead of watching 10min video: No, they're not friendly. I tried petting my pascal triangle once but it bites me really hard.
Shut up. That's what you get for pretending to throw the Fibonacci sequence and then secretly keeping it. Tricking your Pascal's triangle will come back to bite you in the ass.
Another very similar approach would be to find the factor that transforms a choose b into a choose b+1, and then set this factor equal to the desired ratios and solve.
binomial coefficients are famously described by pascal's triangle. you can see this play out if you start with the number 1 and keep multiplying by (x+1), such as with 11 (10+1): 1, 11, 121, 1331, 14641 (it breaks when any digit carries over, but the idea remains)
Great. I first thought its diophante with multiple solutions or no solutions. Also by symmetry (14 choose 10), (14 choose 9) and (14 choose 8) (in reverse order 3:2:1). I could not be sure if there is 2 results as the other one is 3:2:1 ratio, but it could be. Make another exploration with diagonal elements that would be easy. Can you find pi somehow in this Pascals triangle? Maybe no... What about question for ratio x:y:z, so that it appears at least twice in the pascal's triangle (left side only) and solve for x,y,z? No result, I guess... Best case maybe (45 choose 11):(45 choose 10):(45 choose 9) ~ (40 choose 10):(40 choose 9):(40 choose 8) ~ 1:3.6:3.2 = (x:y:z) Not possible, because this system of equations can not give more than one result, no matter what ratio is in concern...
So there's only one occurrence of this thing?! I would have though it would occur more in an infinite triangle. Does the ratio of consecutive entries diverge?
One use would be to expand (x+1)^8 quickly. The coefficients for the terms with powers x^8, x^7, x^6 ... are found on the triangle. Probabilities involving coin tosses live on the triangle. Many uses besides these as well.
He generally takes from higher levels than the P-RMO. Typically National level olympiads the analogue of which would be INMO in India. He even did an INMO problem once. PRMO would be a bit lower level
@@rajvardhansinghsisodiya1095 Yes, and (if it's not cancelled because of the pandemic) I plan to give the PRMO next year (because it is unlikely that it will happen this year, though if it does, I'll give it). Which is why all the olympiad problems that are solved are such a help (Not just that, this channel is a goldmine for anyone who wants to get into competitive maths).
Yes. Yes I am.
Pascal's Triangle your account was only made for this video?
Hi friend :)
@@Thaplayer1209 its not possible? A hacker? Making fake accounts can not be supported in freedom or speech?
3003 is the first number other than 1 to appear in two consecutive rows of Pascal’s triangle. It is in the 14th and 15th row. I knew that, and I knew it was because 1001,2002,3003 appeared in the 14th row, so I had the answer right when you stated the problem. Man, that has never come in handy before!
If you did that in person it would seem like a Ramanujan moment lol
fun extension to try.
Determine conditions for when this is possible for the general ratio a:b:c with a,b,c positive integers
Tiny mistake at 3:30, numerator should have (n-r+1) not (n-k+1).
Yes
it's not a mistake, he chose k when he was talking about the n'th row
@@jaehyunbaig1885 It is a mistake, he's used both k and r when he should have used one or the other. If the numerator is (n-k+1) then the denominator should have been k!. Likewise if the numerator is (n-r+1) the denominator is r!
@@jaehyunbaig1885 no like Ronald said below it has to be both k or both r..otherwise what the heck isbthe relation between k and r..if you stop at k minus 1 in the numerator, then the denominator has to be k factorial NOT r or anything else factorial..unless k equals r of course..
@@ronaldjensen2948 okay I understood ~~
10:43
You are speed
thanks
I like your multi-video with the "multi good places to stop", I can't remember , 6 or 9 videos fusionned?
CM63 9 videos
such a refreshing little problem. easy, but very pleasing to see it unfold.
I liked the answer and I bet that someone saw that part of row 14 and thought "hey, this would be an interesting problem"
Best maths teacher
🥺😍
Damn right
does anyone else constantly admire his biceps?
Not me because I train bodybuilding :p
This guy climbs at an elite level of rock climbing. In his video celebrating 50,000 subscribers, he mentioned he had climbed a few 5.14s which only a tiny handful of the climbers can do. So it makes sense that you admire his biceps since he can probably do one- fingered one-armed pullups.
That final result, in terms of the integers which the binomial coefficients reduce to, is quite surprising and rather cool.
Why would anyone think of defining Pascal triangle in terms of binomials or combinatorics..it's random and not smart..isn't there are a more intuitive logical way..like plugging in multiples of 3 seeing which ones can be made from some multipken of 2 since it has to be an even number and etc...
@@leif1075 Dude, that's literally how it was defined throughout history before Pascal even existed...
Such a satisfying solution. 1001 is quite fascinating
I honestly thought the answer to the math question would be "no"!
@@darreljones8645 You don't know all of my secrets 😏
It's a product of 3 consecutive primes, too.
One fact worth knowing for this question, is that, in row n of Pascal's triangle, C(n,k), successive terms are obtainable from their immediate predecessors by multiplying by factors
n/1, (n-1)/2, (n-2)/3, ...
because
C(n,1) = (n/1) C(n,0) , C(n,2) = ½(n-1) C(n,1) , C(n,3) = ⅓(n-2) C(n,2) , ... , C(n,k+1) = [(n-k)/(k+1)] C(n,k), etc.
This is quickly verified by using: C(n,k) = n!/[k!(n-k)!]
So in order to find consecutive terms that stand in the ratio 1:2:3, requires finding positive integers n and k for which
(n-k+1)/k = 2, and (n-k)/(k+1) = 3/2
This is a 2E2U problem (2 equations in 2 unknowns). And the equations are even linear!
n - k + 1 = 2k; n - 3k = -1
2(n - k) = 3(k+1); 2n - 5k = 3
===> (n, k) = (14, 5) . . . Check:
(n-k+1)/k = (14-5+1)/5 = 10/5 = 2
(n-k)/(k+1) = (14-5)/(5+1) = 9/6 = 3/2 √
[C(n,k-1), C(n,k), C(n,k+1)] = [C(14, 4), C(14, 5), C(14, 6)]
= [14·13·12·11/4!, 14·13·12·11·10/5!, 14·13·12·11·10·9/6!]
= [1001, 2002, 3003]
It works!! And according to this method, it's the only solution.
Fred
Oh no do more Vertex algebras !
is the parker square unfriendly?
I have a problem:
Let a,b,c,d positive integers (0 isn’t include) such that a*b=c*d, show that:
a=gcd(a,c)*gcd(a,d)/gcd(a,b,c,d)
If you had time please I need help with this exercise. Thanks, I admire you a lot.
gcd(a,c)*gcd(a,d) = gcd(gcd(a,c)*a, gcd(a,c)*d) .... distributive property
= gcd(gcd(a², ac), gcd(ad, cd)) .... distributive property
= gcd(a², ac, ad, cd) .... associative property
= gcd(a², ac, ad, ab) .... using ab=cd
=a*gcd(a,c,d,b) .... distributive property
= a*gcd(a,b,c,d) .... commutative property
QED
summary of properties used:
Commutative - gcd(a,b)=gcd(b,a)
Associative - gcd(a,b,c)=gcd(gcd(a,b),c)
Distributive - a*gcd(b,c)=gcd(ab,ac)
Caesar Cipher 👍
If you want the short answer instead of watching 10min video:
No, they're not friendly. I tried petting my pascal triangle once but it bites me really hard.
Shut up. That's what you get for pretending to throw the Fibonacci sequence and then secretly keeping it. Tricking your Pascal's triangle will come back to bite you in the ass.
Pascal's Triangle the fact that you are Pascal‘s triangle made this 100000 times better xD
Another very similar approach would be to find the factor that transforms a choose b into a choose b+1, and then set this factor equal to the desired ratios and solve.
Why would anyone think of applying bnimomials orncombonstorics tonthebtriangle..it's so random
binomial coefficients are famously described by pascal's triangle. you can see this play out if you start with the number 1 and keep multiplying by (x+1), such as with 11 (10+1): 1, 11, 121, 1331, 14641 (it breaks when any digit carries over, but the idea remains)
By symmetry of the choose function n=14,k=10 is the same as n=14,k=5.
.. and the ratios would also reverse 3:2:1 .... but that's a different question😉
Great. I first thought its diophante with multiple solutions or no solutions.
Also by symmetry (14 choose 10), (14 choose 9) and (14 choose 8) (in reverse order 3:2:1).
I could not be sure if there is 2 results as the other one is 3:2:1 ratio, but it could be.
Make another exploration with diagonal elements that would be easy.
Can you find pi somehow in this Pascals triangle? Maybe no...
What about question for ratio x:y:z, so that it appears at least twice in the pascal's triangle (left side only) and solve for x,y,z?
No result, I guess...
Best case maybe (45 choose 11):(45 choose 10):(45 choose 9) ~ (40 choose 10):(40 choose 9):(40 choose 8) ~ 1:3.6:3.2 = (x:y:z)
Not possible, because this system of equations can not give more than one result, no matter what ratio is in concern...
Does this generalize to the case where the top of n choose r has an element from any associative algebra? (complex numbers, etc.)
I think so.
So there's only one occurrence of this thing?! I would have though it would occur more in an infinite triangle. Does the ratio of consecutive entries diverge?
well as it happens, it's two equations with two variables. it only happens once!
One question I wanna ask , is what is the use of Pascal's triangle
One use would be to expand (x+1)^8 quickly. The coefficients for the terms with powers x^8, x^7, x^6 ... are found on the triangle. Probabilities involving coin tosses live on the triangle. Many uses besides these as well.
If you like Poker, Pascal's triangle can be used to calculate the probability of all the different possible hands
Never... will be a good place for you to stop making videos.
Sir please take problem from prermo India.
He generally takes from higher levels than the P-RMO. Typically National level olympiads the analogue of which would be INMO in India. He even did an INMO problem once. PRMO would be a bit lower level
@@ribhuhooja3137 are you an indian
@@rajvardhansinghsisodiya1095 Yes, and (if it's not cancelled because of the pandemic) I plan to give the PRMO next year (because it is unlikely that it will happen this year, though if it does, I'll give it). Which is why all the olympiad problems that are solved are such a help (Not just that, this channel is a goldmine for anyone who wants to get into competitive maths).
Ribhu hooja are you a member of this channel
@@rajvardhansinghsisodiya1095 No. Just subscribed
This must be the only solution, right?
Cute