A small hint: Once you've proved associativity for the * operator, you can leave out all parentheses in your starred products, thus saving you a lot of handwriting.
@@adchayansivakumar1667 it’s the same as induction, any sort of repeated continuation to a pattern is underlying induction. Induction just adds rigour to everything and is not needed when the result is clear, for example showing n! = n(n-1)…1 using n! = n(n-1)! for all naturals n. Same is true here
Rigorously: a*(b*c) = (a+1)(b+1)(c+1) - 1 = (a*b)*c. So we can use operator (*) in any order. Claim: 1*...*n = (n+1)!-1. Indeed: 1*2 = 3! -1 and if 1*...*(k-1) = k!-1, then 1*...*k = (k!-1)k + k! - 1 + k = (k+1)! - 1 which finishes the proof by induction. Therefore 1*(2*...(99*100))...) = 1*...*100 = 101! - 1
Don't overcomplicate the question A simple idea You can just do something like this and avoid induction (as it's not for general case) A*B=(A+1)(B+1)-1 That implies A*B+1=(A+1)(B+1) For (A*B)*C=(A*B+1)(C+1)-1 That equals to (A+1)(B+1)(C+1)-1 The pattern continues So Q=(100+1)(99+1)....(1+1)-1 So, Q=101×100×...×2-1 Rewrite as 101×100×...2×1-1 Thus 101!-1. You can use fictional blackboards and simply solve that
Votre phrase " so we can use... in any order" prête à confusion. Pour changer L'ORDRE des termes il faut que la loi soit COMMUTATIVE. Il serait préférable d'utiliser une phrase du genre " we can begin to ASSOCIATE two éléments WHERE we want" puisque la commutativité n'a pas encore été démontrée, bien qu'elle soit évidente. Ou bien préciser avant d'écrire votre phrase que la loi commutative. Ce n'est qu'un détail.
Unusual application of this operator is to describe the n-dimensional triangle family. With a bit of imagination the star operator definition can be read as a*b = line segment ab + the points a and b. Then (a*b)*c reads as the triangular face abc + the 3 line segments (ab, ac, bc) and 3 corners (a, b, c). And (((a*b)*c)*d) gives the tetrahedra abcd, the 4 faces abc, abd, acd, bcd, the 6 edges ab, ac, ad, bc, bd, cd, and 4 corners a, b, c, d. Continuing with 5 corners, a*...*e, gives the 4-simplex. Etc. , with the operation always giving one of each term for uniqueness. Fun compaction. Thank you for such beautiful teaching.
@PrimeNewtons Aw garsh! Thanks for not wincing at my playing fast and loose with the "=" sign. I love your teaching style. It's as if you're teaching from inside of us, with rich emotional expressions of perplexity, false starts, ah hahs, and fascination.❤
What a cool problem. This is about as close as I can get to abstract algebra without losing my mind! If like me you did not spot the link to factorials, there is another approach with an interesting connection to Vieta's formulas... Let f(x) = (x-a)(x-b) = x^2 - (a+b)x + ab f(-1) = 1 + (a+b) + ab = 1 + a*b ⇒ a*b = f(-1) - 1 (a*b)*c = (a+b+ab) + c + (a+b+ab)c = (a+b+c) + (ab+bc+ca) + abc Let f(x) = (x-a)(x-b)(x-c) ⇒ f(x) = x^3 - (a+b+c)x^2 + (ab+bc+ca)x - abc f(-1) = -1 - (a+b+c) - (ab+bc+ca) - abc = -1 - (a*b)*c ⇒ (a*b)*c = - f(-1) - 1 Similarly ((a*b)*c)*d = (a+b+c+d) + (ab+ac+ad+bc+bd+cd) + (abc+abd+acd+bcd) + abcd Let f(x) = (x-a)(x-b)(x-c)(x-d) ⇒ ((a*b)*c)*d = f(-1) - 1 Let Q[n](x) = (x-1)(x-2)(x-3)...(x-n) ⇒ Q[n](-1) = (-1-1)(-1-2)(-1-3)...(-1-n) = (-1)^n . (n+1)! Let q[1] = 1 and q[n] = q[n-1]*n for n>1 ⇒ q[n] = (...((1*2)*3)*...)*n So the pattern appears to suggest that q[n] = (-1)^n . Q[n](-1) - 1 ⇒ q[n] = (n+1)! - 1 This is proved using induction as in the video.
Mister prime, your videos are so well detailed and your way of teaching is so charismatic, it doesn't feel like a boring math video at all. I wish I had a video for every problem in my daily life with you explaining how to solve it. Keep up with the good work.
Exploit a * b = ab + a + b = ( ab + a + b + 1 ) - 1 = (a+1)(b+1) - 1 by defining f(x) = x + 1. Note that f-inverse(x) = x - 1. Thus a * b = (a+1)(b+1) - 1 = f(a) f(b) - 1 = f-inverse( f(a) f(b) ). Also, applying f to both sides gives f(a*b) = f(a) f(b). Then a * (b * c ) = f-inverse( f(a) f(b *c) ) = f-inverse( f(a) { f(b) f(c) } ) = f-inverse( f(a) f(b) f(c) ). Similarly a * ( b * ( c * d ) ) = f-inverse( f(a) f( b * ( c * d ) ) ) = f-inverse( f(a) f(b) f(c * d) ) = f-inverse( f(a) f(b) f(c) f(d) ). The inductive clearly works (of course, if doing "real math", the proof of the induction should be written out, as was done in the video.) Thus 1 * ( 2 * ( 3 * ( 4 * ( ... * ( 99 * 100 ) )))...) = f-inverse( f(1) f(2) f(3) ... f(100) ) = f-inverse( (2) (3) (4) ... (101) ) = f-inverse( 101! ) = 101! - 1. (If you know some abstract algebra, f(a*b) = f(a) f(b) might make you suspicious of the following, which turns out to be true: That (Reals\{-1}, *, 0) is a group, and f is a group isomorphism from (Reals\{-1}, *, 0) to (Reals\{0}, ordinary-multiplication, 1).)
Without assuming commutation for *, you can observe that (n-1) * n = (n-1)n + (n - 1) + n = (n-1 + 1 + 1)n - 1 = (n + 1)n - 1 = n(n+1) - 1 Then the next * calculation to the left would be (n-2) * [n(n+1) - 1] = (n-2) [ n(n+1) -1 ] + (n - 2) + [n(n+1) - 1] = (n-2) [ n (n+1) ] - (n - 2) + (n -2) + n(n+1) - 1 (distribute) = (n-2 + 1)[ n (n+1) ] - 1 (combine the n(n+1) terms) =(n-1)n(n+1) -1 If you do the same for n-3 then you'll see the pattern as you apply * on the left. n(n+1) -1 ==> (n-1)n(n+1) - 1 ==> (n-2)(n-1)n(n+1) - 1 and so on
i havent watch the full video, just figure out by myself . a*b+1 = (a+1)(b+1)-1 (notice that a turns into a+1, b turns to b+1) ---> a*(b*c) = a * ((b+1)(c+1)-1) = (a+1)(b+1)(c+1) - 1 (use the fact above) ---> general formula: a1*a2*.....*an = (a1+1)(a2+1).....(an+1)-1 replace a1=1, a2=2,.....,a100=100, we have Q = (1+1)(2+1)(3+1).....(100+1)-1 = 101!-1
This is nice, I think a more insightful way of seeing the problem. The version in the video uses a technique that can be applied (or, at least attempted) in more circumstances, but your answer I think says more about “what is actually happening” in the problem. (Namely, the star operation is basically the “conjugation” of the multiplication operation by a bijection from the set of integers to itself, and so the answer is obtained by just applying the bijection in one direction to all the inputs, then doing all the multiplications, and then applying the bijection in the other direction).
My keyboard has '*', '×', and '★'. I use '*' as my normal multiply operator for regular math, symbolic inputs (spreadsheets, WolframAlpha, ...), and computer programming, so I substitute the distinctive '★' character for his new operator to avoid any confusion. We define the ★ operator as: a ★ b = a*b + a + b We find interesting properties: a ★ b = a*b + a + b + 1 - 1 = (a + 1)*(b + 1) - 1 { Eq. 1 } = (b + 1)*(a + 1) - 1 = b ★ a { Eq. 2 } We also find: (c ★ (b ★ a)) = ((b ★ a) ★ c) = (a ★ b) ★ c = ((a + 1)*(b + 1) - 1) ★ c = (a + 1)*(b + 1)*(c + 1) - 1 = a ★ b ★ c We next find: (a ★ b ★ c) ★ d = (a + 1)*(b + 1)*(c + 1)*(d + 1) - 1 = a ★ b ★ c ★ d = (a ★ b) ★ (c ★ d) Composing the function with itself allows the sequence of its parameters to be reordered in any manner. The function increments each parameter by 1, replaces '★' with '*', and decrements the product by 1. Thus, the original problem becomes: 1 ★ 2 ★ 3 ★ 4 ★ ... 99 ★ 100 = 100 ★ 99 ... ★ 4 ★ 3 ★ 2 ★ 1 = (100+1)*(99+1)*...*(4+1)*(3+1)*(2+1)*(1+1) - 1 = 101*100*...*5*4*3*2*1/1 - 1 = 101! - 1
Let A(n), n >= 0, be any sequence of elements in a ring (not necessarily Abelian) with unity 1. In this example, the ring is Z and A(n) = 100 - n. Define K(0) = A(0), K(n+1) = A(n+1)*K(n), for n >= 0, where a*b = ab + a+ b. Lemma. K(n) = [product (for i = n to 0) (A(i) +1)] - 1, for n >= 0. Proof. The Lemma holds for n = 0. Suppose the Lemma holds for some n >= 0. Then K(n+1) = A(n+1)*K(n) = A(n+1)*([product (for i = n to 0) (A(i) +1)] -1) = A(n+1)([product (for i = n to 0) (A(i) +1)] -1)+A(n+1)+ [product (for i = n to 0) (A(i) +1)] - 1 = A(n+1)[product (for i = n to 0)(A(i) +1)] - A(n+1) +A(n+1) + [product (for i= n to 0)(A(i) +1)] - 1 = A(n+1)[product (for i = n to 0) (A(i) +1)] + [product (for i = n to 0) (A(i) +1)] - 1 = (A(n+1)+1)[product(for i = n to 0) (A(i) +1)] - 1 = [product(for i = n+1 to 0) (A(i) +1)] - 1, completing the proof of the Lemma. In this example, we want to calculate K(99) = [product(for i = 99 to 0)(A(i) + 1)] - 1 = [product(for i = 99 to 0)(101 - i)] - 1 = (2)(3)...(100)(101) - 1 = 101! - 1.
Thank you for another of your inspirational and blessed videos, which I love watching. I liked your discussion and advantageous use of associativity to solve this problem. This alternative approach avoids the need for this though and I think is easier. Note that a*b = ab+a+b = (a+1)(b+1)-1 Hence 99*100 = (99+1)(100+1) - 1 = (100)(101) - 1 The -1 at the end of this number ensures that when it is an input to * the bracket (this number + 1) is a clean product Thus 98*(99*100) = (99)(100)(101) - 1 Again there is -1 at the end so the above is again true for the next application of * Repeating this process, the required answer is therefore (2)(3)...(100)(101) -1 = 101! -1. Again, thank you for your lovely videos and i look forward to seeing the next one. God bless you ♥️
my approach was just: a*b = a + (1+a)b Then Q can be expressed 1*(2*(3*(....) --> 1 + 2(2*(3*(...) --> 1 + 2(3 + 4(5 + 6(...98 + 99(99 + 100 times 100)...) Because multiplication is distributive, then for that last 100, I have a chain of 100*99*98...*2*1 as its coefficient For the last 99, similarly I have a 99! as its coefficient For the last 98, I have 98! as its coefficient and so on So I wind up with Sum_{n=1 to 100} of n * Product_{m=1 to n} of m Sum_{n=1 to 100} of n * n! Oh hey, an identity for (n+1)! - 1
Wow i think you are overcomplicating this. My first instinct was to rewrite * in a more compact way. a*b= ab+a+b=ab+a+b+1-1= a(b+1)+b+1-1=(a+1)(b+1)-1 Now for a*b*c it becomes much easier to evaluate as either : (a*b)*c= ((a+1)(b+1)-1)*c=((a+1)(b+1)-1+1)(c+1)-1= (a+1)(b+1)(c+1)-1 and similar for a*(b*c) but going by the same logic the original expression becomes 2x3x...x101-1
pardonne-moi mais ce que tu proposes n'est pas plus simple mais en plus la rédaction n'est pas assez rigoureuse, car la récurrence n'est pas explicitement écrite.
Un exposé clair et soigné ; une bonne méthode de résolution. Exercice intéressant qui montre bien l'utilité de l'associativité. 👍 Une petite remarque bien pratique à propos des parenthèses: quand une loi est ASSOCIATIVE l'écriture des PARENTHÈSES devient INUTILE puisque on peut commencer à associer les éléments là où on veut (sans changer l'ordre, sauf si la loi est commutative, cette loi * est commutative) Ce qui simplifie grandement l'écriture des calculs. a*(b*c) = (a*b) *c = a*b*c donc * est associative on peut écrire a*b*c*d*e= (a*b)*c*d*e = a*(b*c)*d*e= a*b*(c*d)*e = a*b*c*(d*e)
forgot to prove this pattern continues: a * b = (a+1)(b+1) - 1 a * b * c = (a+1)(b+1)(c+1) - 1 assume a * b * c * ... * m = (a+1)(b+1)(c+1)...(m+1) - 1 a * b * c * ... * m * n = ((a+1)...(m+1) - 1) * n = (a+1)...(m+1)(n+1) - 1 and the proof is done: a * b * c * ... * n = (a+1)(b+1)(c+1)...(n+1) - 1
If you realize that b * a = (b + 1)(a + 1) - 1 a recursion pattern is easy to recognize: c * (b * a) = (c+1)((b+1)(a+1)-1+1) - 1 which simplifies to c*(b*a) = (c+1)(b+1)(a+1) - .1 similarly d*(c*(b*a)) = (d+1)(c+1)(b+1)(a+1)-1 Therefore the answer I get is 1*(2*(3*....(99*100)...)) = 101! - 1
пф элементарно. рассмотрев a*b*c можно понять это это все суммы симметрических многочленов, что так же равно произведению (1+ax)(1+bx)... за вычетом единички, и при x=1. получаем очевидно 101!-1. решение занимает пол минуты в голове
The rewriting of parentheses part is flawed. The star’s associative property has only been proven within three numbers, whereas to move the parenthesis from far right end to far left end involves a lot more numbers and is yet to be verified. This can be solved by introducing a number sequence defined as p(n)=(((…(x*(x-1))…)*(x-n), and it is not hard to find its formula by using math induction: p(n)=(x+1)!/(x-n)!-1. Then let n=x-1, (((…(x*(x-1))…)*(x-n)= (((…(x*(x-1))…)*4)*3)*2)*1=(x+1)!-1, and we have the formula for the problem. Let x=100, we have the final answer that is 101!-1.
I did this in a general intuitive way.. I tried a*b*c and it is a+b+c+ab+ac+bc+abc So according to my intuition, a*b*c*......*n (n terms 🥲🥲pls adjust) will contain the sum of the terms taken 1,2,...,and n at a time.(it follows from a simple induction) If we add 1 to it, it will be, (a+1)(b+1)... (n+1) According to the binomial expansion.. Let X(n)=a*b*c*...n* Then X(n)=(a+1)(b+1)...(n+1)-1 So X(100)=101!-1.. Is this right primenewtons? Or any other guys?... Thanks
yes it is true you actually can do it easier a*b = ab + a + b + 1 - 1 a*b = a(b+1) + 1(b+1) - 1 a*b = (a+1)(b+1) - 1 a*(b*c) = a*[ (b+1)(c+1) -1 ] = (a+1)(b+1)(c+1) - 1 (a*b)*c = [ (a+1)(b+1) - 1 ]*c = (a+1)(b+1)(c+1) - 1 which means "*" operator is associative and commutative and you can just add 1 to all the "multiples" multiply them and subtract 1 which is n!-1 for 1*2*3*...*n
Given the numbers 1,2,...,100 on a board. We'll remove 2 of the numbers a,b and replace them with a*b Let S be the product of (each number on the board+1) So initially, S=101! After removing, find S' the same way S'/S=(a*b+1)/((a+1)(b+1))=1 Therefore S doesn't change regardless of how we choose a and b Perform this action until only 1 number is left. It would be 101!-1.
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from Morocco thank you for this wonderful problem and proof
Again, nice video. As always this is a good video, well explained. I always have a lot of sympathy for people, who like me, sometimes sing while doing math! Maybe it lacks an adapted formalism towards the end; for example when defining a sequence u(n), the proof by induction could have been more concise. But, the final result is missing, I expected the video may end up explaining that Q = 94259477598383594208516231244829367495623127947025437683278893534169775993162214765030878615918083469116234900035495995833697063026032639999999999999999999999999. I'm kidding ! 😄🎵🎵🎶
Or, he can use explicit multiplication with the '*' symbol, rather than use implied multiplication, and also use parentheses. The only issue is that he already uses a star symbol for his special operator, which is easily confused with the multiplication symbol. That's why I substituted his with '★' to be very clear. If I had an ASCII-only keypad, then I'd pick a character from the string "#$%&@\~".
first we prove that this operation is associative, this is easy just do it, we then note that 1*2+1=1+2+2+1=6=3!=(2+1)! and so we proceed inductively since a*b+1=(a+1)(b+1) (1*2*...*(n-1)*n) + 1 = (1*2*...*(n-1)+1)(n+1) = (n-1+1)!*(n+1)=(n+1)! so (1*2*...*(n-1)*n)=(n+1)!-1 so this is just 101!-1
Well, if you look at 101! , it has a factor of 5 for the 20 multiples of 5 from 5, 10, 15, ..., 100, plus an extra 5 in 25, 50, 75, and 100. That's 5²⁴ as a factor of 101! . There are a lot more than 24 factors of 2, so you know 2²⁴ is definitely a factor, too. [Probably 50+25+12+6+3+1, or 97 for 2⁹⁷.] That means 101! has 10²⁴ as a factor, and you can't get 10²⁵ out of it because there isn't another factor of 5. Thus, the value in base 10 ends in 24 zeroes. When you subtract 1 to get 101! - 1, those 24 zeroes become 24 nines. I got the number from Wolfram Alpha, and posted the 160-digit result, but Yoo Toob removed it. A version posted as four 40-digit groups is shad:owba:nned -- I can see my post only if I'm signed in.
Well, if you look at 101! , it has a factor of 5 for the 20 multiples of 5 from 5, 10, 15, ..., 100, plus an extra 5 in 25, 50, 75, and 100. That's 5²⁴ as a factor of 101! . There are a lot more than 24 factors of 2, so you know 2²⁴ is definitely a factor, too. [Probably 50+25+12+6+3+1, or 97 for 2⁹⁷.] That means 101! has 10²⁴ as a factor, and you can't get 10²⁵ out of it because there isn't another factor of 5. Thus, the value in base 10 ends in 24 zeroes. When you subtract 1 to get 101! - 1, those 24 zeroes become 24 nines. I got the number from Wolfram Alpha, and posted the 160-digit result, but Yoo Toob cen sored it. A version posted as four 40-digit groups is shadow ßåññëd -- only I can see my post, but I must be signed in.
If you want a fun result, calculate : 1&(2&(3&(...(999999999&1000000000)))) with a&b = (a+b) - a.b It also works with a%b = a.b -(a+b) which is not associative. Enjoy !
@Rando2101: Wait a minute ... The sum minus the product: a & b = a + b - a*b . Great observation for a = 1: 1 & b = 1 + b - 1*b = 1 . Let's prove (a & b) & c = a & (b & c) to collapse the original expression easily. (a & b) & c = (a + b - a*b) & c = (a + b - a*b) + c - (a + b - a*b)*c = (a + b +c) - (a*b + a*c + b*c) + a*b*c a & (b & c) = a & (b + c - b*c) = a + (b + c - b*c) - a*(b + c - b*c) = (a + b + c) - (b*c + a*b + a*c) + a*b*c Thus, (a & b) & c = a & (b & c) . So, we have 1 & 2 & 3 & ... & 10⁹ : (1 & 2) & 3 & ... & 10⁹ = (1 & 3) & ... & 10⁹ = 1 & ... & 10⁹ ... = 1 & 10⁹ = 1 That's faster than my solution.
I checked that you have 9 nines and 9 zeroes, which is hard on a tiny smartphone screen. You have 1&(2&(3&(...((10⁹-1)&10⁹)))) using the '&' operator, where: a & b = a + b - a*b = 1 - (a - 1)*(b - 1) Subtract 1 from each parameter, multiply, subtract product from 1. Note that the parameters are interchangeable. (a & b) & c = 1 - -(a - 1)*(b - 1)*(c - 1) = 1 + (a - 1)*(b - 1)*(c - 1) Note that a sign toggles for the product term, and the parameters are interchangeable. ((a & b) & c) & d = 1 - (a - 1)*(b - 1)*(c - 1)*(d - 1) Note that the sign toggles again ... (-1)ⁿ . Your problem uses consecutive integers from 10⁹ down to 1 with an odd number of '&' operations, so we have: 1 - (10⁹ - 1)*(10⁹ - 2)*...*(2 - 1)*(1 - 1) = 1 - (10⁹ - 1)!*0 = 1 At the point composing from 10⁹ down to 2, the result is: 1 + (10⁹ - 1)*(10⁹ - 2)*...*(2 - 1) = 1 + (10⁹ - 1)! And the next iteration is: 1 & (1 + (10⁹ - 1)!) = 1 - (1 - 1)*(10⁹ - 1)! = 1
That wasn't the part which has (k + 1)! as a factor. What's left after the cancellations is: (k + 1)*(k + 1)! - 1 + (k + 1)! = (k + 1)!*(k + 1 + 1) - 1 = (k + 1)!*(k + 2) - 1 = (k + 2)! - 1
I got to this site accidentally and fell in love with it 😅 then subscribed. At the end, after getting the final answer, I got hungry; therefore, I ate a bowl of SPECIAL K 😂. Thank you!
Nope, you're not. I fell for this at first until I realized that the "*" operator was not supposed to mean multiplication but was to be introduced/defined as a _new_ operator. They could have used a less ambiguous symbol or put the word "let" or "define" on the thumbnail text.
@@oahuhawaii2141 Ah cool. My computer keyboard doesn’t seem to have any non-ASCII characters, and my phone keyboard doesn’t seem to have any of those stars, so I had just copy pasted after searching for Unicode asterisks
He's sloppy with his parentheses. His first Q is missing the open parenthesis before the 4. His second Q has the latter 2 close parentheses in the wrong positions; "*)99*100)" should be ")*99)*100”. By 20:00, he missed the open parenthesis after "Assume" and repeated the mistake twice again. I dislike his notation using '*' because it's the standard multiplication operator. He should've used something more distinctive, such as '★' or '$', and insert '*' instead of using implied multiplication. His operator is better shown as: a ★ b = a*b + a + b . Thus, the product sum is (a + 1)*(b + 1) - 1 . If we substitute b with (b ★ c) , we have: a ★ (b ★ c) = (a + 1)*((b ★ c) + 1) - 1 = (a + 1)*((b + 1)*(c + 1) - 1 + 1) - 1 = (a + 1)*(b + 1)*(c + 1) - 1 Therefore, the problem is: 1 ★ (2 ★ (3 ★ (4 ★ ... (99 ★ 100) ... ))) = 2*3*4*5* ... 100*101 - 1 = 101!/1 - 1 = 101! - 1
I broke it up in chunks of 40-digit blocks to be accepted by You Toob: 9425947759838359420851623124482936749562 3127947025437683278893534169775993162214 7650308786159180834691162349000354959958 3369706302603263999999999999999999999999
This question is so easy. Are you sure you took it from MIT something? Or am I so intelligent because I solved it? Or is there something else? Probably there is something else that I don't know
Here's a challenge idea for you. Prove the commutative property of multiplication under 1. Whole numbers Then maybe later (the true challenge) 2. The reals
I think I spent 40 minutes taking notes on this I might be cooked 💀💀💀 Note: I made a typo in my writing and got confused so I went back on the last 4 minutes trying to figure it out LMAO 😭😭😭
Why not start with the definition, and apply this from the start? a ★ b = a*b + a + b = b ★ a = a*b + a + b + 1 - 1 = (a + 1)*(b + 1) - 1 Thus: (n! - 1) ★ n = n!*(n + 1) - 1 = (n + 1)! - 1 But more importantly: (a ★ b) ★ c = (a + 1)*(b + 1)*(c + 1) - 1 This means the parentheses on the LHS can be removed: (a ★ b) ★ c = a ★ b ★ c . The original problem becomes: (1+1)*(2+1)*...*(99+1)*(100+1) - 1 = 2*3*...*100*101 - 1 = 101! - 1
101! - 1 equals: 9425947759838359420851623124482936749562312794702543768327889353416977599316221476503087861591808346911623490003549599583369706302603263999999999999999999999999 . If you prefer 101! , then add 1 to it.
I broke it up in chunks of 40-digit blocks to be accepted by You Toob: 9425947759838359420851623124482936749562 3127947025437683278893534169775993162214 7650308786159180834691162349000354959958 3369706302603263999999999999999999999999
A small hint: Once you've proved associativity for the * operator, you can leave out all parentheses in your starred products, thus saving you a lot of handwriting.
Oh my! Just realized that.
@@PrimeNewtons I was also wondering why you made a calculation mistake in the middle, i was really comfused how u got it
• a*b=b*a
• a*b = ab+a+b+1-1
= a(b+1)+(b+1) -1
→ a*b+1 = (a+1)(b+1)
• (a*b)*c = (a*b+1)(c+1) -1
= (a+1)(b+1)(c+1) -1
• ((a*b)*c)*d = ((a*b)*c+1)(d+1) -1
= (a+1)(b+1)(c+1)(d+1) -1
And the pattern continues.
So ((100*99)*98)*97.....*3)*2)*1)
= 101×100×99....×2 -1
=101! -1
excellent
It's for all case independent from whether it's associative or not. I think this is better than induction
@@adchayansivakumar1667 i mean, the ... hides an induction somewhere, but it is indeed better than just proving it for 1->100
I also solved it the same way as yours..
@@adchayansivakumar1667 it’s the same as induction, any sort of repeated continuation to a pattern is underlying induction. Induction just adds rigour to everything and is not needed when the result is clear, for example showing n! = n(n-1)…1 using n! = n(n-1)! for all naturals n. Same is true here
Rigorously: a*(b*c) = (a+1)(b+1)(c+1) - 1 = (a*b)*c. So we can use operator (*) in any order. Claim: 1*...*n = (n+1)!-1. Indeed: 1*2 = 3! -1 and if 1*...*(k-1) = k!-1, then 1*...*k = (k!-1)k + k! - 1 + k = (k+1)! - 1 which finishes the proof by induction. Therefore 1*(2*...(99*100))...) = 1*...*100 = 101! - 1
Don't overcomplicate the question
A simple idea
You can just do something like this and avoid induction (as it's not for general case)
A*B=(A+1)(B+1)-1
That implies A*B+1=(A+1)(B+1)
For (A*B)*C=(A*B+1)(C+1)-1
That equals to (A+1)(B+1)(C+1)-1
The pattern continues
So Q=(100+1)(99+1)....(1+1)-1
So, Q=101×100×...×2-1
Rewrite as 101×100×...2×1-1
Thus 101!-1.
You can use fictional blackboards and simply solve that
@@adchayansivakumar1667 "The pattern continues"
You can't "avoid induction", since you _need_ induction to _prove_ that the pattern indeed continues.
@@adchayansivakumar1667 Refer to the first word in my comment to get why it's written that way.
@@bjornfeuerbacher5514To avoid induction, leave it as an exercise to the reader
Votre phrase " so we can use... in any order" prête à confusion. Pour changer L'ORDRE des termes il faut que la loi soit COMMUTATIVE. Il serait préférable d'utiliser une phrase du genre " we can begin to ASSOCIATE two éléments WHERE we want" puisque la commutativité n'a pas encore été démontrée, bien qu'elle soit évidente. Ou bien préciser avant d'écrire votre phrase que la loi commutative. Ce n'est qu'un détail.
Unusual application of this operator is to describe the n-dimensional triangle family. With a bit of imagination the star operator definition can be read as a*b = line segment ab + the points a and b. Then (a*b)*c reads as the triangular face abc + the 3 line segments (ab, ac, bc) and 3 corners (a, b, c). And (((a*b)*c)*d) gives the tetrahedra abcd, the 4 faces abc, abd, acd, bcd, the 6 edges ab, ac, ad, bc, bd, cd, and 4 corners a, b, c, d. Continuing with 5 corners, a*...*e, gives the 4-simplex. Etc. , with the operation always giving one of each term for uniqueness. Fun compaction. Thank you for such beautiful teaching.
Thank you for your input. This added a dimension of practicality to the math.
@PrimeNewtons Aw garsh! Thanks for not wincing at my playing fast and loose with the "=" sign. I love your teaching style. It's as if you're teaching from inside of us, with rich emotional expressions of perplexity, false starts, ah hahs, and fascination.❤
What a cool problem. This is about as close as I can get to abstract algebra without losing my mind!
If like me you did not spot the link to factorials, there is another approach with an interesting connection to Vieta's formulas...
Let f(x) = (x-a)(x-b) = x^2 - (a+b)x + ab
f(-1) = 1 + (a+b) + ab = 1 + a*b
⇒ a*b = f(-1) - 1
(a*b)*c = (a+b+ab) + c + (a+b+ab)c
= (a+b+c) + (ab+bc+ca) + abc
Let f(x) = (x-a)(x-b)(x-c)
⇒ f(x) = x^3 - (a+b+c)x^2 + (ab+bc+ca)x - abc
f(-1) = -1 - (a+b+c) - (ab+bc+ca) - abc = -1 - (a*b)*c
⇒ (a*b)*c = - f(-1) - 1
Similarly ((a*b)*c)*d
= (a+b+c+d) + (ab+ac+ad+bc+bd+cd) + (abc+abd+acd+bcd) + abcd
Let f(x) = (x-a)(x-b)(x-c)(x-d)
⇒ ((a*b)*c)*d = f(-1) - 1
Let Q[n](x) = (x-1)(x-2)(x-3)...(x-n)
⇒ Q[n](-1) = (-1-1)(-1-2)(-1-3)...(-1-n) = (-1)^n . (n+1)!
Let q[1] = 1
and q[n] = q[n-1]*n for n>1
⇒ q[n] = (...((1*2)*3)*...)*n
So the pattern appears to suggest that
q[n] = (-1)^n . Q[n](-1) - 1
⇒ q[n] = (n+1)! - 1
This is proved using induction as in the video.
You are a great teacher. I am not only learning math but also to teach well. Thanks for doing what you do!
Mister prime, your videos are so well detailed and your way of teaching is so charismatic, it doesn't feel like a boring math video at all. I wish I had a video for every problem in my daily life with you explaining how to solve it. Keep up with the good work.
Your solution is COMPLETE, thanks!
Exploit a * b = ab + a + b = ( ab + a + b + 1 ) - 1 = (a+1)(b+1) - 1 by defining f(x) = x + 1.
Note that f-inverse(x) = x - 1.
Thus a * b = (a+1)(b+1) - 1 = f(a) f(b) - 1 = f-inverse( f(a) f(b) ).
Also, applying f to both sides gives f(a*b) = f(a) f(b).
Then a * (b * c ) = f-inverse( f(a) f(b *c) ) = f-inverse( f(a) { f(b) f(c) } ) = f-inverse( f(a) f(b) f(c) ).
Similarly a * ( b * ( c * d ) ) = f-inverse( f(a) f( b * ( c * d ) ) ) = f-inverse( f(a) f(b) f(c * d) ) = f-inverse( f(a) f(b) f(c) f(d) ).
The inductive clearly works (of course, if doing "real math", the proof of the induction should be written out, as was done in the video.)
Thus
1 * ( 2 * ( 3 * ( 4 * ( ... * ( 99 * 100 ) )))...)
= f-inverse( f(1) f(2) f(3) ... f(100) )
= f-inverse( (2) (3) (4) ... (101) )
= f-inverse( 101! )
= 101! - 1.
(If you know some abstract algebra, f(a*b) = f(a) f(b) might make you suspicious of the following, which turns out to be true: That (Reals\{-1}, *, 0) is a group, and f is a group isomorphism from (Reals\{-1}, *, 0) to (Reals\{0}, ordinary-multiplication, 1).)
Without assuming commutation for *, you can observe that (n-1) * n = (n-1)n + (n - 1) + n = (n-1 + 1 + 1)n - 1 = (n + 1)n - 1 = n(n+1) - 1
Then the next * calculation to the left would be (n-2) * [n(n+1) - 1] = (n-2) [ n(n+1) -1 ] + (n - 2) + [n(n+1) - 1]
= (n-2) [ n (n+1) ] - (n - 2) + (n -2) + n(n+1) - 1 (distribute)
= (n-2 + 1)[ n (n+1) ] - 1 (combine the n(n+1) terms)
=(n-1)n(n+1) -1
If you do the same for n-3 then you'll see the pattern as you apply * on the left.
n(n+1) -1 ==> (n-1)n(n+1) - 1 ==> (n-2)(n-1)n(n+1) - 1 and so on
i havent watch the full video, just figure out by myself
.
a*b+1 = (a+1)(b+1)-1 (notice that a turns into a+1, b turns to b+1)
---> a*(b*c) = a * ((b+1)(c+1)-1) = (a+1)(b+1)(c+1) - 1 (use the fact above)
---> general formula: a1*a2*.....*an = (a1+1)(a2+1).....(an+1)-1
replace a1=1, a2=2,.....,a100=100, we have Q = (1+1)(2+1)(3+1).....(100+1)-1 = 101!-1
This is nice, I think a more insightful way of seeing the problem. The version in the video uses a technique that can be applied (or, at least attempted) in more circumstances, but your answer I think says more about “what is actually happening” in the problem.
(Namely, the star operation is basically the “conjugation” of the multiplication operation by a bijection from the set of integers to itself, and so the answer is obtained by just applying the bijection in one direction to all the inputs, then doing all the multiplications, and then applying the bijection in the other direction).
My keyboard has '*', '×', and '★'. I use '*' as my normal multiply operator for regular math, symbolic inputs (spreadsheets, WolframAlpha, ...), and computer programming, so I substitute the distinctive '★' character for his new operator to avoid any confusion.
We define the ★ operator as:
a ★ b = a*b + a + b
We find interesting properties:
a ★ b = a*b + a + b + 1 - 1
= (a + 1)*(b + 1) - 1 { Eq. 1 }
= (b + 1)*(a + 1) - 1
= b ★ a { Eq. 2 }
We also find:
(c ★ (b ★ a)) = ((b ★ a) ★ c)
= (a ★ b) ★ c
= ((a + 1)*(b + 1) - 1) ★ c
= (a + 1)*(b + 1)*(c + 1) - 1
= a ★ b ★ c
We next find:
(a ★ b ★ c) ★ d = (a + 1)*(b + 1)*(c + 1)*(d + 1) - 1
= a ★ b ★ c ★ d
= (a ★ b) ★ (c ★ d)
Composing the function with itself allows the sequence of its parameters to be reordered in any manner. The function increments each parameter by 1, replaces '★' with '*', and decrements the product by 1. Thus, the original problem becomes:
1 ★ 2 ★ 3 ★ 4 ★ ... 99 ★ 100
= 100 ★ 99 ... ★ 4 ★ 3 ★ 2 ★ 1
= (100+1)*(99+1)*...*(4+1)*(3+1)*(2+1)*(1+1) - 1
= 101*100*...*5*4*3*2*1/1 - 1
= 101! - 1
Let A(n), n >= 0, be any sequence of elements in a ring (not necessarily Abelian) with unity 1. In this example, the ring is Z and A(n) = 100 - n. Define
K(0) = A(0), K(n+1) = A(n+1)*K(n), for n >= 0, where a*b = ab + a+ b.
Lemma. K(n) = [product (for i = n to 0) (A(i) +1)] - 1, for n >= 0.
Proof. The Lemma holds for n = 0. Suppose the Lemma holds for some n >= 0. Then
K(n+1) = A(n+1)*K(n) = A(n+1)*([product (for i = n to 0) (A(i) +1)] -1) =
A(n+1)([product (for i = n to 0) (A(i) +1)] -1)+A(n+1)+ [product (for i = n to 0) (A(i) +1)] - 1 =
A(n+1)[product (for i = n to 0)(A(i) +1)] - A(n+1)
+A(n+1) + [product (for i= n to 0)(A(i) +1)] - 1 =
A(n+1)[product (for i = n to 0) (A(i) +1)] + [product (for i = n to 0) (A(i) +1)] - 1 =
(A(n+1)+1)[product(for i = n to 0) (A(i) +1)] - 1 = [product(for i = n+1 to 0) (A(i) +1)] - 1,
completing the proof of the Lemma.
In this example, we want to calculate
K(99) = [product(for i = 99 to 0)(A(i) + 1)] - 1 = [product(for i = 99 to 0)(101 - i)] - 1 =
(2)(3)...(100)(101) - 1 = 101! - 1.
Best Tagline of the year --> "This right here you can see that you have to STOP !!" - This caught be off-guard
Complicated ,but nicely explained,and brilliantly solved.
Thank you for another of your inspirational and blessed videos, which I love watching.
I liked your discussion and advantageous use of associativity to solve this problem.
This alternative approach avoids the need for this though and I think is easier.
Note that a*b = ab+a+b = (a+1)(b+1)-1
Hence 99*100 = (99+1)(100+1) - 1
= (100)(101) - 1
The -1 at the end of this number ensures that when it is an input to * the bracket (this number + 1) is a clean product
Thus 98*(99*100) = (99)(100)(101) - 1
Again there is -1 at the end so the above is again true for the next application of *
Repeating this process, the required answer is therefore (2)(3)...(100)(101) -1 = 101! -1.
Again, thank you for your lovely videos and i look forward to seeing the next one. God bless you ♥️
my approach was just:
a*b = a + (1+a)b
Then Q can be expressed
1*(2*(3*(....) -->
1 + 2(2*(3*(...) -->
1 + 2(3 + 4(5 + 6(...98 + 99(99 + 100 times 100)...)
Because multiplication is distributive, then for that last 100, I have a chain of 100*99*98...*2*1 as its coefficient
For the last 99, similarly I have a 99! as its coefficient
For the last 98, I have 98! as its coefficient
and so on
So I wind up with
Sum_{n=1 to 100} of n * Product_{m=1 to n} of m
Sum_{n=1 to 100} of n * n!
Oh hey, an identity for (n+1)! - 1
❤brilliant
Magnífica questão. Ótima explicação.
You are a wonderful teacher! Thanks for video!
Wow i think you are overcomplicating this. My first instinct was to rewrite * in a more compact way.
a*b= ab+a+b=ab+a+b+1-1= a(b+1)+b+1-1=(a+1)(b+1)-1
Now for a*b*c it becomes much easier to evaluate as either :
(a*b)*c= ((a+1)(b+1)-1)*c=((a+1)(b+1)-1+1)(c+1)-1= (a+1)(b+1)(c+1)-1 and similar for a*(b*c) but going by the same logic the original expression becomes 2x3x...x101-1
This is what i think :)
pardonne-moi mais ce que tu proposes n'est pas plus simple mais en plus la rédaction n'est pas assez rigoureuse, car la récurrence n'est pas explicitement écrite.
Un exposé clair et soigné ; une bonne méthode de résolution. Exercice intéressant qui montre bien l'utilité de l'associativité. 👍
Une petite remarque bien pratique à propos des parenthèses: quand une loi est ASSOCIATIVE l'écriture des PARENTHÈSES devient INUTILE puisque on peut commencer à associer les éléments là où on veut (sans changer l'ordre, sauf si la loi est commutative, cette loi * est commutative) Ce qui simplifie grandement l'écriture des calculs.
a*(b*c) = (a*b) *c = a*b*c donc * est associative
on peut écrire a*b*c*d*e= (a*b)*c*d*e = a*(b*c)*d*e= a*b*(c*d)*e = a*b*c*(d*e)
This is a simplified version of IFYM 2023 here in bulgaria, it was a nice problem. You just prove that * is commutative, and go for induction
ATTEMPT:
Try the first few terms.
1 * 2 = 5
1 * (2 * 3) = 23
1 * (2 * (3 * 4)) = 119
Seems like (n+1)! - 1
PF:
a * b = ab + a + b
= ab + a + b + 1 - 1
= (a+1)(b+1) - 1
Now consider:
a * (b * c) = a * ((b+1)(c+1) - 1)
= (a+1)(b+1)(c+1) - 1
= ((a+1)(b+1) - 1) * c
= (a * b) * c
* is associative, so ignore all brackets.
1 * 2 * 3 * ... * 100
= 2 × 3 × 4 × ... × 101 - 1
= 101! - 1
forgot to prove this pattern continues:
a * b = (a+1)(b+1) - 1
a * b * c = (a+1)(b+1)(c+1) - 1
assume a * b * c * ... * m = (a+1)(b+1)(c+1)...(m+1) - 1
a * b * c * ... * m * n = ((a+1)...(m+1) - 1) * n
= (a+1)...(m+1)(n+1) - 1
and the proof is done:
a * b * c * ... * n = (a+1)(b+1)(c+1)...(n+1) - 1
You used the * to simultaneously mean two different operations. But I get what you mean.
@@Grecks75 oops
If you realize that b * a = (b + 1)(a + 1) - 1
a recursion pattern is easy to recognize:
c * (b * a) = (c+1)((b+1)(a+1)-1+1) - 1
which simplifies to c*(b*a) = (c+1)(b+1)(a+1) - .1
similarly d*(c*(b*a)) = (d+1)(c+1)(b+1)(a+1)-1
Therefore the answer I get is 1*(2*(3*....(99*100)...)) = 101! - 1
пф элементарно. рассмотрев a*b*c можно понять это это все суммы симметрических многочленов, что так же равно произведению (1+ax)(1+bx)... за вычетом единички, и при x=1. получаем очевидно 101!-1. решение занимает пол минуты в голове
I love your videos❤.
Your videos are really addictive.
Note that (99*100 + 99) + 100 = 99*101 + 100 = 100*101 - 1, same if starts with 100, and with no loss of generality, a*b = (a + 1)(b + 1) - 1...
The rewriting of parentheses part is flawed. The star’s associative property has only been proven within three numbers, whereas to move the parenthesis from far right end to far left end involves a lot more numbers and is yet to be verified. This can be solved by introducing a number sequence defined as p(n)=(((…(x*(x-1))…)*(x-n), and it is not hard to find its formula by using math induction: p(n)=(x+1)!/(x-n)!-1. Then let n=x-1, (((…(x*(x-1))…)*(x-n)=
(((…(x*(x-1))…)*4)*3)*2)*1=(x+1)!-1, and we have the formula for the problem. Let x=100, we have the final answer that is 101!-1.
I did this in a general intuitive way..
I tried a*b*c and it is a+b+c+ab+ac+bc+abc
So according to my intuition, a*b*c*......*n
(n terms 🥲🥲pls adjust)
will contain the sum of the terms taken 1,2,...,and n at a time.(it follows from a simple induction)
If we add 1 to it, it will be,
(a+1)(b+1)... (n+1)
According to the binomial expansion..
Let X(n)=a*b*c*...n*
Then X(n)=(a+1)(b+1)...(n+1)-1
So X(100)=101!-1..
Is this right primenewtons? Or any other guys?...
Thanks
yes it is true
you actually can do it easier
a*b = ab + a + b + 1 - 1
a*b = a(b+1) + 1(b+1) - 1
a*b = (a+1)(b+1) - 1
a*(b*c) = a*[ (b+1)(c+1) -1 ]
= (a+1)(b+1)(c+1) - 1
(a*b)*c = [ (a+1)(b+1) - 1 ]*c
= (a+1)(b+1)(c+1) - 1
which means "*" operator is associative and commutative and you can just add 1 to all the "multiples" multiply them and subtract 1 which is n!-1 for 1*2*3*...*n
@@hannoii thanks
yeah. this is super elegant.
no need associativity.
(a*b) = ab + a + b = (a+1)(b+1) - 1
so (99*100) = (99+1)(100+1) - 1 = 100(101) - 1
98*(99*100) = (98 + 1)((100(101) - 1) + 1) - 1 = 99((100(101) + (-1)) + 1) - 1 = 99(100(101) + ((-1) + 1)) - 1 = 99(100(101) + 0) - 1 = 99(100(101)) - 1
97*(98*(99*100)) = 98(99)(100)(101) - 1
...
(1*(2*(3*...*(99*100)...))) = (2)(3)...(98)(99)(100)(101) - 1 = 1(2)(3)...(101) - 1 = 101! -1
@@hannoii nice thats my solution too
😂 exactly I did the same
Using symmetric polynomials you can arrive at 1*2*3*…*100=-x^100 + prod (1
So glad found your videos!
Great video as always, huge respect to you.
So clever - the proof of commutativity justified too!
Given the numbers 1,2,...,100 on a board.
We'll remove 2 of the numbers a,b and replace them with a*b
Let S be the product of (each number on the board+1)
So initially, S=101!
After removing, find S' the same way
S'/S=(a*b+1)/((a+1)(b+1))=1
Therefore S doesn't change regardless of how we choose a and b
Perform this action until only 1 number is left. It would be 101!-1.
from Morocco thank you for this wonderful problem and proof
Again, nice video. As always this is a good video, well explained.
I always have a lot of sympathy for people, who like me, sometimes sing while doing math!
Maybe it lacks an adapted formalism towards the end; for example when defining a sequence u(n), the proof by induction could have been more concise.
But, the final result is missing, I expected the video may end up explaining that Q = 94259477598383594208516231244829367495623127947025437683278893534169775993162214765030878615918083469116234900035495995833697063026032639999999999999999999999999.
I'm kidding ! 😄🎵🎵🎶
notice that * is associative and a1*…*an = (a1+1)(a2+1)…(an+1)-1 (it easily can be checked by induction). So 1*2*…*100=2•3•…•101-1=101!-1
man thats was so nice to watch, i like your videos.
19:51 It's better to write((k+1)+1)(k+1)! b/c it's like the subfactorial function"!n"
Or, he can use explicit multiplication with the '*' symbol, rather than use implied multiplication, and also use parentheses. The only issue is that he already uses a star symbol for his special operator, which is easily confused with the multiplication symbol. That's why I substituted his with '★' to be very clear. If I had an ASCII-only keypad, then I'd pick a character from the string "#$%&@\~".
first we prove that this operation is associative, this is easy just do it, we then note that
1*2+1=1+2+2+1=6=3!=(2+1)!
and so we proceed inductively
since a*b+1=(a+1)(b+1)
(1*2*...*(n-1)*n) + 1 = (1*2*...*(n-1)+1)(n+1) = (n-1+1)!*(n+1)=(n+1)!
so (1*2*...*(n-1)*n)=(n+1)!-1 so this is just 101!-1
With luck and more power to you.
hoping for more videos.
Simply magic! We don't know what Q is but we do know that it enda with a whole bunch of 9. Yes but how many? ☺
Never stop searching ...
Well, if you look at 101! , it has a factor of 5 for the 20 multiples of 5 from 5, 10, 15, ..., 100, plus an extra 5 in 25, 50, 75, and 100. That's 5²⁴ as a factor of 101! . There are a lot more than 24 factors of 2, so you know 2²⁴ is definitely a factor, too. [Probably 50+25+12+6+3+1, or 97 for 2⁹⁷.] That means 101! has 10²⁴ as a factor, and you can't get 10²⁵ out of it because there isn't another factor of 5. Thus, the value in base 10 ends in 24 zeroes. When you subtract 1 to get 101! - 1, those 24 zeroes become 24 nines.
I got the number from Wolfram Alpha, and posted the 160-digit result, but Yoo Toob removed it. A version posted as four 40-digit groups is shad:owba:nned -- I can see my post only if I'm signed in.
Well, if you look at 101! , it has a factor of 5 for the 20 multiples of 5 from 5, 10, 15, ..., 100, plus an extra 5 in 25, 50, 75, and 100. That's 5²⁴ as a factor of 101! . There are a lot more than 24 factors of 2, so you know 2²⁴ is definitely a factor, too. [Probably 50+25+12+6+3+1, or 97 for 2⁹⁷.] That means 101! has 10²⁴ as a factor, and you can't get 10²⁵ out of it because there isn't another factor of 5. Thus, the value in base 10 ends in 24 zeroes. When you subtract 1 to get 101! - 1, those 24 zeroes become 24 nines.
I got the number from Wolfram Alpha, and posted the 160-digit result, but Yoo Toob cen sored it. A version posted as four 40-digit groups is shadow ßåññëd -- only I can see my post, but I must be signed in.
[13:33] Souka Nayo! a*b = (b+1)! - 1 "under some conditions": if a = sum[1*...*(b-1)]
I love this question and you solution .
Very good 👍
If you want a fun result, calculate : 1&(2&(3&(...(999999999&1000000000)))) with a&b = (a+b) - a.b
It also works with a%b = a.b -(a+b) which is not associative.
Enjoy !
Haha.
It reminds me of the old joke of multiplying out (a-x)(b-x)(c-x)...(z-x)
1&b=1
bruh 😭😭😭
@Rando2101: Wait a minute ...
The sum minus the product: a & b = a + b - a*b .
Great observation for a = 1: 1 & b = 1 + b - 1*b = 1 .
Let's prove (a & b) & c = a & (b & c) to collapse the original expression easily.
(a & b) & c = (a + b - a*b) & c
= (a + b - a*b) + c - (a + b - a*b)*c
= (a + b +c) - (a*b + a*c + b*c) + a*b*c
a & (b & c) = a & (b + c - b*c)
= a + (b + c - b*c) - a*(b + c - b*c)
= (a + b + c) - (b*c + a*b + a*c) + a*b*c
Thus, (a & b) & c = a & (b & c) .
So, we have 1 & 2 & 3 & ... & 10⁹ :
(1 & 2) & 3 & ... & 10⁹
= (1 & 3) & ... & 10⁹
= 1 & ... & 10⁹
...
= 1 & 10⁹
= 1
That's faster than my solution.
I checked that you have 9 nines and 9 zeroes, which is hard on a tiny smartphone screen. You have 1&(2&(3&(...((10⁹-1)&10⁹)))) using the '&' operator, where:
a & b = a + b - a*b
= 1 - (a - 1)*(b - 1)
Subtract 1 from each parameter, multiply, subtract product from 1. Note that the parameters are interchangeable.
(a & b) & c = 1 - -(a - 1)*(b - 1)*(c - 1)
= 1 + (a - 1)*(b - 1)*(c - 1)
Note that a sign toggles for the product term, and the parameters are interchangeable.
((a & b) & c) & d = 1 - (a - 1)*(b - 1)*(c - 1)*(d - 1)
Note that the sign toggles again ... (-1)ⁿ .
Your problem uses consecutive integers from 10⁹ down to 1 with an odd number of '&' operations, so we have:
1 - (10⁹ - 1)*(10⁹ - 2)*...*(2 - 1)*(1 - 1)
= 1 - (10⁹ - 1)!*0
= 1
At the point composing from 10⁹ down to 2, the result is:
1 + (10⁹ - 1)*(10⁹ - 2)*...*(2 - 1)
= 1 + (10⁹ - 1)!
And the next iteration is:
1 & (1 + (10⁹ - 1)!) = 1 - (1 - 1)*(10⁹ - 1)! = 1
Great. As always
👍👍👍
20:07 why can we factor out (k+1)! When we have -1 as itself? Wouldn’t it result in … -1/(k+1)! …?
That wasn't the part which has (k + 1)! as a factor. What's left after the cancellations is:
(k + 1)*(k + 1)! - 1 + (k + 1)!
= (k + 1)!*(k + 1 + 1) - 1
= (k + 1)!*(k + 2) - 1
= (k + 2)! - 1
Beautiful😆
the star operator?? is that a convolution??
I got to this site accidentally and fell in love with it 😅 then subscribed. At the end, after getting the final answer, I got hungry; therefore, I ate a bowl of SPECIAL K 😂. Thank you!
first to view and comment
Love from India 🇮🇳🇮🇳❤️❤️
Am i the only one who thought a*b=ab+a+b was a+b=0
Nope, you're not. I fell for this at first until I realized that the "*" operator was not supposed to mean multiplication but was to be introduced/defined as a _new_ operator. They could have used a less ambiguous symbol or put the word "let" or "define" on the thumbnail text.
@@Grecks75 agreed
@@Grecks75it isn’t supposed to be “*” so much as something more like “✱” or maybe “✶” .
@drdca8263: My keyboard has '*', '×', and '★'. I use '*' as my normal multiply operator, and substituted '★' for his new operator.
@@oahuhawaii2141 Ah cool. My computer keyboard doesn’t seem to have any non-ASCII characters, and my phone keyboard doesn’t seem to have any of those stars, so I had just copy pasted after searching for Unicode asterisks
P.N. says : "i have seen the future", lol. . nice voice.
19:48 Writing the open parenthesis inside and the closing parenthesis outside hurt my brain.
He's sloppy with his parentheses. His first Q is missing the open parenthesis before the 4. His second Q has the latter 2 close parentheses in the wrong positions; "*)99*100)" should be ")*99)*100”. By 20:00, he missed the open parenthesis after "Assume" and repeated the mistake twice again.
I dislike his notation using '*' because it's the standard multiplication operator. He should've used something more distinctive, such as '★' or '$', and insert '*' instead of using implied multiplication. His operator is better shown as: a ★ b = a*b + a + b . Thus, the product sum is (a + 1)*(b + 1) - 1 . If we substitute b with (b ★ c) , we have:
a ★ (b ★ c) = (a + 1)*((b ★ c) + 1) - 1
= (a + 1)*((b + 1)*(c + 1) - 1 + 1) - 1
= (a + 1)*(b + 1)*(c + 1) - 1
Therefore, the problem is:
1 ★ (2 ★ (3 ★ (4 ★ ... (99 ★ 100) ... )))
= 2*3*4*5* ... 100*101 - 1
= 101!/1 - 1
= 101! - 1
so whats the answer? i wanna see big number
9425947759838359420851623124482936749562312794702543768327889353416977599316221476503087861591808346911623490003549599583369706302603263999999999999999999999999.
9425947759838359420851623124482936749562312794702543768327889353416977599316221476503087861591808346911623490003549599583369706302603263999999999999999999999999
9425947759838359420851623124482936749562312794702543768327889353416977599316221476503087861591808346911623490003549599583369706302603263999999999999999999999999
I listed the full number, but Yoo Toob cen sors answers for no good reason.
I broke it up in chunks of 40-digit blocks to be accepted by You Toob:
9425947759838359420851623124482936749562
3127947025437683278893534169775993162214
7650308786159180834691162349000354959958
3369706302603263999999999999999999999999
This question is so easy. Are you sure you took it from MIT something? Or am I so intelligent because I solved it? Or is there something else? Probably there is something else that I don't know
Here's a challenge idea for you.
Prove the commutative property of multiplication under
1. Whole numbers
Then maybe later (the true challenge)
2. The reals
Tf you mean you just saw that it was 1 less than the next factorial no shot
I brute forced it. Letting a_n=na_{n-1}+a_{n-1}+n and then applyied exponential generating function. And then I read the comments...😂
Easiest solution:
a*b = ab+a+b = (a+1)(b+1)-1
99*100 = (100)(101)-1
98*(99*100)= (99)(100)(101)-1
.
.
.
1*(3.4.5......101-1) = 101! -1
It's just me, or drawing a 9 with two strokes is very strange?
once associativity is proven you don't need to write the parentheses
احسنت
Its so nice when math enthusiasts share ideas .....thank prime Newtons for such a community
I think I spent 40 minutes taking notes on this I might be cooked 💀💀💀
Note: I made a typo in my writing and got confused so I went back on the last 4 minutes trying to figure it out LMAO 😭😭😭
Audio is too low, always (just a suggestion)
Then crank up the volume.
Mama mia!!!!
I could solve this in my head within a minute, was just wondering why the video was over 20min long
This video was probably not for you.
@@PrimeNewtons i prefer the full tour which is more important than the answer itself
98*99*100=999899
99*100=10099
a*b=b*a
(n!-1)*n=(n!-1)(n)+n!-1+n= n(n!)-n+n!-1+n= n!(n+1)+n-n-1=(n+1)!-1 which is what you want to prove.
Why not start with the definition, and apply this from the start?
a ★ b = a*b + a + b
= b ★ a
= a*b + a + b + 1 - 1
= (a + 1)*(b + 1) - 1
Thus:
(n! - 1) ★ n = n!*(n + 1) - 1
= (n + 1)! - 1
But more importantly:
(a ★ b) ★ c = (a + 1)*(b + 1)*(c + 1) - 1
This means the parentheses on the LHS can be removed: (a ★ b) ★ c = a ★ b ★ c .
The original problem becomes:
(1+1)*(2+1)*...*(99+1)*(100+1) - 1
= 2*3*...*100*101 - 1
= 101! - 1
Now you just have to work out what 101! Is, quick 5 minute job 😂
101! - 1 equals:
9425947759838359420851623124482936749562312794702543768327889353416977599316221476503087861591808346911623490003549599583369706302603263999999999999999999999999 . If you prefer 101! , then add 1 to it.
I listed the full number, but Yoo Toob cen sors answers for no good reason.
I broke it up in chunks of 40-digit blocks to be accepted by You Toob:
9425947759838359420851623124482936749562
3127947025437683278893534169775993162214
7650308786159180834691162349000354959958
3369706302603263999999999999999999999999
Hello hi
101!-1, where 101! is 101 factorial
99(100)+99+100=10099
nope, * in this case is not multiplication, a*b=ab+a+b
@namanhnguyen7933 Yes. This is a crafty problem, with which you have to be very careful.
99*100 = 10099, in this case. Now fold in the others. Warning: It's getting huge! 😅
99(100)+99+100=10099
99(100)+99+100=10099
Would help if you wrote the question down correstly. Subscribe ? No thanks
Those who stop learning, stop living, but existing 🏸
逻辑的奥义！ 你滴明白？🤣
98(10099)+98+10099=999899
98*99*100 = 999899
@@Grecks75 Correct