Elegant way of finding ONE of the solutions. But there is more than one solution. The mistake made in the video is discounting 10 and 1 (and 0, since 0! = 1) as possible values for m and n. 10! x 1! = 10! => m = 10, n = 1 or m = 1, n = 10 => mn = 10 x 1 = 10 is also a correct possible solution. 10! x 0! = 10! => m = 10, n = 0 or m = 0, n = 10 => mn = 10 x 0 = 0 is also a correct possible solution.
You need to state some restriction on m,n in the initial problem since clearly m=10, n=0, or m=10,n=1 for example also work as 10!1!=10! and 10!0!=10!.
@@jacobgoldman5780 Anyone watching this and then reading the comments will think, if math people can’t agree, how am I supposed to understand it. “Content creators” should make sure their content is correct. I follow another creator who puts out videos every day. Several each month have errors. It’s not like 20 years ago, where you had to have a publisher put out your content that was reviewed for errors. Even then a few errors snuck through. It’s great that anyone can reach an audience, but if errors are posted, eventually nothing in the internet will be trusted.
Nice problem and nice solution. The mathematician in me says you needed to discuss the values of 10 and 0 (10! and 1!, and 10! and 0!), but the entertainer in me says that this was done perfectly. Well done !!!
Alternate solution: We know immediately that either m or n>=7, since no factorial less than 7 can generate 7 as a factor of 10!. From there it is trivial to check; just take a factorial between 7 and 10 and divide 10! by it and check if it's a factorial. 10!=3628800 3628800/(7!)=720=6!, so we have a solution (6; 7) and 6*7=42. 3628800/(8!) is not a factorial, and neither is 3628800/(9!). 3628800/(10!)=1, and we know 1!=1 and 0!=1, so our answers are 42, 10, and 0.
In my opinion is better to start from primes, from the biggest one. 7 is prime, so one of them is 7 or more . 5 is prime 10! has 5*5 so if both are less then 10 second must be 5 or more.
@@JossWainwright I imagine it like it was shown in that video but in a different order: It is easy to see that in the prime factorization of 10!, there is one 7 and two 5s. First, in the formula for m!, we insert 7 because there is only one and it has to be somewhere. Before 7, we sequentially insert 6, 5, 4, 3, 2, 1. Now, we have one 5 inserted and we need to decide about the second one. Case 1: We multiply by 2 and get 10. We cannot insert into the formula for n! because we don't have a second 7, so we insert 10 into the formula for m! between 7 and 10 filling in the formula for m! with numbers 8, 9 and get 10!. Therefore, n! = 1. so n=0 or n=1 Case 2: We insert the second 5 into the formula for n! and fill in to the left: n! = 1 * 2 * 3 * 4 * 5... we have 2 and 3 left, and 2 * 3 = 6. So we insert 6 into the formula for n! and we have m! = 7, n! = 6
Yes, the larger has to be at least 7, then with a little thinking you can conclude that it cannot be 8 as 3, 3, 2 and 5 cannot make a factorial. Then just test the 7.
I actually solved this a different way, by listing the first few factorial numbers up to 6 (1, 2, 6, 24, 120, 720), then multiplying from the top of 10! (10, 10x9 = 90, 90x8 = 720). If m! is missing a multiplier of 720 to become 10!, then we can make n compensate for this, so n = 6! = 720 and 10-3 = 7, so m= 7. Therefore mn = 42. Very nice question, and a, great video! (Sorry if the explanation is a bit off)
Assume m>n Consider that 7 is prime and only once in 10! therefore m is 7 or more and n m=7 n=6 Then of course m=10 also works with n=0 or 1 Answer 42 or 10 or 0 😁 Finally 8 or 9 do not work.
What I did before watching video. Write 10 to 1 in prime factorisation. I then multiplied them. Then since I know m,n are both factorial I started with 7 being highest prime and then just went down crossing out each prime like 6 cross out 2,3. I was then left with n! Fun to see my strategy was different than this video
This is a nice systematic approach, but there is a shortcut. The largest prime less than 10 is 7. Either m! or n! has to have a factor of 7. So let us try m=7, so m!=7!. Then n!= 10 times 9 times 8. Has to have a 5 in it (because of the 10), so try 5!= 120, nope, so try 6!. This is a little trial and error but you get the answer in a few seconds.
I went at it from the other end. I knew that the smaller factorial needed to be equal to one of these: 10, 10*9, or 10*9*8. The factors of 10*9 are {2,3,3,5}, so it couldn't be that. That left the factors of 10*9*8, which are {2,2,2,2,3,3,5}, which gives the answer.
The presentation in the video managed to find a solution, but didn't perform a rigorous analysis or search for all solutions. You can do it a bit more rigorous if you let m be the larger and n the smaller of the two natural numbers and then formally reason about the ranges in which they must lie. For example, m must be at least 7 and at most 10, etc.
You are great. Please come up with any problem on Convolution Theorem and solution thereof. Some 60 years ago I read this theorem in my Engineering college and have forgotten now.
There are 2 mistakes here. 1. 1! * 10! = 10!, and the same idea about 0! * 10!. 2. The equation is symmetrical to M and N. So if M=A and N=B are the solution-pair, then M=B and N=A fit either. It's inaccurate, to claim, M=this and N=that without a note. E.g. a prior assumption: "let M be > N, for a while, but the opposite will be good too" (I have, but omit the proof, why this is a complete list) Roots, (m,n): (0,10), (10,0), (1,10), (10,1), (6,7), (7,6) Thus final answers are: 0, 10, 42
I solved it by loocking at the factors of the factors. I said, "10 is 2×5 and 5 is 5×1. We need two 5s. The only way to get 2 5s is if both M and N are at least 5. Then I looked for the highest prime. I saw 7, and said that exactly one of the numbers can have a 7. Meaning M >= 7 and N < 7. Then I divided 10! By 7! And 5! And was left with 6. So we needed one more 6. So bump the 5! Up to a 6!" Checked the math and it was correct. M!×N! = 7!×6! = 10! M×N = 7×6 = 42.
I approached this the other way around: spotting that 7 is the biggest prime factor of 10!, and that it's not a repeated factor, meaning that m is 7! or higher and n is 6! or lower. Then just breaking down the factors of 10x9x8 in a similar manner to what you did. (I actually prime factorised everything first, but same premise)
Interestingly enough… M and N are my initials. So I sometimes use them instead of “x” and “y” so that I illuminate the risk of misconstruing the variable “x” for the multiplications sign. Yes I know there are OTHER signs for it but that “x” sign for multiplication is a huge peg peeve for me. Idk 🤷♂️ y
My approach. Let m >= n. 1. m=10, n=0/1, perfectly OK. 2. 7 is prime, and it should be at least in m. As m≠10, prime number of 5 should be in n → n! >= 120 → at least 10·9·8 should be in n!. Just check whether 10·9·8 is factorial. Answer: 0, 10, 42
We have two trivial solutions: m=10 and n=1 or m=10 e n=0 WLOG disregarding permutation So mn=10 or mn=0 m=10 10! =7*5^2*3^4*2^8 * m=9 9! = 7*5*3^4*2^7. So to correct the exponent of the prime 5 we need that n>=5 so the exponent of 3 will be wrong, contradiction. m=8 8! = 7*5*3^2*2^7, So to correct the exponent of 5 we need n>=5 so the exponent of 2 becomes grater than 8, no way. m=7! 7! = 7*5*3^2*2^4 We need that n>= 5 to correct the exponent of the prime 5. But n needs be less than 7 to not disturb the correct exponent of the prime 7. So n has two possibilities 5 or 6. As five does not correct the exponent of the prime 2. We only have m=6 as a possibility. Checking m=6 6!=5*3^2*2^4 so m! n!= 7*5^2*3^4*2^8= 10! m=7 and n=6 is also a solution. And we have no more solutions as m!*n! will decrease for m=m*
What you just did can also be done by simply listing all the prime factors for each of the numbers up to and including ten (10). In fact, this is exactly what you did.
Well it isn't specified that m and n are natural numbers (even if they are than 1 and 10 are also solutions) so there are infinitely many solutions (if you use gamma function) one example is m=3.6 n=8.87.... as 3.6!=13.3812858709 and 8.87!=271032.053904 so by multiplying both we get 3626757.39347 which is error of 2000 but i can't really find an accurate value of n and still it is only 0.055114638% error
Generally a question of this type is looking for exact solutions. It's been a few years since I delved into number theory but I believe that the only known non-trivial decomposition is 10!= 7!6!
One way to determine how long one of the sequence needs to end or go longer is take the largest prime number (7), then figure out if the rest can be factored and form a factorial.
Well a!= a*(a-1)! and b!= b*(b-1)! so a!b!= (ab)(a-1)!(b-1)! = (ab)^2 and then divide (ab) on both sides you get (a-1)!(b-1)!=ab so (a-1)! = a and (b-1)!= b in other words, the factorial of (a number minus 1) equals itself and the only solutions possible for that is think Is (1) because 0!=1 so if a=b=1 and 1!1!=(1*1)^2 --> 1=1.... I think
@@SalmonForYourLuck try with a=3 and b=4. I found this relation when I was trying to find another pair with the same relation than in the video. I have no idea how this can be solved but your reasoning is a good start point!
@@alipourzand6499 A not so elegant way but it works: For a > 4 we have a! > a^2, so if both a,b > 4 the equality cannot hold, now one has to check the cases if just one of them is > 4 and the other =< 4, then it wont work either, so both have to be =< 4, so now we just have to check a few cases Hm so i guess now it comes down to actually show that a! > a^2, for this one could consider a prime factorization a = p1 * … * pn, now the right hand sight has prime factors p1^2 * … * pn^2, and if we consider the left side, these pj must occur in the product somewhere, but between every two prime numbers there is atleast one non-prime number, so on the left side we will have something like p1 * (p1 + 1) * … * p2 * (p2 + 1) * …, and each product pj * (pj + 1) is greater than pj^2, hence the left side must be bigger (here we need a > 3 so that the left side has enough terms). Ok this needs some fine tuning maybe since there is no number between 2 and 3 although they are prime, but it should still work
Well... Lot of solutions we can have here, but 2 of them: m*n=0 and m*n=10 Solution: m!*n! =10!=10!*1 It happened if... m!*n! =10!*1=10!*1! or m!*n! =10!*1=10!*0! So... m=10 and n=1; m*n=10 or m=10 and n=0; m*n=0
don't do it that way because for each number you have to pick what to do. It is not mathematical. Take it the other way around. 10!=10*9*8*7!. Since 7 is a prime number you know you need m (assuming m>n) to be either 7, 8, 9 or 10. If m=7, n!=10*9*8 which can reorders as 6!. If m=8, n!=10*9 which is not possible as 4!
A very incomplete solution, there are a total of 6 solutions: n=0, m=10 n=1, m=10 n=6, m=7 And their symmetric opposites. Additionally, even if you add the restriction that the numbers should start at 2 and that (without loss of generality) n
@@PrimeNewtons No, you have the three solutions I wrote, and I added "their symmetric opposites". That's a total of 6 solutions. If you want a full list here it is: n=0, m=10 n=10, m=0 n=1, m=10 n=10, m=1 n=6, m=7 n=7, m=6 Additionally, one needs to prove that these are the only solutions.
If you meant the product mn, that has three solutions indeed, which are 42, 10 and 0, but you have to go through the 6 options mentioned above to see that.
@@danielleza908Basically yes, but it was asked for the product mn which is commutative, so order of m,n doesn't matter, and we're back at three. Just nitpicking. But I agree with you, a rigorous analysis would look different. 😊
You take prime numbers so it is proper way to split factors to prime. Write it down right way, then use em. Solution is tru, but way, it is not lear path, just whacking through bushes.
The creators of this question probably intended for m and n to be integers. But, in Mathematics it is critical to be very specific. So it should have included that m and n are integers and that m, n must be elements of {2, 3, ..., 10-1}.
I used a semi-iterative solution. 7 is the largest prime, so I assumed m=7. Eliminating from 10! gives 10,9,8. 5 is the next prime, so I tried n=5!, leaving 6 unaccounted for, giving a result of 10! = 7!6! or mn =42.
Solution: First, m and n are interchangeable, there there are always solution pairs. m!n! = 10! |:n! m! = 10!/n! if n! = 1, which is the case if n = 0 or n = 1, then we get m! = 10! m = 10 First two solution pairs are 0|10 and 1|10 m * (m - 1) * (m - 2) * ... * 1 = 10 * 9 * ... * (n + 1) Left side: 1! = 1 2! = 2 3! = 6 4! = 24 5! = 120 6! = 720 7! = 5040 Right side: 10!/8! = 10 * 9 = 90 10!/7! = 10 * 9 * 8 = 720 10!/6! = 10 * 9 * 8 * 7 = 5040 Therefore a third solution pair exists at 6|7 In total, we have: m, n = {0, 10 | 1, 10 | 6, 7 | 7, 6 | 10, 1 | 10, 0 }
. В комментариях справедливо указано о потере двух тривиальных решений. Замечание о том,что актуальны только нетривиальные решения - чушь и невежество.
Full solution. Let m >= n it's easy to check that (1) 10! / (6!)² = 7 It means 1)10 >= m >= 7 2) m = 7, n = 6 is one solution (see (1)) 3) If m = 8 (10! / 8!) = 9 * 10 isn't factorial of any number 0! = 1 1! = 1 2! = 2 3! = 6 4! = 24 5! = 120 4) if m = 9 10! / 9! = 10 - no solution too 5) if m = 10 10! / 10! = 1 => n = 0 or n = 1 All solutions are (m >= n) m = 10, n = 0 => mn =0 m = 10, n = 1 => mn = 10 m = 7, n = 6 => mn = 42
7 is the largest prime number which is less than 10, so m or n is 7 at least. You need only try whether 10!/7!, 10!/8!, 10!/9!, 10!/10! is a factorial or not.
I'm quite disappointed : it is written as an algebra problem (i.e. 'find all possible m,n, such as...'), and ends as an arithmetic problem, just a guess and try. So not any demonstration of the existence (or inexistence) of other solutions. Many add (1,10) and (0,10) as solutions in N2, but even the working domain (N or R using Gamma extension) was not stated here.
I started looking at the high numbers and then seeing which set made a factorial number. 10 is not a factorial number, 10 x 9 is 90 and not a factorial. 10 x 9 x 8 is 720 and that is a factorial. So m is 7, and n! makes 720, so n is 6. m x n is 42.
If you have some familiarity with factorials you should be able to solve this easily in your head. In order for m!n! = 10! to be true, one and only one of them must have 7 as a factor, because there is only one 7 in the factors of 10! Which means one of them must be at least 7! (or more) Starting with the lowest option of m = 7 you get m! x (8 x 9 x 10) = 10! Which would mean m! x (72 x10) =10! or m! x 720 = 10! Thus n! = 720. And as I said, if you are familiar with factorials you will recognize straightaway that 720 = 6! So mn = 42. (This is disregarding the trivial answer of 0 that people keep bringing up.)
My starting point was 7 | 10! so the bigger of m,n (wlog m) must be greater than or equal to 7. Then since 10! = 7 * 5^2 * 3^4 * 2^8 we can just look at the prime factorizations of 7!, 8!, 9!, 10! and see which one leaves a factorial. And if m
Watching the thumbnail i just went for 1!•10! or 0!•10!. The answer can be 10 or 0. Since it didn't say "find all solutions", i guess that's as good as any? Love seeing there is one more though.
Hi, there! You talk about two numbers in the video, so by the way you approach the resolution you're just considering only natural numbers. Well, given the fact that 1 is the first natural one, that implies that taking m=10 and n=1 (or samely making m=1, n=10)→10!1!=10!→ mn=10, thus you've missed this solution.... On the other hand, given the fact that 0!=1, if you consider m=10, n=0 (samely m=0, n=10)→10!0!=10!, thus mn=0 is another solution as you talk in your video about two unspecific numbers. Finally, it would be utterly interesting to consider the same problem extending the solutions to R, using the gamma function, don't you think so? 😊 ... Blessings!
Now that I've seen you solve it I see that there might be a faster way. Start working backward from 9! * 10. Factor 10 to see if it can be made into a ! If not, try 8! * (9*10) and factor (9*10) to see if they can. If not, try 7! * (8*9*10) and factor (8*9*10) to see if they can be a ! 2*2*2*3*3*2*5 works
Before watching: Alright, let's get the *trivial cases* out of the way. 0!=1!=1, so mn = 0 and mn = 10 are acceptable solutions. However, the third answer requires knowledge of factorization. Let's look at the first numbers in the factorial expansion. Counting down from 10, we'll stop when we hit a prime number. 10! = (10)(9)(8)(7!) We can further decompose (10) into (2)(5), 9 into (3)(3), and 8 into (2)(2)(2). Thus we can rewrite this as: (2)(5)(3)(3)(2)(2)(2)(7!) Now we mess with the factors of our 10, 9, and 8. We know it must be at least 5!, because 5 is our largest prime. However... (2)(5)(3)(3)(2)(2)(2) = (5)(2)(2)(3)(2)(2)(3) = (5!)6. ...we have a leftover 2 and 3. Fortunately, since 2*3 = 6, we can rewrite our expression as 6! (aka 6*5!) Then 10! = (6!)(71) And mn = 42. Summary: Trivial case answers: mn = 0, mn = 10. Non-trivial: mn = 42.
Sir Can you solve this nice double factorial problem with the thrill of cube. The problem is , n^{3} = (n+2)!! Sir please find all values of ' n ' for which the equation is true such that the solution "n" is an integer and "n" is greater than equal to 0 .(i.e. n >= 0 ) Please sir solve this problem !!! I ❤ your methods and techniques as well as teaching how easliy you teach the topics and deal with mathematical problems 🫡🫡 Love from India 🇮🇳
There are at least two trivial answers to this. Those are 10 and 0. Those are for m=10, n=1 and m=10, n=0 respectively (or the same transposed). Now I will watch the video. Ok. I've barely started the video, and he has already made an incorrect statement in saying that m and n must both be less than 10. That's not true; they must both be less than or equal to 10 as with the two (or four if we count the transpositions).
10! contains 7 exactly once, so m, assuming that it is the larger number, must be at least 7. 10! contains 5 exactly twice, so n must be at least 5. Thus you can start the procedure immediately at 7! and 5!.
What happened to the 3rd 3 of the 9? And what about the first 3 (on the top of the board? Why you did not strike it out? Still failed to see your methodology!
The way I tackled this is by assuming that m! Was 10 at first, and slowly counting down until the product of numbers I went down from coukd be the product of consecutive numbers. Really helped when I considered the numbers as nothing but the product of powers of primes It went like this: 10! Went to 9! But 10 can only be written as 2*5 9*10 for 8! Can only be written as 2*3*3*5 in low terms anyway 8*9*10 can be written as 2*3*4*5*6 which is 6! And since we went down to 7! To get those terms out of the 10!, we got our answer as 6 and 7
