agreed, sometimes he gets solutions by inspection in addition to calculations
@PS-mh8ts
Ай бұрын
@@davefried 😏 admit that you missed those solutions! 😜
@marscience7819
Ай бұрын
@@davefried I don't how 0,10 and 1,10 are trivial solutions
@bharath__100
Ай бұрын
It was simple... the answer to everything in this universe 42
@boguslawszostak1784
Ай бұрын
or 10 or 11
@wernerviehhauser94
Ай бұрын
@@boguslawszostak1784 11? Its either 0 or 10.
@boguslawszostak1784
Ай бұрын
@@wernerviehhauser94 You are absolutely right, I added instead of multiplying.
@Aryan-tm1sm
Ай бұрын
Wth with this 42 i'm seeing it everywhere
@boguslawszostak1784
Ай бұрын
@@Aryan-tm1sm 42=6*7 6!*7!=10!
@davek6415
Ай бұрын
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.
@WitchidWitchid
Ай бұрын
If this were a test question I would have to mark your answer correct since the problem did not specifically state "find a non trivial solution".
@yiutungwong315
Ай бұрын
@@davek6415 And 6! × 7! = 10! So 6 × 7 = 42
@jacobgoldman5780
Ай бұрын
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!.
@davefried
Ай бұрын
only non-trivial solutions are relevant.
@Alians0108
Ай бұрын
He didn't mention that anywhere so no, they're not @@davefried
@evr0.904
Ай бұрын
@@Alians0108Yes he did.
@bobh6728
Ай бұрын
@@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.
@samyaspapa
Ай бұрын
@@davefried What about n=m where the final answer of n*m=42.369189103967302951...? That's definitely non-trivial.
@Jon60987
Ай бұрын
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 !!!
@kaiserquasar3178
Ай бұрын
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.
@PS-mh8ts
Ай бұрын
😎method! 💯
@boguslawszostak1784
Ай бұрын
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.
@robertpearce8394
Ай бұрын
Yes. I think your solution is clearer. Start with 7!, then look at 10*9*8.
@boguslawszostak1784
Ай бұрын
@@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
@okaro6595
Ай бұрын
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.
@elieabourjeily8520
Ай бұрын
What about (0! And 10!) or (1! And 10!) so mn could also be 0 or 10!
@suleymancanan
Ай бұрын
Let s start with primes 7!.5!.3!=10! We may see 5!.3!=6! Then 7!.6! m=7 n=6 m.n=42 tata
@jf3518
Ай бұрын
I looked at 10! and it has only a single factor of 7. This implies that one number needs to be at least 7! and the other needs to be less than 7!.
@CalculusReviser
Ай бұрын
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)
@rogerkearns8094
Ай бұрын
Douglas Adams clearly knew that.
@dougaugustine4075
Ай бұрын
I sure do enjoy watching you work.
@eurocouto
Ай бұрын
A magical- beautiful problem with a stunning answer! Bravo!
@NikTechmag
Ай бұрын
Пропустили как минимум 2 решения: 10!*1! И 10!*0!
@nox5282
Ай бұрын
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
@voidmxl8473
Ай бұрын
Good strategy
@childrenofkoris
13 күн бұрын
I NEVER STOP WATCHING UR VIDEOS.. so good and satisfying
@apriljohnson6191
Ай бұрын
I wish I had you as a college math professor. This is awesome!
@smalin
Ай бұрын
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.
@nzeches
Ай бұрын
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.
@jayofray
21 күн бұрын
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.
@Grecks75
Ай бұрын
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.
@a0z9
26 күн бұрын
Muy interesante. Esas preguntas son las que merece la pena responder.
@marsdeat
Ай бұрын
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)
@nooruddinbaqual7869
Ай бұрын
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.
@bryandata6658
8 күн бұрын
There were multiple possible solutions: (6, 7) (10, 1) (10, 0)
@hassanalihusseini1717
Ай бұрын
The question should be more precise to exclude the more obvious solutions as (m;n) = (0;10) or (1;10) or other way round.
@xislomega242
Ай бұрын
This man makes even the more complicated mathematics look so primitive and still entertaining! I am happy to have discovered his channel!
@xyz9250
Ай бұрын
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.
@mikeburns6603
Ай бұрын
There is an infinite number of solutions when you realize that you can use the gamma function and that m and n were not required to be integers.
@WitchidWitchid
Ай бұрын
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}.
@user-bk6gx7sg3j
18 күн бұрын
Two unknowns, 1 equation. By definition, the answer will be speculative.
@musicnumbers128
Ай бұрын
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
@mikefochtman7164
Ай бұрын
Always a joy to watch your videos. An interesting way to find the solution here.
@kushagrasharma5983
Ай бұрын
Hello sir , could you please make some videos about permutations and combinations? I love your teaching style by the way.
@niconiconichan2098
Ай бұрын
I recommend the permutations and combinations playlist by ExamSolutions.
@niraj3872
Ай бұрын
10! = 3628800 We have 6! = 720 and 7! = 5040, 720 x 5040 = 3628800 So, it's pretty simple, without using complexities of algebra, we get : 6 x 7 = 42
@parthhooda3713
Ай бұрын
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
@WitchidWitchid
Ай бұрын
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!
@michaeledwards2251
Ай бұрын
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.
@antonionavarro1000
Ай бұрын
interesting exercise!
@Warriormon87
Ай бұрын
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.
@UrievJackal
Ай бұрын
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
@christiannavarro3556
Ай бұрын
Woah, definitely confusing at the start. Thank you for making this video❤
@AbouTaim-Lille
Ай бұрын
You should replace the factorial by the Gamma function to go Out of Number theory and give the equation more sense.
@wuchinren
Ай бұрын
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.
@krwada
Ай бұрын
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.
@hvnterblack
Ай бұрын
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.
@deltalima6703
Ай бұрын
Life, the universe and everything = 42
@gadysam
Ай бұрын
This guy likes factorial so much 😭😭😭
@alipourzand6499
Ай бұрын
Brilliant! Another nice problem: a!b! = (ab)^2 ☺
@SalmonForYourLuck
Ай бұрын
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
@alipourzand6499
Ай бұрын
@@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!
@ilprincipe8094
Ай бұрын
@@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
@Mercury13kiev
Ай бұрын
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
@tarciso21claudia28
Ай бұрын
Never stop teaching!!!!!!!!
@pedrojose392
Ай бұрын
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*
@tamiratsolomon4655
Ай бұрын
The problem is very interesting . But not the solution. What will be your step by step solution if 10 was 90 or other higher number other than 10 ?
@florianbasier
Ай бұрын
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!
@jamesharmon4994
Ай бұрын
The answer to Life, The Universe, and Everything
@gadysam
Ай бұрын
Please do this question 2^x=4x Find the value of x , it has been disturbing me for a long time now 😭😭
@danielleza908
Ай бұрын
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
Ай бұрын
You meant to say a total of 3 solutions.
@danielleza908
Ай бұрын
@@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.
@PrimeNewtons
Ай бұрын
@danielleza908 I see
@danielleza908
Ай бұрын
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.
@Grecks75
Ай бұрын
@@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. 😊
@Misteribel
Ай бұрын
Who's "he" in that last quote? I assume it's meant to stand for mathematics?
@valeriykotlov2763
Ай бұрын
I didn’t see somebody offered other answers. m and n could be 0 and 10, 1 and 10, products will be 0 and 10 respectively!
@laobernadus5743
Ай бұрын
According to me: M!N! = 10! . All sides divided with ( ! ). And the result is MN = 10. Make it simple.
@salimahmad7414
Ай бұрын
Of course, This requires a lot of thinking and imagination.
@stevenwilson5556
Ай бұрын
You also have 10! and 0! and 10! and 1! so other possible products are 10 and 0. Your answer is the maximum value tho.
@omkumarsingh7
Ай бұрын
When I saw this my mind immediately said that one of them is 10 and the other is 1 for the values of m & n . Because I am not so good at maths sorry.
@lumina_
Ай бұрын
you're correct too
@davefried
Ай бұрын
only non-trivial solutions are relevant.
@thaerthaer1120
24 күн бұрын
Great . Do u have some lectures about fractions of calculs like d^1/2(y) and some things about special function like Biesl function
@GuiveChafai
Ай бұрын
this is done with such lack of rigour.
@pataplan
Ай бұрын
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.)
@eduardionovich4425
13 күн бұрын
. В комментариях справедливо указано о потере двух тривиальных решений. Замечание о том,что актуальны только нетривиальные решения - чушь и невежество.
@DarkBoo007
Ай бұрын
A problem I can do LMAO I got 6 & 7 but also got 1 & 10 and 0 & 10 (trivial cases). Therefore, my products were 42, 10, and 0
@johnplong3644
19 күн бұрын
That was cool Yeah I was trying to overthink this problem
@KarlDeux
Ай бұрын
m = 1 and n = 10 also works, as m = 0 and n = 10. So mn = 10 or mn = 0.
@cyborg8992
Ай бұрын
This was an awesome problem, the way you did it didn't even occur to me
@bertblankenstein3738
Ай бұрын
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.
@gonado
Ай бұрын
That is not the only solution though. Without stating that m and n are greater than 1 we cannot take that as the only solution of the problem
@Metal_dead
Ай бұрын
I quickly find 0 and 10. Can we find more?
@LawCharlotte-ev9bk
Ай бұрын
!(mn)=10! mn=10
@SalmonForYourLuck
Ай бұрын
😂
@sunil.shegaonkar1
Ай бұрын
Very interesting. I guess, the two of obvious solution would be m=10 n=0 or 1. then m.n= 10 or 0. But that is no brainer solution.
@dneary
Ай бұрын
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
@antoinegrassi3796
11 күн бұрын
Joli petit problème. Vous avez trouvé une solution (m, n) 👍, mais est-ce la seule (exceptée n, m)? Il y a aussi 1!.10! = 10!😉. Encore une autre ?
@mohamedali2858
Ай бұрын
Thank you 🙏
@7th_dwarf542
24 күн бұрын
that's genius!
@elquesohombre9931
Ай бұрын
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
@alainmosnier
Ай бұрын
Definitely not a rigorous proof, and it indeed misses solutions (0 and 10, at least). Warning: this is not proper math!
@wolfgangkoller4495
Ай бұрын
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!.
@fCauneau
Ай бұрын
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.
@stefankocic744
25 күн бұрын
How do we know there arent any other solutions?
@GAGAN_RAO
Ай бұрын
Three solutions of this problem (m,n) = (10,0) , (10,1) , (7,6) m×n = 0 , 10 , 42
@TheCktulhu
7 күн бұрын
Распишем 10! Учитывая что в разложении на множители 10! есть только одна цифра 7, то есть число минимум 7! Далее, чисел кратных 5 должно быть хотя бы 2. Иначе было бы 10! и 1! Следовательно, большее число не превышает 10! Поэтому у нас хотя бы 7! и 5! Далее т.к. у нас цифры 3 всего 3 штуки, то есть только 2 варианта. Либо одно число 9! а второе 10 - но это неверный случай. А значит 7! и 6! И этих чисел как раз хватает чтобы получить 10!. Значит наш ответ 6*7=42
@musicalsawpetewentworth6102
Ай бұрын
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
@Nikioko
2 күн бұрын
The answer is 42. What else?
@m.h.6470
Ай бұрын
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 }
@m.h.6470
Ай бұрын
Forgot to add the result of m*n: m*n = { 0, 10, 42 }
@rwrobson42
Ай бұрын
Does anyone know any other non-trivial x for which m!n!=x! has integer solutions? Or is 10 the only one?
@simmmr.9040
Ай бұрын
2:49 you forgot to cross out three
@GGANGRI-
13 күн бұрын
7은 소수이기때문에 m= 7!자체로 바로 생각한 후 뒤에 8 9 10을 n!형식으로 소인수분해 후 재조합하면 6! Mn 42
@SGuerra
Ай бұрын
Que bela questão. Parabéns pela sua escolha. Brasil Julho 2024. What a beautiful question. Congratulations on your choice. Brazil July 2024.
@souzasilva5471
Ай бұрын
Tell me if there is another example other than m!n! = x!, m and n different from 0 and 1 and x different from 10.
@Misteribel
Ай бұрын
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.
@m.j.a6002
5 күн бұрын
0,10 or 1,10 are also answers i.e. mn=0 or mn=10 or mn= 42
@victorchoripapa2232
Ай бұрын
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
@ashrafabdelalem494
Ай бұрын
Very very good 👍
@rahulsanjay8500
Ай бұрын
Hello sir , i got another solution m = 1 , n =10 such that m!=1!=1 and n!=10! therefore m!n!=10! . Am i wrong ?
@lumina_
Ай бұрын
no you're not wrong, question should've had some clarifications
@davefried
Ай бұрын
only non-trivial solutions are relevant.
@Grecks75
Ай бұрын
42 is the answer (here at least for the nontrivial solution). Solved without pen and paper. I also missed the trivial solutions, I must admit.
@maruthasalamoorthiviswanat153
Ай бұрын
Excellent solution sir.
@rbbl_
Ай бұрын
Divide both sides by factorial and then the answers are (0,10),(2,5)
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