Can you solve this tricky factorial problem?

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[@Paddy-ip7qk]

¿Por qué la gran mayoría de gente que viene aquí a comentar piensa que TODOS los videos de matemática en KZbin están dirigidos únicamente a grandes eruditos que ya conocen todos los métodos y que son capaces de resolverlos mentalmente? Existen personas que recién se inician en la matemática a quienes les resulta útil una explicación paso a paso y con mucho detalle. Si ese no es tu caso, cierra el video y vete a ver otro, pero no dejes tu mala vibra aquí ni critiques que no lo haya resuelto en 20 segundos. Si en verdad tienen gusto por la matemática, sepan que las tonterías que dicen no ayudan en nada a que esta bella ciencia se siga difundiendo.

[@stevenmayhew3944]

Here's another one: 5!x3!=6! because 3!=6 and 5!x6=6!. So, that's two cases: 6!x7!=10! and 3!x5!=6!. Put them together and you get that 3!x5!x7!=10!.

[@DerekRoss1958]

Also 0!x1!=1!, 1!x2!=2!, 5!x3!=6!, 23!x4!=24!, 119!x5!=120!, 719!x6!=720!, 5039!x7!=5040!, and so on.

[@pranaypallavtripathi2460]

great observations guys 😅

[@2019inuyasha]

If you already know 6!×7! = 10! Then you dont need any further steps...

[@emanuellandeholm5657]

@@DerekRoss1958 (n! - 1)! n!= (n!)!. Follows from the recursive definition of the factorial.

[@DerekRoss1958]

@@emanuellandeholm5657 Yes, I pointed that out in another comment.

[@myfyrmadocjones]

9:25 Not the only case where a factorial is a multiple of two factorials - you also have 6! = 3! X 5!

[@rainerzufall42]

Also 2! = 1! x 2! 🤣

[@rainerzufall42]

Might be interesting to expand the problem. 2! x 2! x 3! = 4! 3! x 5! = 6! 3! x 4! x 5 = 6! 5! x 9! x 11 = 12! 6! x 7! = 10! 7! x 6! = 10! 8! x 6! / 8 = 10! 2! x 3!! = 3! 7! x 4!! = 8! 4! x 5!! x 2 = 6! 4! x 7!! x 2 = 7! 6!! x 7!! = 7! (okay, that last one was easy: (n-1)!! x n!! = n!)

[@ericwiddison7523]

There are an infinite number of cases like this of the form n!*(n!-1)!=(n!)! The first few are 1!0!=1!, 2!1!=2!, 3!5!=6!, 4!23!=24!, 5!119!=120!... There are also an infinite number of trivial solutions that 0!n!=n! and 1!n!=n! Non-trivial solutions to a!b!=c! where a

[@oahuhawaii2141]

@rainerzufall42: What is this "n!!" operator? It seems to use only odd or even integers in the product, rather than all the integers, by matching with n being odd or even. If a parser doesn't recognize the double "!", it'll do a factorial of a factorial: (n!)! . Yikes!

[@quigonkenny]

There are an infinite number of them. (n!-1)!•n! = (n!)! For example, 3! = 6, so 5!•3! = 6!, or 5! = 120, so 119!•5! = 120! Every factorial is an integer, so naturally every factorial will equal the final term of a larger factorial. 6!•7! = 10! is the only one I know of where both multiplier factorials make up multiple terms of the product factorial, though I wouldn't be too surprised if there are others.

[@tamirerez2547]

DID YOU KNOW? The number of seconds in 6 weeks is 10!

[@拗拗拗拗]

Well I didn’t know that

[@xl000]

This is why time seems to pass so fast sometimes !

[@simmmr.9040]

There's no need to factorise the 7! as a product: x!=7!×6!=7!×6×5×4×3×2×1=7!×(4×2)×6×5×3=7!×8×3×2×5×3=8!×(3×3)×(2×5)=8!×9×10=9!×10=10! → x=10

[@hippophile]

Here's my solution method: Assuming there is an answer, x must have two factors of 5, as both 6! and 7! have obviously both only one factor of 5, so for the 5-factors to balance, you need two in x!. Thus it must be at LEAST 10! However, it cannot have a factor of 11 (as 7! has not got far enough for that). So the only possible asnwe is x=10. Then it is a simple matter of checking.

[@dbly5460]

You mean 11 being prime, I assume.

[@jpdemer5]

The "simple matter of checking" amounts to the solution presented here. (Or else a whole lot of multiplying.)

[@prashnaprashant4770]

Can you make videos on IIT JEE ADVANCE OR MAINS question?

[@prashnaprashant4770]

Nice warm up as well as easy question.

[@RyanLewis-Johnson-wq6xs]

10!=3628800

[@KingGisInDaHouse]

Multiply the 6! To the other side and you recursively divide until you get one to reverse the factorial.

[@astrolad293]

11 is the smallest prime > 7. So the answer must be 8!.,9!, or 10!. And 6! must be equal to 8, or 8*9 (72), or 8*9*10 (720). 6! is 720, so the answer is 10!.

[@nichtvonbedeutung]

You can also multiply 7*8*9*10 at the end and then divide the result by 1*2*3*4*5*6*7 and the result should be 1.

[@xyz9250]

A little silly to expand 7! It has to grow from 7!, first thing is to find a 8 from the expansion of 6!

[@markmajkowski9545]

IF there’s an answer it must be at least 7 to get a 7 in the product and at least 10 to get a 5 and not be more than 10 or it has an 11. Nothing wrong with Guess and Check.

[@krrishrohilla2945]

i did it in first attempt

[@vaclavkrpec2879]

This wasn't tricky, I didn't even need to write it down. Take 2 x 4 from the 6!, that gives you 8! on the right hand side. You're left with 3 x 5 x 6. That's 90 = 9 x 10. That gives you 10! on the right hand side, equal to x!, ergo x = 10.

[@andrewwhitehead2002]

Really enjoyed this; thank you…

[@manuelgonzales2570]

Excellent. Thank you!

[@krwada]

It is not that difficult. 7! = 7 x 6! ... You just need to do a prime factorization of 6! and work your way up. Prime Factorization of (6!) = 2x3x5x2x2x3x2 Start working way up x8 = cross out three (2) x9 = cross out two (3) x10 = cross out two (2) and five (5) You will see that at this point all factors of 6! are crossed out. That is 8x9x10 = 6! Therefore, x = 10

[@jollyjoker6340]

6! = 2*3*4*5*6 needs to divide into numbers starting from 8; Remove 2 and 4 leaving 3*5*6, 9; remove 3 and half the 6, leaving 5*2=10. Edit: First time one of these videos has the same solution I came up with!

[@oahuhawaii2141]

Actually, he built up on 6! by taking away increasing factors from the expansion of 7! by starting with 7. That's a slow way to solve it. It's better to build up on 7! by taking successive factors from 6! by starting with 8.

[@archangecamilien1879]

Not that tricky, actually, lol...x! (x being an integer, no gamma function stuff, lol) will be a factorial, only certain numbers can be "factorials", like 1, 2, 6, 24, 120, 720, if I'm not mistaken...try to see which of these is equal to 7*8*...*n for some n, etc...as soon as you "hit" one that is equal to such a product, etc, bingo!, lol...Just try each of them, 7*8, then 7*8*9 until you hit a "factorial" number...

[@AlephThree]

X! will include 7! as it is greater than 7!. So you know 6!=8*9*…*”x”. 6! Is 720 which is 8*9*10 so x=10.

[@BN-hy1nd]

Brilliant 👍🏿

[@EduardoPierreMG]

Outstanding question!

[@oahuhawaii2141]

x!/6! = 7! x! = 6!*7! We see 7! as a good starting point to increase it to 8!, 9!, ... by converting 6! to 8*9*... We do it here: 6! = 720 = 8*9*10 x! = 10*9*8*7! = 10! x = 10

[@ManojkantSamal]

X=10, because 10!=6! × 7×8×9×10 7×8×9×10=7×2×4×3×3×2×5 =1×2×3×4×5×(3×2)×7 =1×2×3×4×5×6×7 =7! As per question X!/6!=7! (6!×7!)/6!=7!=R. H. S There are plural methods to get the answer

[@thomasvronik6517]

Haha the fly on cam😂😂😂

[@RyanLewis-Johnson-wq6xs]

(10!)/(6!)=5040=7! x=10

[@AffiliateXAJFunVideos]

You can’t use a calculator in a math competition

[@저녘놀]

Cause 6! mod 5 = 0 and 7! mod 5 = 0, x mod 5^2 = 0 and x mod 5^3 != 0 Therefore, x >= 10 and x < 15 Cause 7! mod 11 != 0, x < 11 So x >= 10 and x < 11 Therefore, x = 10 Maybe the fastest proof

[@lucretiusag1767]

Thx a lot

[@robert.eduard]

4! * 23! = 24! and the list goes on... 5! * 119! = 120!

[@user-qp6sy9fe4m]

can you make a general formula for this?😂

[@aidenbooksmith2351]

Not all cases apply here, but every implementation of my formula will result in this happening: if x! =y-1 y! = x!y-1!

[@robert.eduard]

@@aidenbooksmith2351 there’s a mistake in your “formula”. I’ll let you figure it out on your own!

[@v.volynskiy]

@@user-qp6sy9fe4m x!=(x!)!/(x!-1)!

[@oahuhawaii2141]

@robert.eduard: Well, he has 2 errors. For an arbitrary x ∈ ℕ1, we can compute x! and set it to y , and use: y! = y*(y - 1)! = x!*(y - 1)! = x!*(x! - 1)! = (x!)! 1!*0! = 1! = (0!)! = (1!)! 2!*1! = 2! = (2!)! 3!*5! = 6! = (3!)! 4!*23! = 24! = (4!)! 5!*119! = 120! = (5!)! 6!*719! = 720! = (6!)! 7!*5039! = 5040! = (7!)! 8!*40319 = 40320! = (8!)! ... This formula doesn't generate: 6!*7! = 10! or 3!*5! = 6! Note that I wrote (x!)! instead of x!! because there's a weird factorial-like operator "!!" that multiplies only the odd or even integers, depending upon x being odd or even. [Check it out on WolframAlpha!] If you scan through a table of factorials and can rearrange the factors into a product of a sequence of consecutive integers, then you have a number to use for this form. For example, 5! is 120, which can be factored as 4*5*6 and can follow on 3! to get 3!*5! = 6! . Likewise with 6! , we see it's 720, which can be factored as 8*9*10 and can follow on 7! to get 7!*6! = 10! . This 7! is factored as 7*8*9*10 and follows 6! to get 6!*7! = 10! . Anyway, with this pattern of thinking, I found 2 sets of numbers with interesting properties. But rather than show how I created them, I'll pose one of them as a math problem for you to solve. Find x, where x! = 3!*6!*22!/11! .

[@Математиканапять]

Интересно, есть те, кто не смог это решить устно за несколько секунд? Как же сильно у нас отличантся уровень математического образования..

[@bookert2373]

x! = x(x-1)…(x-k)(6)(5)(4)(3)(2)(1) =7! for some k so, after dividing x! by 6!, the (x-k) must be 7, so x =7+k. As many have observed, 11 can’t appear in 7!, so k must be 1, 2, or 3. k=1 and 2 are immediately eliminated (since neither 8! nor 9! can equal 7!). This leaves k=3 (ie, x=10) as only option.

[@arikahn3907]

Note that answer must be less than 11 since 11 is prime and not a factor in 7!

[@rainerzufall42]

Why do you do the long way? 6 < 7. Use 7! as the base and see that 6! = 8 x 9 x 10 (as a continuation of 1 x 2 x 3 x 4 x 5 x 6 x 7: 6! has 3 even factors, giving the 8 = 2x2x2. It has two factors of 3, so 9 = 3x3. What's left (!= 1)? The second 2 from 4 and the 5, so we have a 10. Thus x = 10. It was not possible to continue this product otherwise. We knew, that x! starts with 1 x ... x 7. You can't expand it to the left, only to the right: x 8 x 9 x 10... Problem solve with x = 10.

[@oahuhawaii2141]

Yeah, it makes sense to build up successively from 7! by taking increasing factors from 6! -- beginning with 8 -- until the remaining factor is 1 (solution found) or the next successive value isn't available (no solution).

[@oahuhawaii2141]

I created a fun case for you. Find x, where: x!*11! = 3!*6!*22! .

[@KaivalyaChess]

very nice sir

[@paulgoodman5187]

This was too easy. 6! is 720 (you get to know most of these lower factorial values), which is also equal to 8x9x10 and that stacks nicely on 7! We readily wind up with X=10.

[@winterswarrior2064]

Stupid Longway X!/6! = 7! X! = 7!6! = 7 x (6!)² = 3,628,800 Since 3,628,800 is a factorial product and we know that X! = X•(X-1)•(X-2)•....•2•1, we can simply divide the X! = 3,628,800 repeatedly by natural numbers starting at 2 untill the quotient is 1. At which point we can count the steps and find that X = 10, and 10! = 3,628,800

[@collegemathematics6698]

I solved it mentally before i run the video. Thanks for tha challenge ❤

[@abhishekpatil1063]

I solved it by calculating 7*(6!)^2which is same as 10 ! .So X=10.

[@pranaypallavtripathi2460]

I sort of cheated and did hit and trial using calculator starting from 9! and got the answer. But nice question.

[@DerekRoss1958]

By definition y! = y * (y - 1)!, for any positive integer y. So if y = x! then (x!)! = x! * (x! - 1)! for any positive integer x. Which all implies that there are an infinite number of cases where a! = b! * c!

[@peterweusten4251]

6! = 720 which is 10x9x8 x! =6! 7! so x! = 10x9x8x7! done

[@ОлегПолканов-д1н]

Every second simpliest question marked as Olympiad. Really? Could you provide a link to what Olympiad exactly?

[@Kaurenjalalifarahani]

Just do 6 factorial multiplied by 7! And that gives you the factorial of 10

[@creanero]

I just multiplied up 6! to get 720 and divided it by 8 and 9 to get 10 which told me it went to 10!

[@KennethDuda]

This is an olympiad question?! I solved it in 30 seconds. For there to be any answer, x! = 7! * 6!, so 6! must be 8 times 9 times ... up to something. Is 6! = 8? no. Is 6! = 8*9? no. Is 6! = 8*9*10? Well, yes. So x = 10. Gees

[@molgera3]

Anyone else nearly get jumpscared by the sudden appearance of a fly right as he was writing "7! = 7!"?

[@neerajkumarjha1417]

x!= x(x-1)! = 7!×(7-1)! Comparing x=7

[@user-yw3it2wj5u]

👌👍

[@fibrahim4171]

🎉🎉🎉🎉

[@markotesla2016]

can you just remove the factorial and multiply 6x7 to find x?

[@joetyson3216]

fun trick math question: If 6a x 7a = 10a, what is a?

[@oahuhawaii2141]

a = 0, 5/21

[@KpoXa12]

👍

[@bogdangarkusha8727]

6! = 720. 720/10=72. 72/9=8. it means x! = 7!*8*9*10 = 10! therefore, x=10

[@kpdywo848]

Easy, 7!*6! =x= 3628600 you divided by 2 then by 3....up to 10 where the result is 1. so x=10.

[@maxaafbackname5562]

There are 10! seconds in a week.

[@madankundu6035]

x!= 7! 6! = 7! 6 x 5 x 4 x 3 x 2 = 7! 8 x 6 x 5 x 3 = 8! 9 x 10 = 10!

[@jean-claudecaillot1290]

X = 10

[@shernettejohnson3804]

42

[@dirac17]

This literally took me about 5 seconds to solve

[@stottpie]

It took me only two seconds. You're slow.

[@RWBHere]

@@stottpie I solved it almost instantly, by skipping to the end of the video. 🤪 🤦🏼‍♂

[@igoranisimov6549]

Sometimes fast is not good. And it is not what you are thinking. The best one is to be CORRECT.

[@nspheadshot9419]

Me who find it pretty easily by a method mostly similar to trial and error method

[@Bizmyurt]

You can also say 6!^2 x 7

[@mandolinic]

You can solve this iteratively in a minute or two without needing factors (or even a calculator): 7! = 5040 (which most maths students know by heart, but if not then it's a quick calculation) By inspection: X!/6! = 7 . 8 . 9 . ..... . X So, just start multiplying: 7 x 8 = 56 ; 56 x 9 = 504 ; 504 x 10 = 5040 and the answer is found! X = 10. And that's it. Job done. Question answered. No need for anything more complex. If in an exam, use the time saved to answer another question.

[@ReasonableForseeability]

Yes! I solved it similarly: 7! * 720 = x! so 720 = 8*9*...*x and x = 10.

[@ahmedmasir7278]

It's a math Olympiad question you have to show the way you solve it , it is not about the result it's about the method

[@khaledalsaoub6760]

Great

[@thakurfamily9669]

It's obv x = 10

[@ankeunruh7364]

The only tricky thing: x is not 10!, but 10.

[@ЭнтониГон-п7п]

I found solution in 2 minutes

[@verites1491]

6!=2x3x4x5x6=8x3x5x6=8x3x5x2x3 =8x9x2x5=8x9x10 so x!=10! and x=10

[@jwil4286]

Is there an inverse gamma function?

[@davidlu1003]

10!

[@remoteadministrator4036]

10

[@sergeyvarin8434]

10 ясно сразу. Просто 5 должно быть кратно. Следующее за 5 10. Все.

[@vishalmishra3046]

x! / 6! = 7! so x! / 7! = 6! so x! / 8! = 6 x 5 x 3 so x! / 9! = 2x5 so x! / 10! = 1, so x = 10.

[@boguslawszostak1784]

Easy peasy. It is obvious that in the prime factorization of 6!⋅7!, there are two factors of 5, one factor of 7, and there is no factor of 11. Therefore, the only candidates for the solution is x=10 So it is enough to check. 6!*7!=10!

[@oahuhawaii2141]

You have a typo: 5!*7! ≠ 10! .

[@boguslawszostak1784]

@@oahuhawaii2141 Thank You

[@spicymickfool]

X is smaller than 11 since 11 doesn't divide 7!. 5 divides 7! and 6! So 25 must divide X! That requires x=10.

[@CharlesChen-el4ot]

6 ! = 10 * 9 * 8 X ! = 10 !

[@darmasalla8015]

An easy one. Takes 3 minutes to solve.

[@maryjay7833]

Thanks

[@SanePerson1]

Since 7! has 7 as a prime factor and 14! is far too large to be x!, x! has to be (6!)×7×8×9×10, or there is no solution ... all the prime factors in 6! are there: one needs four prime factors of 2 - three of them are in 8 and one is in 10, two prime factors of 3 come from 9, and the prime factor of 5 comes from 10.

[@ericwiddison7523]

I thought of it as 7*8*...*x=7! (this is x!/6!, or the remaining factors of x! after the 6! factors are removed). Looking at it this way, the product must contain a multiple of 5 and cannot contain 11, so it must be at least 10 and less than 11. The only possible solution is therefore x=10.

[@guitartommo2794]

x! = 7! x 6! 6! = 1 x 2 x 3 x 4 x 5 x 6 2 x 4 = 8 3 x 5 x 6 = 3 x 5 x 2 x 3 = 9 x 10 So 6! = 8 x 9 x 10 So x = 10

[@ToanPham-wr7xe]

😮

[@sfbefbefwfvwfvsf2722]

why india math olympiad so easy? or is this for primary school level?

[@marklyles3335]

Beautiful! Be careful about writing the quantity directly under the numbers. You give the impression that you are dividing.

[@xl000]

yeah.. people won´t fall for this. This is not 4th grade This is like PEMDAS in advanced math and physics books. They understand what the author is meaning

[@yazakariae]

هذا اسمه رياضيات الخشيبات.

[@animadas6785]

It took me Just 20 sec

[@sajidrafique375]

You are WRONG ! There are other factorial products too . look below at comments

[@danh2368]

PLEASE stop making x’s like that.

[@nitinsinghmouni-ru8dp]

im from INDIA sir

[@andypandy6063]

Darn you are slow...

[@agentkosticka17]

x = 14*6!

[@akmmonirulislam3961]

X!/6! = 7! X! = 7!.6! 6! = 2.3.4.5.6 = 6.4.30 = 24.30 = 720 720 / 8 = 90 90 / 9 = 10 10 / 10 = 1 So, 720 = 8.9.10 So, 7!.6! = 7!.8.9.10 = 10! So, X = 10 Why should one takes eleven minutes to do this math?

[@Paddy-ip7qk]

Because he's not solving; he's explaining and solving, so this takes more time than just solving.

[@d.bpatil6361]

V X = 10

[@WilliamWizer]

it spends a lot of time to just do the same thing twice as a supposed proof that the first time was right. as a lot of people pointed, it's a blatand lie that 6!*7! = 10! is the only case. 0!*1!=1!, 1!*2!=2!, 5!*3!=6!, 23!*4!=24!... but, let's it make even more extreme and prove there's infinite combinations. ∀X∈ℕ;N=X!→(N-1)!*X!=N! in case I wrote it wrong, and for those that can't read that small piece of madness, for each X that is part of the set of natural numbers, let N=X!. this implies that (N-1)!*X!=N! why? because (N-1)!*X!=N! is, by definition, (N-1)!*N=N! and this is the recursive definition of a factorial. here are the first few: 0!*1!=1! 1!*2!=2! 5!*3!=6! 23!*4!=24! 119!*5!=120! 719!*6!=720! 5039!*7!=5040! goes without saying that this implies you can also get A!*B!*C!=D! an example would be: 719!*5!*3!=720! but 0!*1!*2!=2! is, maybe, the most special one. three consecutive factorials that multiplied together are equal to a factorial. this is a lot more special than 6!*7!=10! and, yet, it has to be proven if it's unique or not.

[@santokhsidhuatla7045]

X=6! * 7! =720*720*7 =3628800

[@oahuhawaii2141]

The relationship is: x!/6! = 7! What is x?

[@user-qy8re3yx3d]

10!

[@kylehines3185]

42

[@youssefelyousfi4929]

x=10

[@jkbrahim9017]

10!