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.
¿Por qué la gran mayoría de gente que viene aquí a comentar piensa que TODOS los videos de matemática en UA-cam 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.
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.
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!
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.
I tried to do the same way, ,but extending 7! with 6! - I got the 2*4 = to make 8! and gave up. I did not think to convert 6 to 2*3 and continue - as you did. Good example
14 днів тому
42 is in here, too! 10! seconds is exactly 6 weeks or 42 days... So beautiful
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!)
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
@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!
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.
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!.
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.
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).
I did it as (x)P(x-6) then i had to find arithmetic sequence of few numbers that can give me the same value of 7! which is 5040, i started with 10 so i had to take 4 numbers coz of the (x-6), so 10 x 9 x 8 x 7 = 5040 therefore X=10
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.
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...
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
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
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
@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! .
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.
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
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.
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.
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!
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!
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
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
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.
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.
you didn't really "SOLVE" this though... You just said "OK, let's try 10 as an example..." and then proved that it works... But that was not really a "SOLUTION" ... -_- .
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
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!.
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.
great observations guys 😅
If you already know 6!×7! = 10! Then you dont need any further steps...
@@DerekRoss1958 (n! - 1)! n!= (n!)!. Follows from the recursive definition of the factorial.
@@emanuellandeholm5657 Yes, I pointed that out in another comment.
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.
You mean 11 being prime, I assume.
The "simple matter of checking" amounts to the solution presented here. (Or else a whole lot of multiplying.)
As a jee aspirant I solved this problem in 1.3 minutes
DID YOU KNOW?
The number of seconds in 6 weeks is 10!
Well I didn’t know that
This is why time seems to pass so fast sometimes !
and 11! = 10! × 11
Not surprising, as both 3600 seconds in a hour and 24 hours in a day have a ton of divisors
@@gmdFrameThe fact that it has the exact right prime factors to the exact right powers entirely by coincidence is astounding.
A little silly to expand 7! It has to grow from 7!, first thing is to find a 8 from the expansion of 6!
Really enjoyed this; thank you…
¿Por qué la gran mayoría de gente que viene aquí a comentar piensa que TODOS los videos de matemática en UA-cam 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.
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.
Nice warm up as well as easy question.
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.
Multiply the 6! To the other side and you recursively divide until you get one to reverse the factorial.
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!
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.
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.
I tried to do the same way, ,but extending 7! with 6! - I got the 2*4 = to make 8! and gave up. I did not think to convert 6 to 2*3 and continue - as you did. Good example
42 is in here, too! 10! seconds is exactly 6 weeks or 42 days... So beautiful
9:25 Not the only case where a factorial is a multiple of two factorials - you also have 6! = 3! X 5!
Also 2! = 1! x 2! 🤣
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!)
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
@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!
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.
Excellent. Thank you!
Outstanding question!
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!.
6! = 720. 720/10=72. 72/9=8.
it means x! = 7!*8*9*10 = 10! therefore, x=10
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.
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).
I created a fun case for you. Find x, where: x!*11! = 3!*6!*22! .
Just do 6 factorial multiplied by 7! And that gives you the factorial of 10
Can you make videos on IIT JEE ADVANCE OR MAINS question?
I did it as (x)P(x-6) then i had to find arithmetic sequence of few numbers that can give me the same value of 7! which is 5040, i started with 10 so i had to take 4 numbers coz of the (x-6), so 10 x 9 x 8 x 7 = 5040 therefore X=10
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.
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...
I don't think anything is "left" after you get to 10, lol...
Wonderful
Seeing so many of the same problem floating around in FB, UA-cam, I can even remember the answer immediately as 10!
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
Haha the fly on cam😂😂😂
Brilliant 👍🏿
Marvellous
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
Thx a lot
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
6! = 720 which is 10x9x8
x! =6! 7! so
x! = 10x9x8x7!
done
Thats the clearest and easiest solution.
This literally took me about 5 seconds to solve
It took me only two seconds. You're slow.
@@stottpie I solved it almost instantly, by skipping to the end of the video. 🤪 🤦🏼♂
Sometimes fast is not good. And it is not what you are thinking. The best one is to be CORRECT.
Do you thought as me?
6!×7! must have two multiples of 5 as factors (so we know it's less then 15!), but do not have 11 as a factor. So it must be 10!
10!=3628800
4! * 23! = 24! and the list goes on... 5! * 119! = 120!
can you make a general formula for this?😂
Not all cases apply here, but every implementation of my formula will result in this happening:
if x! =y-1
y! = x!y-1!
@@aidenbooksmith2351 there’s a mistake in your “formula”. I’ll let you figure it out on your own!
@@SzabolcsHorváth-r1z x!=(x!)!/(x!-1)!
@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! .
How often is the ratio of two factorials itself a factorial? That seems exceedingly unlikely. Are there other non-trivial examples?
I solved it mentally before i run the video. Thanks for tha challenge ❤
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.
very nice sir
i did it in first attempt
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!
It's obv x = 10
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
(10!)/(6!)=5040=7! x=10
You can’t use a calculator in a math competition
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.
Yes! I solved it similarly: 7! * 720 = x! so
720 = 8*9*...*x and x = 10.
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
Another fun fact is that 10! seconds is exactly 6 weeks.
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.
Anyone else nearly get jumpscared by the sudden appearance of a fly right as he was writing "7! = 7!"?
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.
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!
You can also say 6!^2 x 7
Note that answer must be less than 11 since 11 is prime and not a factor in 7!
Got this with hit and trial
can you just remove the factorial and multiply 6x7 to find x?
Did 7! × 6! then I "built" the numbers 8, 9, 10 from what was on the page
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!
You have a typo: 5!*7! ≠ 10! .
@@oahuhawaii2141 Thank You
6!=2x3x4x5x6=8x3x5x6=8x3x5x2x3 =8x9x2x5=8x9x10 so x!=10! and x=10
Thanks
fun trick math question: If 6a x 7a = 10a, what is a?
a = 0, 5/21
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!
Is there an inverse gamma function?
I sort of cheated and did hit and trial using calculator starting from 9! and got the answer. But nice question.
I solved it by calculating 7*(6!)^2which is same as 10 ! .So X=10.
x!= x(x-1)! = 7!×(7-1)!
Comparing
x=7
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
Great
👌👍
Me who find it pretty easily by a method mostly similar to trial and error method
10 ясно сразу. Просто 5 должно быть кратно. Следующее за 5 10. Все.
Интересно, есть те, кто не смог это решить устно за несколько секунд? Как же сильно у нас отличантся уровень математического образования..
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
6 ! = 10 * 9 * 8
X ! = 10 !
Every second simpliest question marked as Olympiad. Really? Could you provide a link to what Olympiad exactly?
You just need to learn how to write the number seven by hand so that it looks different from the number one.
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
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.
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.
10!
X = 10
Video is twice as long as needed, but cool math puzzle.
👍
🎉🎉🎉🎉
1! = 1!*1!
Easy, 7!*6! =x= 3628600 you divided by 2 then by 3....up to 10 where the result is 1. so x=10.
42
10
There are 10! seconds in a week.
The only tricky thing: x is not 10!, but 10.
I found solution in 2 minutes
😮
It took me Just 20 sec
good work. ( I got it wrong).
An easy one. Takes 3 minutes to solve.
why india math olympiad so easy? or is this for primary school level?
هذا اسمه رياضيات الخشيبات.
you didn't really "SOLVE" this though... You just said "OK, let's try 10 as an example..." and then proved that it works...
But that was not really a "SOLUTION" ... -_-
.