A Factorial Equation | n! = 6!7!
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- Опубликовано: 7 фев 2025
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It is easy to solve the problem by showing that 6! = 8*9*10. But there is a more elegant solution. Since 6!7! cannot contain the prime factor 11, we know that n < 11. Furthermore, 6!7! contains 5^2, which implies that n >= 10. Therefore, 10
🤓☝️you need to verify it works(just leaft it as a reminder, since i fucked this part up way too often)
Vous démontrez fort bien que si n existe il vaut 10 mais existe-t-il? Peut être bien que oui peut être bien que non.
Il faut donc alors en plus appliquer votre première méthode par regroupement des facteurs de 6! ou sortir la calculette.
La logique appliquée à ces problèmes sur les entiers est souvent élégante pourvu qu'elle soit vraiment logique.
Votre seconde solution parait séduisante mais elle est boiteuse en l'état.
Quick way is to combine digits from 6! as follows: 2x4=8, 3x5x6=90=9x10, so 6!=8x9x10 In conclusion 7! x 6! = 7! x8x9x10 = 10!, so n=10
Maybe leave 7! alone. Then work with 6! to make 8, 9 and 10.
Easy: 11 is not a factor of the LHS, so n
👍
n!=6!7! 6!7!=3628800 10!=3628800 n=10
I wanna learn about that alternative method to find how many powers of a prime divides n! Which you mentioned at 4:18 I think I knew this before but I have forgot about it. What's the formula named please tell me so I can search about it on Wikipedia you know when you sum floor(75/2)+floor(75/4)+..... So on etc I was talking about that not the floor and ceiling functions.
You just have to recombine the factorization of 6!; or in a general case, the factorization of the minor factor.
An interesting question whether there is a general case, or indeed any other case, where the product of successive integers greater than 2 is itself a factorial.
A general case would be n! = k! p! where n, k and p are integers and k>p.
To find n we just have to recombine the factorization of p to add consecutive factors to k: (k + 1), (k +2), ... => n = (k + k') => n! = (k + k')(k + k' -1)···(k + 1)k!
Notice that we must use all the factors in the factorization of p, and we have some limitations, as said i the video:
n must be greater than k
n can't be prime
😮
N=10, 1:23 said n can be 12, but it could not or 11 as a prime would be missing in the factorial.
So: N>7, n
You had solved this exact same before
i wanna learn more about what you said at 4:18 any sources or videos???
Check these first because my videos on floor function (aka greatest integer function) are not that basic:
en.wikipedia.org/wiki/Floor_and_ceiling_functions
ruclips.net/video/q68Hl7DhsuQ/видео.html
Let us know of any questions
@@SyberMath I want to know about the one you used to find how many powers of 2 divide 75!
@@SyberMath I want to know about the one you used to find how many powers of 2 divide 75!
since it's not a general solution for n! = a!b!, IMHO solving the reverse problem n!*m! = 10! is a more interesting one (because we need to find 2 variables)
Good idea!
OK, so n!/7! = 6!, which is an integer. I start by noting n < 11, as 11 does not divide 6!. On the other hand, 5 divides 6!, so 5 must also divide n!/7!. So n must be 10. Sure enough, 10*9*8 = 720 = 6!.
n!=6!7! --> n>7
6!=6×5×4×3×2×1
=(3×2)×5×(4×2)×3
=(2×5)(3×3)(4×2)
=10×9×8
Therefore n!=10×9×8×7!
=10! --> n=10
Good Morning，Sir
Starting from 8 if divided by 7!，then 8*9*...*n = 6! = 720
Now 8*9 = 72，then n = 10
Good Luck
James 08-14-2024
There's another alternative here，since n must be less than 11 because 11 is a prime number
But 8 and 9 are less than 100，therefore the only possible outcome is 10，if without calculation
6!7! is 100 x m, m is integer. So the closest 100th integer multiple is 10!. Number theory arguments are better, I feel.
N cannot be 12! bc there will be an 11. N cannot be bigger than 10. After 7, there are only composite numbers.
6! X 7!= (1x2x3x4x5x6) X (1×2×3×4×5×6×7) = (1x2x3x4x5x6) X (1x2x3x(2x2)x5x(2x3)×7) = (1x2x3x4x5x6)x(7×(2×2×2)×(3×3)×(2×5)) =10!
12! includes 11 which is not in 6! or 7!, so n cant be 12, and in fact must be less than 11
n = 10
N=10
I think this is stupid n leanthy way ... lets dig in quickly
n (n-1)(n-2) (n-3)! = 720 (7!)
n = 10 on simple comarison
This is a 30-second problem expanded into a 9-minute video.
The trivial cases are:
n! (n!-1)! = n!!
Examples of the trivial cases:
2!1! = 2!
3!5! = 6!
4!23! = 24!
5!119! = 120!
6!719! = 720!
The movie contains a non-trivial case
6!7! = 10!
I found no other non-trivial examples (with a Python program, but I didn't get very far...). Do any other exist?
Note another non-trivial fact:
3! weeks = 10! seconds
Wow! That's amazing. Thanks for sharing
10
I got n=10.
I got 10! by guess and check with the aid of a calculator.
i made this recently, n=10
Good evening sir ji
It could be simpler.
Trivial. n!/6!=7*8*9*10*....n , n can not be more than 10 because RHS is not divisible by 11.and can not be less than 10 because it needs 5 as a divider , Hence 10 is the only candidate.
Not trivial. are there any other m,n that n!=m!*(m+1)!. Or more general n1=m1*k1.: 6!=5!*3! - Something else?
n! = m!(m+1)!
The only solutions are (n, m) = (10, 6) or (2, 1).
Proof:
Starting from the equation n! = m!(m+1)!, we derive that n(n-1)...(m+2) = m! (Equation 1).
Let q be the largest prime number such that q ≤ n. Since q cannot be a factor on both sides of Equation 1, it follows that q = m + 1.
Next, let p be the largest prime number such that 3p ≤ n. If p > 3, then p, 2p, and 3p cannot all be factors of either side of Equation 1. Thus, p ≤ 3, and n < 15. Consequently, q can be one of 2, 3, 5, 7, 11, or 13.
Given that q = m + 1, we have n! = q!(q-1)! (Equation 2).
Now, consider the following cases:
‣Case 1: If n = q, then q-1 = 1, giving us (n, q) = (2, 2).
‣Case 2: If q = 3, 5, 11, 13, then by the definition of q, n = q + 1. In this case, q + 1 = (q-1)! (Equation 3). If q > 3, then q + 1 = (q-1)! > (q-1) * 2 = 2q - 2, which contradicts q > 3. Therefore, the only candidate is q = 3, but in this case, Equation 3 does not hold.
‣Case 3: If q = 7, the possible values for n are n = 8, 9, 10 based on the definition of q. The right-hand side of Equation 2 is divisible by 5^2, so 10 ≤ n must hold. Therefore, the only candidate is n = 10, and in this case, Equation 2 is satisfied.
Thus, the possible pairs are (n, q) = (10, 7) or (2, 2). Therefore, the solutions are (n, m) = (10, 6) or (2, 1).
@@イキリスト教教祖 More interesting than the original on from Syber indeed.
Nice!
I agree
good!
Oh··· as a programer, at the first view, it reads as "n not equals 6!7!". 😅,Sorry if being rude.
No! You're fine 😄
8 minutes to do something that should take 30 seconds…
thanks