A Factorial Equation | n! = 6!7!
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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
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!=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
👍
😮
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
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! 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!
n!=6!7! 6!7!=3628800 10!=3628800 n=10
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!.
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!
N cannot be 12! bc there will be an 11. N cannot be bigger than 10. After 7, there are only composite numbers.
12! includes 11 which is not in 6! or 7!, so n cant be 12, and in fact must be less than 11
6!7! is 100 x m, m is integer. So the closest 100th integer multiple is 10!. Number theory arguments are better, I feel.
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.
You had solved this exact same before
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
N=10
n = 10
I got n=10.
It could be simpler.
I got 10! by guess and check with the aid of a calculator.
Oh··· as a programer, at the first view, it reads as "n not equals 6!7!". 😅,Sorry if being rude.
No! You're fine 😄
Good evening sir ji
i made this recently, n=10
10
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!
8 minutes to do something that should take 30 seconds…
thanks