A Factorial Equation | n! = 6!7!
Вставка
- Опубліковано 10 вер 2024
- 🤩 Hello everyone, I'm very excited to bring you a new channel (aplusbi)
Enjoy...and thank you for your support!!! 🧡🥰🎉🥳🧡
/ @sybermathshorts
/ @aplusbi
❤️ ❤️ ❤️ My Amazon Store: www.amazon.com...
When you purchase something from here, I will make a small percentage of commission that helps me continue making videos for you.
If you are preparing for Math Competitions and Math Olympiads, then this is the page for you!
CHECK IT OUT!!! ❤️ ❤️ ❤️
❤️ A Differential Equation | The Result Will Surprise You! • A Differential Equatio...
❤️ Crux Mathematicorum: cms.math.ca/pu...
❤️ A Problem From ARML-NYSML Math Contests: • A Problem From ARML-NY...
❤️ LOGARITHMIC/RADICAL EQUATION: • LOGARITHMIC/RADICAL EQ...
❤️ Finding cos36·cos72 | A Nice Trick: • Finding cos36·cos72 | ...
⭐ Join this channel to get access to perks:→ bit.ly/3cBgfR1
My merch → teespring.com/...
Follow me → / sybermath
Subscribe → www.youtube.co...
⭐ Suggest → forms.gle/A5bG...
If you need to post a picture of your solution or idea:
in...
#radicals #radicalequations #algebra #calculus #differentialequations #polynomials #prealgebra #polynomialequations #numbertheory #diophantineequations #comparingnumbers #trigonometry #trigonometricequations #complexnumbers #math #mathcompetition #olympiad #matholympiad #mathematics #sybermath #aplusbi #shortsofsyber #iit #iitjee #iitjeepreparation #iitjeemaths #exponentialequations #exponents #exponential #exponent #systemsofequations #systems
#functionalequations #functions #function #maths #counting #sequencesandseries
#algebra #numbertheory #geometry #calculus #counting #mathcontests #mathcompetitions
via @UA-cam @Apple @Desmos @NotabilityApp @googledocs @canva
PLAYLISTS 🎵 :
Number Theory Problems: • Number Theory Problems
Challenging Math Problems: • Challenging Math Problems
Trigonometry Problems: • Trigonometry Problems
Diophantine Equations and Systems: • Diophantine Equations ...
Calculus: • Calculus
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
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! --> 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 cannot be 12! bc there will be an 11. N cannot be bigger than 10. After 7, there are only composite numbers.
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
12! includes 11 which is not in 6! or 7!, so n cant be 12, and in fact must be less than 11
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
n!=6!7! 6!7!=3628800 10!=3628800 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.
6!7! is 100 x m, m is integer. So the closest 100th integer multiple is 10!. Number theory arguments are better, I feel.
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!
😮
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.
I got 10! by guess and check with the aid of a calculator.
It could be simpler.
Good evening sir ji
i made this recently, n=10
Oh··· as a programer, at the first view, it reads as "n not equals 6!7!". 😅,Sorry if being rude.
No! You're fine 😄
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).
@@user-ki4hm3sg9f More interesting than the original on from Syber indeed.
Nice!
I agree
good!
8 minutes to do something that should take 30 seconds…
thanks