A Factorial Equation | n! = 6!7!

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PLAYLISTS 🎵 :

[stephenshefsky]

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 <= n < 11 which has one integer solution, n = 10.

[peshepard]

Maybe leave 7! alone. Then work with 6! to make 8, 9 and 10.

[Beathan]

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

[YAWTon]

Easy: 11 is not a factor of the LHS, so n<11. The LHS has the factor 25 (5 from 6! and 5 from 7!), so n must be 10 which is the only n<11 that is divisible by 25. Legendre' formula...

[aashsyed]

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 So on etc I was talking about that not the floor and ceiling functions.

[FisicTrapella]

You just have to recombine the factorization of 6!; or in a general case, the factorization of the minor factor.

[jarikosonen]

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<11.

[RyanLewis-Johnson-wqxs]

n!=6!7! 6!7!=3628800 10!=3628800 n=10

[aashsyed]

i wanna learn more about what you said at 4:18 any sources or videos???

[nasrullahhusnan]

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

[wonghonkongjames]

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

[jesusalej]

N cannot be 12! bc there will be an 11. N cannot be bigger than 10. After 7, there are only composite numbers.

[malabikasaha]

6!7! is 100 x m, m is integer. So the closest 100th integer multiple is 10!. Number theory arguments are better, I feel.

[rickdesper]

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!.

[adamrussell]

12! includes 11 which is not in 6! or 7!, so n cant be 12, and in fact must be less than 11

[MarsRover]

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)

[Quest]

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

[yedemon]

Oh··· as a programer, at the first view, it reads as "n not equals 6!7!". 😅, Sorry if being rude.

[gregwochlik]

I got 10! by guess and check with the aid of a calculator.

[MrSivilla]

6! X 7!= (1x2x3x4x5x6) X (1×2×3×4×5×6×7) = (1x2x3x4x5x6) X (1x2x3x(2x2)x5x(2x3)×7) = =10!