A Nice And Perfect Factorial Equation

🤩 Hello everyone, I'm very excited to bring you a new channel (aplusbi)
Enjoy...and thank you for your support!!! 🧡🥰🎉🥳🧡

When you purchase something from here, I will make a small percentage of commission that helps me continue making videos for you. ❤️ ❤️ ❤️

If you need to post a picture of your solution or idea:
#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 #sequence
via @YouTube @Apple @Desmos @NotabilityApp @googledocs @canva

PLAYLISTS 🎵 :

[Chrisoikmath_]

Since we solve this over integers d can be either positive or negative integer. So, (a, b) = (5d^2, +-60d), d€Z

[FisicTrapella]

Notice that we can write the equality as 6! = b · (b/a); so a must divide b; and this put some restrictions on d.
Taking the pair (5d^2, 60d) => b/a = 12/d; so d can't be anyone as d must divide 12. So,

d = 1, 2, 3, 4, 6, 12 => a= 5, 20, 45, 80, 180, 720. So, we get the pairs (a, b)

(5, 60) (20, 120) (45, 180) (80, 240) (180, 360) (720, 720)

[GirishManjunathMusic]

a·6! = b²
a and b are integers

there is of course the trivial solution of a = b = 0; ignoring that:
6! = 720 = 12²·5

Therefore, for the above equation to be true:
a is of the form 5n²; and b is of the form ±60n; where n is a non-zero integer not divisible by 5.
Or, a is of the form n²/5; and b is of the form ±12n; where n is a non-zero integer divisible by 5.

[wtspman]

6! = 2^4 * 3^2 * 5^1 => a must have an odd power of 5 as a factor to get a perfect square. a must also have even powers of 2 and 3 as factors. For any other prime factors that are not factors of 720, must also be present as even powers. a can be {5, 20, 45, 80, 125, 180, 245, 405, 605, 625, 720…}

[tixanthrope]

6! = 2^4 * 3^2 * 5

so a = 5 * c^2 where c is any integer
then b = 60 * c

[LoggyWD]

D can be positive or negative interger

[johns.]

Now do a! = b^2 for positive integers a and b.

[popodori]

i got a, b = 6!, 6! , your method is obviously better :)

[subversively]

Or set a=5^(2n+1)*d^2
Where n € N
AND d € Z😊