At first glance: 6! is 2^4 * 3^2 * 5^1. All perfect squares have all prime factors to even exponents. So any number of the form 2^(2k ) * 3^(2m) * 5^(2n-1) * p^2 (where k, m, n and p are non-negative integers) should accomplish this. (Edit: actually, we can simply say 5^(2n-1) * p^2, because the 2 and 3 can be rolled into the p.)
It is much simpler .. every digit in the faculty must be compensated by something to give a square. The 6 by 2*3 to give 6*6, the 4=2x2 is a square already so stripe away 2,3,4 and 6 (2*3*4*6 is square) on the faculty site and they need not be compensated anymore , the 5 cannot be compensated, so it must have factor 5 to give a square 5x5. So a = 5 and b=sqrt(4) *5*6 =60 . I did it without paper ,by head.
B squared is equal to b x b So a x 6! = b x b Idk how to explain my working out but you can see 6! = a is a solution, which would also mean that b is equal to both a and 6! Therefore 6! which equals 720 is a solution
6!=1*2*3*4*5*6=6*4*5*6=2*2*6*6*5=12*12*5 So if a*6!=b^2 with natural numbers a and b, a must be dividable by 5, because the lleft side of the equation is diviidable by 5, so b must be diviidable by 5 and b^2 must be dividable by 5^2. So a=5 and b=60 is a solution, also a=5 and b=--60. The others solutions are a=5*c^2 with a whole number c and b=60*c. There is no solution with a negative whole number,because the right side of the equationis can never be smmaler than zero.
Well, actually that is not quite true. a must be a multiple of 5, agreed. But it might be a multiple of any prime number greater than 5 raised to the power of 2n, where n is an integer. Every prime number might appear or not as a multiple and n can be different for every prime number...
a•6! = b^2
a•(3^2 • 4^2 • 5) = b^2
Let a= 5(x^2)
(3•4•5•x)^2 = b^2
(60x)^2 = b^2
So b=60x, a=5(x^2)
The exponents of the prime factors of a perfect square are even !
Very cool and original ! Thank you !
5a x 12^2 = b^2
a = 5^(2n-1), b=12x5^n, non negative int n
At first glance: 6! is 2^4 * 3^2 * 5^1. All perfect squares have all prime factors to even exponents. So any number of the form 2^(2k ) * 3^(2m) * 5^(2n-1) * p^2 (where k, m, n and p are non-negative integers) should accomplish this. (Edit: actually, we can simply say 5^(2n-1) * p^2, because the 2 and 3 can be rolled into the p.)
It is much simpler .. every digit in the faculty must be compensated by something to give a square. The 6 by 2*3 to give 6*6, the 4=2x2 is a square already so stripe away 2,3,4 and 6 (2*3*4*6 is square) on the faculty site and they need not be compensated anymore , the 5 cannot be compensated, so it must have factor 5 to give a square 5x5. So a = 5 and b=sqrt(4) *5*6 =60 . I did it without paper ,by head.
B squared is equal to b x b
So a x 6! = b x b
Idk how to explain my working out but you can see 6! = a is a solution, which would also mean that b is equal to both a and 6!
Therefore 6! which equals 720 is a solution
6!=1*2*3*4*5*6=6*4*5*6=2*2*6*6*5=12*12*5
So if a*6!=b^2 with natural numbers a and b, a must be dividable by 5, because the lleft side of the equation is diviidable by 5, so b must be diviidable by 5 and b^2 must be dividable by 5^2.
So a=5 and b=60 is a solution, also a=5 and b=--60. The others solutions are a=5*c^2 with a whole number c and b=60*c. There is no solution with a negative whole number,because the right side of the equationis can never be smmaler than zero.
I read "Dopamine equation" and wondered what kind of property has this expresion that makes you so happy lol
Equations always make me happy hehehe 😜
If a = (1/20) the b would be ±6 OR if a=(1/5) b= ±12, so I think there will be infinitely many solutions.
😊😊😊👍👍👍🎉🎉🎉
a solution is a=720, b=720
Nice!
i did this in my head in 1 minute as i looked at the thumbnail
😲
Very nice!
Thanks!
8:14 I can hear Any Math saying "Nice".
yes!!!!! 👌
6 !
= 2 * (3!) * (3! ) * 4 * (5 !) ^2 * 6
= ( 4 * (3)! * (5!)) ^2 * 3
Hereby
a * (6!) = b^2 implies
a * 3 * ( 4 * (3!) * (5!)) ^2 = b^2
Hereby
b = 4 * (3!) * (5!) * 3 = (2!) * (3!) * (6!)
a = 3
Nice
Thanks
Actually any a of the form 5^(2m+1)*c² will work.
Same as my solution
6!=720=5*144...a=5,a=20,a=45...a=80...
❤
a = 5, b = 60
Why is c positive?
Well, actually that is not quite true. a must be a multiple of 5, agreed. But it might be a multiple of any prime number greater than 5 raised to the power of 2n, where n is an integer. Every prime number might appear or not as a multiple and n can be different for every prime number...
a=b=6! Surely?
That’s a good one!
you are very intense
What does that mean?
Bravo. It's not difficult, but it's interesting how you parametrize the solutions with the integers
Why do you ramble on
What do you mean?
@@SyberMath i mean that you don’t really get to the point of your problems