It's great seeing solutions that are a mix of combinatorics (isomorphism to a different, similar problem) and number theory ((n-1)!/n is an integer if n has 2 distinct factors). Awesome video!
Hey does anyone know how he’s determining if the number is an integer? Like he’s just saying how many twos the numerator and denominator have. How do you do that?
@@denny8529I have an idea. It’s cause when they have the same factors on top and bottom [say you have (3*2^4)/(2^2)], then by counting the amount of 2’s you can determine if it’s an integer since the denominator cancels out. Hope it helps! (and was the right idea 😅)
Tejasv Bhatia I had to wrap my head around that too. For example, (5^2-1)! = 5 x (5x2) x (5x3) x (5x4) x some integer. This integer has a 5^4 x some value M, whereas (5!)^5 has a 5^5 x some value N, and thus (5^4 x M)/(5^5 x N) = 1/5 x M/N. Because we’ve already accounted for any multiples of 5 within the 24! and (5!)^5, we have no 5s in M or N. This means M/N can’t possibly be a multiple of 5, and thus 1/5 x M/N cannot be an integer. With composite values of n, such as n=6 for (6^2-1)!, there are more than 5 ways (as in 6x12x18x24x30) to get some multiple of 6 (take for example 2x3, 4x9, ...), as well as more than six 6s in (6!)^6 (6! = (1x4x5) x (2x3) x 6 = 20 x 6^2 => (6!)^6 = 20^6 x 6^12). In short, thanks to the prime factorization of composites such as 6, (n^2-1)! and (n!)^n have higher powers of n than just (n-1) and n respectively. This leads to the question of whether or not these values of n work out, and like he said, any composite value except for 4 does work out to be an integer.
Aneesh Saripalli to clarify, because you’ve taken care of all the powers of n in this case (Ex: n=5 => 24! has 5^4 x M and (5!)^5 only has 5^5 x N, with no other ways to produce more 5s (as with primes), 5^4xM/(5^5xN) = 1/5 x M/N cannot be an integer because M/N cannot be a multiple of 5)
Thanks for your question, Thats because in this case (order matters as we are arranging) also we are dividing by n² ,which doesnt help So we change our approach and take a case (where order doesnt matter) and we get a crisp equation which has (n!)⁴ in the denominator. Hope it helps
As he said "you're taking AMC, you would suppose the primes and 4 fail", then you'd count and circle your answer hoping you were right, tbh not many people would even think that this Is the case so you're probably right, you'd have to take a wild guess
A quiet moment for all those who did all the work but didn't count up the primes at the end correctly.
@@SomeRandomGtaDude-zl3us f
F
@@SomeRandomGtaDude-zl3us F
f
@@johntan-aristy3019 Oh hi
It's great seeing solutions that are a mix of combinatorics (isomorphism to a different, similar problem) and number theory ((n-1)!/n is an integer if n has 2 distinct factors). Awesome video!
Why does he make such a complicated problem look sooooooooo simple
the richard effect
Hey does anyone know how he’s determining if the number is an integer? Like he’s just saying how many twos the numerator and denominator have. How do you do that?
@@denny8529I have an idea. It’s cause when they have the same factors on top and bottom [say you have (3*2^4)/(2^2)], then by counting the amount of 2’s you can determine if it’s an integer since the denominator cancels out. Hope it helps! (and was the right idea 😅)
1:46 “never gonna make this an integer” yeah just like these problems are never as easy as they seem
NICE! Finally a solution that properly explain why composites work, except for 4
HOT SOLUTION RUSCZYK
very nice and helpful to also see approaches that dont work, cause they might work in another problem! Thank you
Thank you for all your math videos. It helps and inspires me.
I really enjoy these math videos! 😂
The guy that disliked forgot 4 :)
Ryan Chen didn’t think I’ll ever meet a follow comrade here
Exactly 4 likes lmao
M. Chen 5 now
@@alexwang982 Actually, 7 now
@@M_Chen333 Actually 425 now
Hi. How to and what to read/study to be like you?
I still don't get how he excluded the primes in the beginning.
Aneesh Saripalli now he doesn’t get it even more
Tejasv Bhatia I had to wrap my head around that too. For example, (5^2-1)! = 5 x (5x2) x (5x3) x (5x4) x some integer. This integer has a 5^4 x some value M, whereas (5!)^5 has a 5^5 x some value N, and thus (5^4 x M)/(5^5 x N) = 1/5 x M/N. Because we’ve already accounted for any multiples of 5 within the 24! and (5!)^5, we have no 5s in M or N. This means M/N can’t possibly be a multiple of 5, and thus 1/5 x M/N cannot be an integer.
With composite values of n, such as n=6 for (6^2-1)!, there are more than 5 ways (as in 6x12x18x24x30) to get some multiple of 6 (take for example 2x3, 4x9, ...), as well as more than six 6s in (6!)^6 (6! = (1x4x5) x (2x3) x 6 = 20 x 6^2 => (6!)^6 = 20^6 x 6^12). In short, thanks to the prime factorization of composites such as 6, (n^2-1)! and (n!)^n have higher powers of n than just (n-1) and n respectively. This leads to the question of whether or not these values of n work out, and like he said, any composite value except for 4 does work out to be an integer.
Aneesh Saripalli to clarify, because you’ve taken care of all the powers of n in this case (Ex: n=5 => 24! has 5^4 x M and (5!)^5 only has 5^5 x N, with no other ways to produce more 5s (as with primes), 5^4xM/(5^5xN) = 1/5 x M/N cannot be an integer because M/N cannot be a multiple of 5)
Shouldnt all the odd numbers not work since the numerator would be even and denominator would be odd?
@@AlexandrBorschchev how denominator will be odd since all n! Are even and if x is even then x^a does not change the parity
My brain can’t comprehend factorials.
I'm lost at 8:20 why doesn't it help?
Thanks for your question,
Thats because in this case (order matters as we are arranging) also we are dividing by n² ,which doesnt help
So we change our approach and take a case (where order doesnt matter) and we get a crisp equation which has (n!)⁴ in the denominator.
Hope it helps
Can someone please explain why they cannot be primes?
Or u could guess D on the amc like i did
@@SomeRandomGtaDude-zl3us guess 069
How are we expected to solve this problem with so little time ughhh
As he said "you're taking AMC, you would suppose the primes and 4 fail", then you'd count and circle your answer hoping you were right, tbh not many people would even think that this Is the case so you're probably right, you'd have to take a wild guess
You just borrow Richard's brain for a minute...
@@notspaso6644the answer choices make it very plausible
too difficult to solve the last 5 problems in the short time......
Using de Polignac's formula would have simplified this a bit
@@siddharthamondal4346 could you explain please 😊?
solved this in 5 seconds, 5 answers all in 30 range, 20%
I dont understand why n cant be prime not sure what he meant when he went over that part
You’re back
does this use the sticks and stones method?
Can we use some variation of induction to solve this?
probably
where can i find all released problems? can't find them on maa website
@Edward Cha thanks for replying. but i only see up to 2018 though. maybe im missing something. do you have the link?
@@harryhong6540 They take a day or two to be posted. Most of the 2019 questions are up now. Solutions are still being added though.
@@raghavnarula1057 Where can I find AMC 10B?
The AMC 10b hasn’t happened yet.
Lin Sun Lin stfun
too hard
Indonesia?
I tried this problem b4 but i forgot 4
I feel you
Hơi khó hình dung vấn đề bạn đang trình bày..
what is a factorial and how did you get the 3 i am so lost and confused
He really resembles Drew Brees a little bit.
All these numbers and still miserable and depressed
Self note: 8:17
tfw you're busy counting up the two's but then you forget 6 exists 🙃
Hello, can you make russian subs Please?)
Reply please
d
I am in 9th standard from india amc 12 and aime, is very very easy test i think i can do it in half of the time that is alloted
IN 4 MINUTES. HOWWWWWWW. I feel stoopid
K
Can u teach. I'm in 9th and I can only get like a 90 on this test
I am a baby from New Zealand. This test is nowhere close to the word hard. I can even do this in 10 seconds.