12:30 there will exist a prime p>n then p does not divide n! suppose p lies b/w divisors Dk and Dk+1, this prime p will make a gap b/w these divisors of more than 1 16:00 inequality can be written as 3=< d+(d-1)k and for k>0 it is true for d>=2 and for k=0 true for d>=3 hence works for small d
Sure. However as a college student you may want to learn more exciting stuff related to research. You may join the college math program or research program. Additionally you may also attend olympiad classes for fun.
another solution:by Bernards postulate we can say that there is a prime number between n and 2n,now if we find another pair of consecutive integers greater than n we will be done,so case 1)n is odd,we can take (n-1)sq,n-2(n),now you might be wondering how does n-1 sq work,n-1 sq =2* n-1/2 *n-1,now you can see why,for j even you can take n-2 whole square with similar reasoning and (n-3)(n-1),hopefully my solutions is correct,this problem becomes so simple with Bernards postulate
12:30 there will exist a prime p>n then p does not divide n! suppose p lies b/w divisors Dk and Dk+1, this prime p will make a gap b/w these divisors of more than 1
16:00 inequality can be written as 3=< d+(d-1)k and for k>0 it is true for d>=2 and for k=0 true for d>=3 hence works for small d
Which years usamo problem is this
USAMO 2024 Problem 1
3:20 what did u said here
For number theory, combinatorics, algebra, diagram IS the example
For number theory , combinatories, Algebra this ''sort of'' stuff, diagram means example...
15:56
for k=0, id d>=3 works
What if k is not 0?
Hey Cheenta, I am in my college now, but I wanna learn Olympiad Mathematics after my graduation. Will I be eligible in your program?
Sure. However as a college student you may want to learn more exciting stuff related to research. You may join the college math program or research program. Additionally you may also attend olympiad classes for fun.
Sir i want to learn geometry from you how i can i did
You may check cheenta.com for program related information.
another solution:by Bernards postulate we can say that there is a prime number between n and 2n,now if we find another pair of consecutive integers greater than n we will be done,so case 1)n is odd,we can take (n-1)sq,n-2(n),now you might be wondering how does n-1 sq work,n-1 sq =2* n-1/2 *n-1,now you can see why,for j even you can take n-2 whole square with similar reasoning and (n-3)(n-1),hopefully my solutions is correct,this problem becomes so simple with Bernards postulate
Instead of 2n-2 I just wrote 2n,it doesn't make any differenc3
sir i am a dropper , i feel very bad to have missed the olypiads are there anything i am eligible for
You can attemp Promys india