Search found 10 matches
- Mon Mar 15, 2021 1:04 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National Junior 2020 P4
- Replies: 10
- Views: 7527
Re: BdMO National Junior 2020 P4
I think its 10! or 3628800 we have 10 positions and we dont know their initial positions .each of them has the same potential to obtain each position . so the ans should be 10! Not quite! The guy in rank #1 can never win prizes other than 1st and 2nd. So, each of them DOES NOT have the same potenti...
- Sun Mar 14, 2021 8:32 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National Higher Secondary 2020 P8
- Replies: 8
- Views: 9446
Re: BdMO National Higher Secondary 2020 P8
Let us try to count the number of kawaii permutations in $1,2,3,4, \cdots (n+1)$ Now let us take a choose any number(When in ascending order). Then we can find a kawaii permutation by exchanging it with any number less than itself. All the other numbers must stay in their same position or else we w...
- Sun Mar 14, 2021 8:10 pm
- Forum: Secondary Level
- Topic: Very cute question
- Replies: 9
- Views: 8162
Re: Very cute question
well this sequence I found .
and the best approach of mine to solve it still now .
( its not a proof of course )
±(1-2-3+4+5-6-7+8.......2017-2018-2019+2020) =0
and here total numbers should be a product of 4.
but 2019 isnt a product of 4 .
then what ?
can you share your approaches ?
and the best approach of mine to solve it still now .
( its not a proof of course )
±(1-2-3+4+5-6-7+8.......2017-2018-2019+2020) =0
and here total numbers should be a product of 4.
but 2019 isnt a product of 4 .
then what ?
can you share your approaches ?
- Sun Mar 14, 2021 7:13 pm
- Forum: Secondary Level
- Topic: Very cute question
- Replies: 9
- Views: 8162
Re: Very cute question
I got a easy solution for this math . well , we have 1008 even jumps and 1009 odd jumps. any kinds of sum of 1008 even numbers = random even number any kind 1009 odd numbers = random odd numbers random even number + random odd number ≠ 0 so , he cant back to his initial position in his 2019th jump ...
- Sun Mar 14, 2021 6:52 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National Junior 2020 P4
- Replies: 10
- Views: 7527
Re: BdMO National Junior 2020 P4
I think its 10! or 3628800
we have 10 positions and we dont know their initial positions .each of them has the same potential to obtain each position
. so the ans should be 10!
we have 10 positions and we dont know their initial positions .each of them has the same potential to obtain each position
. so the ans should be 10!
- Sun Mar 14, 2021 6:48 pm
- Forum: Primary Level
- Topic: Short questions
- Replies: 4
- Views: 13025
Re: Short questions
that was so funny xD .
- Sun Mar 14, 2021 6:46 pm
- Forum: Secondary Level
- Topic: Very cute question
- Replies: 9
- Views: 8162
Re: Very cute question
I got a easy solution for this math . well , we have 1008 even jumps and 1009 odd jumps. any kinds of sum of 1008 even numbers = random even number any kinds of sum of 1009 odd numbers = random odd number random even number + random odd number ≠ 0 so , he cant back to his initial position in his 201...
- Thu Mar 11, 2021 8:20 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National Junior 2020 P3
- Replies: 2
- Views: 3031
Re: BdMO National Junior 2020 P3
assume x and y are the two outcomes . 0<x<21 and 0<y<21 . so ,1< x + y <41 we can move forward like this , x + y = sum >> (x+1)+(y-1)= sum and finally at a point >> (x+(y-1))+(y-(y-1)) = sum so , if we take the minimum x and maximum y , we can get the sum in maximum way . so , 20+1=21 (solved :) )
- Thu Mar 11, 2021 6:51 pm
- Forum: Number Theory
- Topic: Total Number of digits
- Replies: 1
- Views: 9557
Re: Total Number of digits
well ,the number of digits in an exponential number n^k is floor[1+{kln(n)/ln(10)} ]( I don't have proof for this algorithm ) . So I found the solution to be 6629 .
- Thu Mar 11, 2021 1:00 am
- Forum: Combinatorics
- Topic: Again PHP
- Replies: 5
- Views: 6559
Re: Again PHP
well how can it satisfy n=2 ? I think we can select exactly n+1 points maximum that wont make a right triangle . hence for all n> 2 , 2n-1>n+1 . and so I think its done .