A frog can jump either right or left from his position.In his 1st jump he can go 1cm,2nd jump 2cm and in n-th jump it goes n cm. Prove that he can't back to his initial position in his 2019th jump.
Re: Very cute question
Posted: Sun Mar 14, 2021 6:46 pm
by Zafar
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 2019th jump (proved )
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 (proved )
The problem with this argument is we actually have $1009$ even jumps and $1010$ odd jumps. $1+2+3+\dots+2019=\frac{2019\cdot2020}{2}$ which is even.
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 (proved )
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 (proved )
Note he can also jump right or left.
I dont know but when I read I saw 1009 even jumps . any kinds of sum means subtraction too . Well let me think now how to solve it now .
Re: Very cute question
Posted: Sun Mar 14, 2021 8:10 pm
by Zafar
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 ?
Re: Very cute question
Posted: Sun Mar 14, 2021 10:43 pm
by Mehrab4226
The question doesn't seem correct.
Let, the frog makes $3$ jumps,
He can come back by $1,2$ jumps right and $3$ jumps left.
Let,
The frog can come back to its initial position when he takes n jumps, where $n $is an odd integer.
So, we claim that the frog can also come back to its initial position when it takes $n+4$ steps,
Notice,
$(n+1)+(n+4)=(n+2)+(n+3)$so we can get to the initial position after $n$ steps and then by taking the last 4 steps like this $(n+1)-(n+2)-(n+3)+(n+4)$ we can get back to the initial position again, notice $n+4$ is also odd.
Now,
$3 \equiv -1(mod4)$
$2019 \equiv -1 (mod4)$
So it is possible to get back in the initial position by $2019$ steps.
Re: Very cute question
Posted: Mon Mar 15, 2021 10:12 pm
by hriditapaul
Won't the frog change its direction after every jump so that it jumps on a rectangular or square path?
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