Very cute question

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Asif Hossain
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Very cute question

Unread post by Asif Hossain » Sun Mar 14, 2021 6:01 pm

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.
Hmm..Hammer...Treat everything as nail

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Zafar
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Re: Very cute question

Unread post by Zafar » Sun Mar 14, 2021 6:46 pm

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 :) )
Last edited by Zafar on Sun Mar 14, 2021 7:14 pm, edited 1 time in total.

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Anindya Biswas
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Re: Very cute question

Unread post by Anindya Biswas » Sun Mar 14, 2021 6:55 pm

Zafar wrote:
Sun Mar 14, 2021 6:46 pm
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.
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
John von Neumann

Asif Hossain
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Re: Very cute question

Unread post by Asif Hossain » Sun Mar 14, 2021 7:09 pm

Zafar wrote:
Sun Mar 14, 2021 6:46 pm
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.
Hmm..Hammer...Treat everything as nail

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Zafar
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Re: Very cute question

Unread post by Zafar » Sun Mar 14, 2021 7:13 pm

Asif Hossain wrote:
Sun Mar 14, 2021 7:09 pm
Zafar wrote:
Sun Mar 14, 2021 6:46 pm
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 .

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Zafar
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Re: Very cute question

Unread post by Zafar » Sun Mar 14, 2021 8:10 pm

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 ?

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Mehrab4226
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Re: Very cute question

Unread post by Mehrab4226 » Sun Mar 14, 2021 10:43 pm

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.
The Mathematician does not study math because it is useful; he studies it because he delights in it, and he delights in it because it is beautiful.
-Henri Poincaré

hriditapaul
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Re: Very cute question

Unread post by hriditapaul » Mon Mar 15, 2021 10:12 pm

Won't the frog change its direction after every jump so that it jumps on a rectangular or square path?

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Anindya Biswas
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Re: Very cute question

Unread post by Anindya Biswas » Wed Mar 17, 2021 7:27 pm

hriditapaul wrote:
Mon Mar 15, 2021 10:12 pm
Won't the frog change its direction after every jump so that it jumps on a rectangular or square path?
Not neccessarily
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
John von Neumann

nahidhasan
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Re: Very cute question

Unread post by nahidhasan » Mon Mar 22, 2021 5:01 pm

i think something is wrong
here

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