BdMO National 2013: Primary 4

Discussion on Bangladesh Mathematical Olympiad (BdMO) National
BdMO
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Joined: Tue Jan 18, 2011 1:31 pm

BdMO National 2013: Primary 4

Unread post by BdMO » Fri Jan 10, 2014 1:14 am

The English alphabets are arranged in $3$ rows in a Keyboard. Now somebody presses one key in the first row in such a way that there are same number of keys on both sides of that key in that row. Now a second person presses a key in the second row in the same way and a third person also does the same in the third row. Show that it is impossible.

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atiab jobayer
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Joined: Thu Dec 19, 2013 7:40 pm

Re: BdMO National 2013: Primary 4

Unread post by atiab jobayer » Thu Jan 30, 2014 11:10 pm

The patern of keyboard is 10:9:7.the term is when a word selected then there must be same number word beside 2side of row. So though the 2nd and 3rd row agree with this term but 1st don't . :twisted:
math champion

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samiul_samin
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Re: BdMO National 2013: Primary 4

Unread post by samiul_samin » Sun Dec 31, 2017 8:22 am

26 is an even number.The situation stated in the question is only possible if 3 of them are odd numbers.But the total of 3 odd numbers is also odd.So,it is impossible.☺☺☺

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SA'AD
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Re: BdMO National 2013: Primary 4

Unread post by SA'AD » Wed Jan 16, 2019 10:40 pm

Let,
in first row, the number of keys is '2k+1' (cuz excluding 1 key, if there are k no. of keys in one side, then total no. of keys is 2k+1)
in second row, the number of keys is '2n+1'
in third row, the number of keys is '2m+1',
/*k,m,n are natural numbers*/
then 2k+1+2m+1+2n+1=26
which implies, (k+m+n)=23/2
But this isn't possible cuz sum of any three natural numbers cannot be 23/2.
Hence, it's impossible..

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