## BdMO National 2013: Primary 4

### BdMO National 2013: Primary 4

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.

- atiab jobayer
**Posts:**23**Joined:**Thu Dec 19, 2013 7:40 pm

### Re: BdMO National 2013: Primary 4

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 .

math champion

- samiul_samin
**Posts:**1007**Joined:**Sat Dec 09, 2017 1:32 pm

### Re: BdMO National 2013: Primary 4

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.☺☺☺

### Re: BdMO National 2013: Primary 4

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..

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..