IIT JEE , AIEEE Forum - Mathematics , Algebra

rajatsen91

there are n students standing in a row.

every second student is asked to sit down.Then every third student is asked to change their positions from sitting to standing or vice versa.then every fourth

fifth, sixth and so on.finally the nth student is asked to change his position.Find the students who r sitting.

nivedh_89

if n = even,n/2 r sitting.................
if n = odd,(n+1)/2 r sitting.........................

spideyunlimited

Ans) ( n - 2 ) children are finally sitting

*PLZ RATE*

rini

every integer starting from 1 which is a perfect square of an integer will be standing at the end and all others are sitting...

so if the process is repeated n times, then the total numbers between 1 to n which are squares of an integer will be standing and remaining will be sitting...

im sure abt this... but it is a sort of question which is required to be done for exams like CAT.. i dont think jee would ever contain such sort of ques...

spideyunlimited

rini see my answer.. it is correct :) u can even check it by putting values

nivedh_89

spidey u r wrong................chk for 6 people
person 1,3,5 will remain standing,
2,4,6 will sit..................

spideyunlimited

no...2,3,5,6 will be sitting

nivedh_89

no....
see 1st one will b standing,2,3,4,5,6 will be sitting....

then,
3,4,5,6 will stand.........
then 4,5,6 will sit...............read the question properly....
then 5,6 will stand.......
then 6 will sit..................!!!!!!!!!!!

spideyunlimited

no 1st
1 stands , 2sits, 3 stands, 4 sits, 5 stands, 6 sits..

the question says that first every second person sits down
then in next round every 3rd person reverses wht his position currently is and so on

nivedh_89

see even u got 3 persons sitting..............then how can u say tht n-2 will sit............!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

spideyunlimited

ohho..
first 1,2,3,4,5,6, are standing
then
1 up, 2 down , 3 up , 4 down , 5 up , 6 down
then
1 up, 2 down, 3 down, 4 down , 5 up, 6 up,
then
1 up, 2 down, 3 down, 4 up, 5 up, 6 up,
then
1 up, 2 down , 3 down, 4 up, 5 down ,6 up
then
1 up, 2 down , 3 down, 4 up, 5 down ,6 down

now see n-2 ie. 6 - 2 ie. 4 ppl are sitting.
got it?

rini

spideyunlimited,

ur ans is wrong...

it seems right for smaller values of n.

luk, i have applied this logic:

first all of them are standing..

the process of reversing their condition starts from the second person..

so. the 1st person is not disterbed at all..

the 2nd person is disturbed once,

3rd - disturbed once..

4th-disturbed twice ( in 2nd and 4th round..)

it is obvious that if number of disturbance is even the person will remain standing and vise versa...

now,

the number of disturbance caused to a perticular no.= total factors of that no.excluding 1(as process started frm 2nd person).

you can check by putting values that every no. which is the square of an integer have even no. of factors(excluding 1) and all other no. have odd no. of factors excluding 1...

spideylimited, put n=10 and check... ur logic will fail..

plz rate!!!!

rajatsen91

Rini is right , that is the answer.

spideyunlimited

got it :)... yes.we have to see which numbers are affected by other numbers so their postion will change a number of times.

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