subsets - interesting

Does the set {1;2;...;3000} contain a subset A consisting of 2000 numbers
such that x belongs to A implies 2x does not belong to A.

17 Answers

9
Celestine preetham ·

hmm b555 tell me do u have the solution for this ?

i donno how u accepted nishant sirs solution !

oopsy sir u dint see 3,6 sitting together ?

9
Celestine preetham ·

urs ofc

dude b4 this u were postin much tuffer Qs
so no one had time to solve it

9
Celestine preetham ·

yes understood now

terrific solution :)

9
Celestine preetham ·

i cudnt understand that can u explain a bit more b555 :)

9
Celestine preetham ·

note :
#9 method to find the summation shud read as [3000/r.4Sr]
its inelegant

see below

another way for S1 + S2 + ......

6 X ( no of odds < 3000/1024)
+
5X(no of odds <3000/256 - odds<3000/256)
+
....
...

= noof odds <3000/1024 +odds <3000/256 + ..... + odds<3000

= 1+6+23+94+375+1500 = 1999 !

39
Dr.House ·

hope everything is cleared now.

and thanks for pointing out celestine !!

39
Dr.House ·

Consider 1001 nonoverlapping pairs of form (a,2a)

1500,3000

1499,2998
...
751,1502

375,750
...
188,376

93,186
...
47,94

23,46
...
12,24

5,10
4,8
3,6

1,2

There are 998 numbers from {1,...,3000} which is not in pairs ==> for any 2000 numbers no less then 1002 lay in pairs ==> two numbers in the same pair

62
Lokesh Verma ·

oh yeah i got excited too fast :P

i thought i did solve another bhargav toughie :P

but not to be...

See the new set of numbers

39
Dr.House ·

wait , wait , wait

yes even i missed 3,6.

you are right.

wait i am working on it right now

62
Lokesh Verma ·

Lets take the numbers from 1501 to 3000

then from 376 to 750

then from 94 to 187

then from 24 to 46

then from 6 to 11

then from 1

1500+375+94+23+6+2 = 1999

Proved by construction of the set that this is possible. :)

Mistake pointed out by celestine

9
Celestine preetham ·

sir how does 3 and 6 both come !

ans is most probably impossible

method of working

let r be odd <3000

now there are no issues in including that in A

now u find that u can always get max no in set A if u include r,4r,16r,64r,256r,1024r whenever feasible ( can u see why ? )

now come to the evens any even of form 2r,8r,32r,128r,512r,2048r with r odd neednt be included from above reasoning

so our nos are

S1 + S3 + S5 +... Sr +S2999

where Sr is minimum value for [ 2000/r.4Sr ] = 0

39
Dr.House ·

yup , a little time every day would be great .

62
Lokesh Verma ·

:D

yeah I am trying to spednd sometime here again..

life's suddenly become very busy.. there has been a lot of work..

And unfortunately i have become a bit money oriented... (I have to earn someday :P)

39
Dr.House ·

ya anyways fabulous work.

AND HAPPY TO HAVE U BACK NISHANT `SIR` [58]

39
Dr.House ·

not air lightning sir , but your genious awakened

( i dont mean it always keeps sleeping :D)

62
Lokesh Verma ·

First i got lost in all the pigeon hole arguements and god knows what all...

Then from the air lightning struck me and i got the solution :D

341
Hari Shankar ·

fantastic!! :D

edit: admiration expressed w/o proper investigation :D

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