Omer S.

asked • 06/07/15

prove that for every 8 choosen numbers from 10 to 36 you can always make equalities.

number can be used once. examples. let say that the choosen numbers are 10, 11, 12, 15, 18, 25, 32, 36 you can write 11+25=36 or 10+12+18=15+25. i tried to prove for summation for every 2 numbers $$\binom{8}{2}=28 $$ but there are 51 different summation that's the closest i got

Mark M.

tutor
Do you mean that at least one of the eight chosen can be expressed as the sum of some of the other 7?
Or must all of the eight be able to be expressed as the sum of some combination of the other 7?
Not too sure how the"pigeon hole principle" relates to this question.
Report

06/07/15

Omer S.

yes 
you dont have to use pigeon hole principle
Report

06/07/15

Omer S.

yes to the first question
Report

06/07/15

Omer S.

any help will be very very appreciated
Report

06/07/15

1 Expert Answer

By:

Mark M. answered • 06/07/15

Mathematics Teacher - NCLB Highly Qualified

Omer S.

thanks for the help, but you are not the one who choose, for any choosen number
Report

06/08/15

Omer S.

is't for every choosen number
Report

06/08/15

Mark M.

tutor
The problem states prove "for every 8 numbers. The phrase "for every" means "for any." If a counter example can be presented, which I have done, the proof is not possible no matter who gets to choose.
Report

06/08/15

Still looking for help? Get the right answer, fast.

Ask a question for free

Get a free answer to a quick problem.
Most questions answered within 4 hours.

OR

Find an Online Tutor Now

Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.