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Question Number 209746 by MM42 last updated on 20/Jul/24

Q)Choose at least some members  frome the set A={14,15,...,20,22,23,...,28}  so that in includes three consecutive  members?

Q)ChooseatleastsomemembersfromethesetA={14,15,...,20,22,23,...,28}sothatinincludesthreeconsecutivemembers?

Commented by MM42 last updated on 20/Jul/24

for example   n=select  numbers  n=3 ; 14,15,16  ✓  or  14,15,17  ×⇒not true  n=4 ; 17,22,23,24  ✓  or  15,17,18,20 ×⇒not true  n=6  ;15,17,18,19,20,27 ✓ or 18,19,22,24,25,28 ×⇒not true  min(n)=?  ; a_1 ,a_2 ,a_3 ,...,a_n    ⇒∃ 1<i,j,k <n ; a_k =a_j +1=a_i +2

forexamplen=selectnumbersn=3;14,15,16or14,15,17×nottruen=4;17,22,23,24or15,17,18,20×nottruen=6;15,17,18,19,20,27or18,19,22,24,25,28×nottruemin(n)=?;a1,a2,a3,...,an1<i,j,k<n;ak=aj+1=ai+2

Answered by MM42 last updated on 20/Jul/24

The least number of choices can be 11.  because in the set {14,15,17,18,20,22,24,25,27,28} there are no three   consecutive numbers.  if we put any other number  in the set,we will have   three consecutive numbers in the collection.★

Theleastnumberofchoicescanbe11.becauseintheset{14,15,17,18,20,22,24,25,27,28}therearenothreeconsecutivenumbers.ifweputanyothernumberintheset,wewillhavethreeconsecutivenumbersinthecollection.

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