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Q-Choose-at-least-some-members-frome-the-set-A-14-15-20-22-23-28-so-that-whith-confidence-includes-three-consecutive-members-




Question Number 209746 by MM42 last updated on 21/Jul/24
Q)Choose at least some members  frome the set A={14,15,...,20,22,23,...,28}  so that whith confidence  includes three consecutive  members?
$$\left.{Q}\right){Choose}\:{at}\:{least}\:{some}\:{members} \\ $$$${frome}\:{the}\:{set}\:{A}=\left\{\mathrm{14},\mathrm{15},…,\mathrm{20},\mathrm{22},\mathrm{23},…,\mathrm{28}\right\} \\ $$$${so}\:{that}\:{whith}\:{confidence}\:\:{includes}\:{three}\:{consecutive} \\ $$$${members}? \\ $$
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
$${for}\:{example}\:\:\:{n}={select}\:\:{numbers} \\ $$$${n}=\mathrm{3}\:;\:\mathrm{14},\mathrm{15},\mathrm{16}\:\:\checkmark\:\:{or}\:\:\mathrm{14},\mathrm{15},\mathrm{17}\:\:×\Rightarrow{not}\:{true} \\ $$$${n}=\mathrm{4}\:;\:\mathrm{17},\mathrm{22},\mathrm{23},\mathrm{24}\:\:\checkmark\:\:{or}\:\:\mathrm{15},\mathrm{17},\mathrm{18},\mathrm{20}\:×\Rightarrow{not}\:{true} \\ $$$${n}=\mathrm{6}\:\:;\mathrm{15},\mathrm{17},\mathrm{18},\mathrm{19},\mathrm{20},\mathrm{27}\:\checkmark\:{or}\:\mathrm{18},\mathrm{19},\mathrm{22},\mathrm{24},\mathrm{25},\mathrm{28}\:×\Rightarrow{not}\:{true} \\ $$$${min}\left({n}\right)=?\:\:;\:{a}_{\mathrm{1}} ,{a}_{\mathrm{2}} ,{a}_{\mathrm{3}} ,…,{a}_{{n}} \: \\ $$$$\Rightarrow\exists\:\mathrm{1}<{i},{j},{k}\:<{n}\:;\:{a}_{{k}} ={a}_{{j}} +\mathrm{1}={a}_{{i}} +\mathrm{2}\: \\ $$
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.★
$${The}\:{least}\:{number}\:{of}\:{choices}\:{can}\:{be}\:\mathrm{11}. \\ $$$${because}\:{in}\:{the}\:{set}\:\left\{\mathrm{14},\mathrm{15},\mathrm{17},\mathrm{18},\mathrm{20},\mathrm{22},\mathrm{24},\mathrm{25},\mathrm{27},\mathrm{28}\right\}\:{there}\:{are}\:{no}\:{three}\: \\ $$$${consecutive}\:{numbers}. \\ $$$${if}\:{we}\:{put}\:{any}\:{other}\:{number}\:\:{in}\:{the}\:{set},{we}\:{will}\:{have}\: \\ $$$${three}\:{consecutive}\:{numbers}\:{in}\:{the}\:{collection}.\bigstar \\ $$$$ \\ $$

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