Question Number 103908 by bobhans last updated on 18/Jul/20

Find the least positive integer n for  which there exists a set { s_1 ,s_2 ,s_3 ,...,s_n  }  consisting of n distinct positive integers  such that (1−(1/s_1 ))(1−(1/s_2 ))(1−(1/s_3 ))...(1−(1/s_n ))  = ((51)/(2010)) .

Commented byRasheed.Sindhi last updated on 18/Jul/20

((51)/(2010))=((17)/(670))

Commented bybobhans last updated on 18/Jul/20

for n =?? sir

Answered by mr W last updated on 18/Jul/20

i got:  n_(min) =9  s_1 =67  s_2 =25  s_3 =18  s_4 =11  s_5 =s_6 =s_7 =s_8 =s_9 =2  but they are not distinct!

Answered by john santu last updated on 18/Jul/20

suppose that for n there exist  the desired numbers; we may  assumse that s_1 <s_2 <s_3 <...<s_n .  surely s_1 >1 since otherwise  1−(1/s_1 ) = 0 . so we have 2≤s_1 ≤s_2 −1≤...≤s_n −(n−1),  hence s_i  ≥i +1 for each  i=1,2,...,n. Therefore ((51)/(2010)) =  (1−(1/s_1 ))(1−(1/s_2 ))...(1−(1/s_n )) ≥ (1−(1/2))(1−(1/3))...(1−(1/(n+1)))=(1/2).(2/3).(3/4)...(n/(n+2))  which implies n+1 ≥ ((2010)/(51))=((670)/(17))>39  so n≥ 39. Now we are left to  show that n=39 fits. consider  the set { 2,3,4,...33,34,...,67 }  which contains exactly 39  numbers. we have (1/2).(2/3)...((32)/(33)).((33)/(34))...((39)/(40)).((66)/(67))  = (1/(33)).((34)/(40)).((66)/(67)) = ((17)/(670)) = ((51)/(2010)).  hence for n = 39 there exist a  desired example . (JS ⊛)

Commented bybobhans last updated on 18/Jul/20

thank you

Commented bymr W last updated on 18/Jul/20

very nice!