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!