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Sum Inverse LCMs (Posted on 2008-10-03) Difficulty: 3 of 5
Each of P, Q, R, S and T are positive integers with P < Q < R < S < T. Determine the maximum value of the following expression.

[P, Q] -1 + [Q, R] -1 + [R, S] -1 + [S, T] -1

Note: [x, y] represents the LCM of x and y.

No Solution Yet Submitted by K Sengupta    
Rating: 3.5000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re: a stab (proof) | Comment 3 of 7 |
(In reply to a stab by Leming)

I can show that (1,2,4,8,16) is maximum.

with x<y for 1/[x,y] to be maximized we want [x,y] to be minimized and obviously [x,y] can't be smaller than y and this is only true if y=x*k for some k

thus we want
Q=P*k1
R=Q*k2=P*k1*k2
S=R*k3=P*k1*k2*k3
T=S*k4=P*k1*k2*k3
for intergers k1,k2,k3,k4>1


then
[P,Q]=Q=P*k1
[Q,R]=R=P*k1*k2
[R,S]=S=P*k1*k2*k3
[S,T]=T=P*k1*k2*k3*k4

thus we want to minimize
1/[P,Q] + 1/[Q,R] + 1/[R,S] + 1/[S,T]
substituting and and extracting the (1/P) common factor we get
(1/P)*(1/k1 + 1/(k1k2) + 1/(k1k2k3) + 1/(k1k2k3k4))
now this is maximized when P is minimal so P=1
it is also maximized when each of k1,k1k2,k1k2k3,k1k2k3k4 are minimized thus
k1=k2=k3=k4=2

putting this all togeather we get
P=1
Q=2
R=4
S=8
T=16 and the maximal value is
(1/2 + 1/4 + 1/8 + 1/16)=15/16=0.9375


  Posted by Daniel on 2008-10-03 14:24:29
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