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 Further Harmonic Integers (Posted on 2006-10-27)
(A) Three distinct positive integers belong to a harmonic sequence. One of these integers equals the product of the other two, and their sum is 1983. Determine the three integers.

(B) Determine three distinct positive integers, belonging to a harmonic sequence, whose product is 3600.

 See The Solution Submitted by K Sengupta Rating: 4.0000 (1 votes)

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 computer solution--spoiler Comment 1 of 1
`list   10   Lim=sqrt(1983)   20   A=1   22   while A<Lim   24   B=A+1   25   while B*A+A+B<=1983   38   C1=(1//B-(1//A-1//B))   39   if C1>0 then   40     :C=1//(1//B-(1//A-1//B))   50     :if C=int(C) then   60      :if C=A*B then   70       :if A+B+C=1983 then print   80   B=B+1   90   wend   95   A=A+1  100   wend  110   Lim=3600^(1/3)  120   A=1  122   while A<Lim  124   B=A+1  125   while B*A<=Lim*Lim  138   C1=(1//B-(1//A-1//B))  139   if C1>0 then  140     :C=1//(1//B-(1//A-1//B))  150     :if C=int(C) then  160      :if A*B*C=3600 then  170       :print A,B,C  180   B=B+1  190   wend  195   A=A+1  200   wendOK`

The double slashes keep the fractions as rational numbers, rather than dividing to a decimal (actually binary) approximation, in this language, UBASIC.

The solutions are:

`run 31      61      1891 12      15      20OK`

 Posted by Charlie on 2006-10-27 11:42:05
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