All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Just Math
Further Harmonic Integers (Posted on 2006-10-27) Difficulty: 2 of 5
(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)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution 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   wend
OK

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      20
OK

  Posted by Charlie on 2006-10-27 11:42:05
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (7)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information