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

Home > Just Math
Two Geometric Series (Posted on 2003-07-14) Difficulty: 2 of 5
Find a geometric series of 3 or more positive integers, starting with 1, such that its sum is a perfect square.

See if you can find another such series.

See The Solution Submitted by Brian Smith    
Rating: 3.6667 (6 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re(2): Programming Solution | Comment 9 of 10 |
(In reply to re: Programming Solution by exoticorn)

You're right; I wrote the program quickly, expecting to find relatively small, workable numbers, then failed to consider the precision involved since all the numbers are integers.

I rewrote the program as follows to show the entire expansion and its sum, as well as the square of the located root for comparison. I used rlim==1 and nlim=30, so as to only show the first two 'accepted' solutions and the lowest 'big' solution found previously:

function display(r, n, root) {
  var sum=0;
  for (a=0; a<=n-1; a++) {
    term=Math.pow(r,a);
    document.write(term);
    if (a<n-1)
      document.write(" + ");
    sum+=term;
  }
  var square=root*root;
  document.write(" = " + sum + "<br>" + root + "&sup2; = "
  document.write(square + "<br><br>");
}

var rlim=11
var nlim=30
for (r=2; r<=rlim; r++) {
  for (n=3; n<=nlim; n++) {
    test=Math.sqrt((Math.pow(r,n)-1)/(r-1));
    if (test)==(Math.round(test)) {
      display(r,n,test);
    }
  }
}

This yielded:

1 + 3 + 9 + 27 + 81 = 121
11 = 121

1 + 7 + 49 + 343 = 400
20 = 400

1 + 11 + 121 + 1331 + 14641 + 161051 + 1771561 + 19487171 + 214358881 + 2357947691 + 25937424601 + 285311670611 + 3138428376721 + 34522712143931 + 379749833583241 + 4177248169415651 + 45949729863572160 + 505447028499293760 + 5559917313492231000 + 61159090448414550000 + 672749994932560100000 + 7.400249944258161e+21 + 8.140274938683976e+22 + 8.954302432552373e+23 + 9.84973267580761e+24 + 1.0834705943388372e+26 + 1.191817653772721e+27 + 1.310999419149993e+28 + 1.4420993610649922e+29 + 1.5863092971714916e+30 = 1.7449402268886408e+30
1320961856712237 = 1.7449402268886406e+30

So indeed, the default precision of 16 places in javascript should account for the insane number of solutions the first program found, as well as the steady decline in the number of terms.

I don't know how to prove that those two soltions are unique, but it is certainly shown that they are the only sets of 'workable' numbers that work.

I really shouldn't tackle problems like this at 2am...
  Posted by DJ on 2003-07-15 05:41:05

Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (1)
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 - 2020 by Animus Pactum Consulting. All rights reserved. Privacy Information