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

Home > Just Math
Looking for a square result (Posted on 2018-06-14) Difficulty: 3 of 5
For each positive integer n, let Mn be the square matrix (nxn) where each diagonal entry is 2018, and every other entry is 1.

Determine the smallest positive integer n (if any) for which the value
of det(Mn) is a perfect square.

No Solution Yet Submitted by Ady TZIDON    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
re(2): derivation of the formula Comment 18 of 18 |
(In reply to re: derivation of the formula by Daniel)

so i use your analysis to check my result that M7 works for 2019. 

(Like a politician, if you can't answer the question, make up your own question.)

Was my answer n=7 correct or was it just roundoff error? 

det Mn = 2018^(n-1)*(2018+n)

if n-1 is even then 2018+n is square

(I immediately know my answer n=7 works as 2018+7 = 45^2,

but, maybe there is a smaller even n)

anyway - finishing the n-1 even case: 

thus 2018+n=k^2 or n=k^2-2018

for n-1 to be even, n is odd which then means k is odd

so for odd k we have n=k^2-2018 as a solution

the smallest integer value for n this gives is when k=45, giving n=7


if n-1 is odd then 2018*(2018+n) is square

if 2018(2018+n) is square then 2018 must divide 2018+n

thus if n is even, n is a multiple of 2018 and thus is greater than 7.

So, yes M7 minimizes n for 2019 being a square.

(I feel better now being Wolfram-Alpha-free)

Edited on June 16, 2018, 11:11 pm
  Posted by Steven Lord on 2018-06-16 23:10:17

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

Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (3)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Copyright © 2002 - 2019 by Animus Pactum Consulting. All rights reserved. Privacy Information