1998 JBMO Problems/Problem 1
Revision as of 21:37, 14 August 2018 by Rockmanex3 (talk | contribs) (Solution to Problem 1 — hidden perfect square trinomial)
Problem
Prove that the number (which has 1997 of 1-s and 1998 of 2-s) is a perfect square.
Solution
The number can be rewritten as This number has a few geometric series and can be written as Simplifying results in Notice that and That means we can "factor" the numerator, and doing so results in Since is divisible by 3, we conclude that the number is a perfect square.
See Also
1998 JBMO (Problems • Resources) | ||
Preceded by First Problem |
Followed by Problem 2 | |
1 • 2 • 3 • 4 | ||
All JBMO Problems and Solutions |