Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1978 IMO Problems/Problem 6"

(Solution (In Progress))
Line 10: Line 10:
  
 
for all <math>1\leq k\leq 6</math>, then, by the Pigeonhole Principle, there is at least one member whose number is the sum of the numbers of two other members from his own country. It is easy to verify by hand that <math>(*)</math> holds for all <math>k</math> within the desired range, and hence, our proof is complete. <math>\blacksquare</math>
 
for all <math>1\leq k\leq 6</math>, then, by the Pigeonhole Principle, there is at least one member whose number is the sum of the numbers of two other members from his own country. It is easy to verify by hand that <math>(*)</math> holds for all <math>k</math> within the desired range, and hence, our proof is complete. <math>\blacksquare</math>
 +
 +
== See Also == {{IMO box|year=1978|num-b=5|after=Last Question}}

Revision as of 16:06, 29 January 2021

$\emph{Problem:}$ An international society has its members from six different countries. The list of members has 1978 names, numbered $\color[rgb]{0.35,0.35,0.35}1, 2, \ldots, 1978$. Prove that there is at least one member whose number is the sum of the numbers of two (not necessarily distinct) members from his own country.

$\emph{Proof:}$ Lets consider the members numbered $1,2,3,\ldots 989$. If these members belong to $1\leq k\leq 6$ countries then, by the Pigeonhole principle, there must exist a country to which a minimum of $\left \lfloor \frac{988}{k}\right \rfloor +1$ members belong. Let the numbers assigned to these members be $a_1,a_2,\ldots, a_{\left \lfloor \frac{988}{k}\right \rfloor +1}$, where we assume WLOG that

\[1\leq a_1<a_2<a_3<\cdots<a_{\left \lfloor \frac{988}{k}\right \rfloor +1}\leq 989.\]

Assuming that these numbers satisfy the conditions of the problem, then there exists no $(x,y,z)$ triple (not necessarily distinct), such that $a_x+a_y = a_z$. Since every pair of the above $a_i$ correspond to a unique sum less than or equal to 1978, then there are minimum of $\dbinom{\left \lfloor \frac{988}{k}\right \rfloor +1}{2}$ numbers that must be assigned to members of the other countries. Note that this doesn't account for the case when $x=y$. Consequently, if we can show that

\[\dbinom{\left \lfloor \frac{988}{k}\right \rfloor +1}{2}>1978-\left (\left \lfloor \frac{988}{k}\right \rfloor +1\right ) (*)\]

for all $1\leq k\leq 6$, then, by the Pigeonhole Principle, there is at least one member whose number is the sum of the numbers of two other members from his own country. It is easy to verify by hand that $(*)$ holds for all $k$ within the desired range, and hence, our proof is complete. $\blacksquare$

See Also

1978 IMO (Problems) • Resources
Preceded by
Problem 5 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Last Question
All IMO Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1978_IMO_Problems/Problem_6&oldid=143972"