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

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<math>emph{Problem:}</math> An international society has its members from six different countries. The list of members has 1978 names, numbered <math>\color[rgb]{0.35,0.35,0.35}1, 2, \ldots, 1978</math>. 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.
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<math>\emph{Problem:}</math> An international society has its members from six different countries. The list of members has 1978 names, numbered <math>\color[rgb]{0.35,0.35,0.35}1, 2, \ldots, 1978</math>. 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.
  
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<math>\emph{Proof:}</math> Lets consider the members numbered <math>1,2,3,\ldots 989</math>. If these members belong to <math>1\leq k\leq 6</math> countries then, by the Pigeonhole principle, there must exist a country to which a minimum of <math>\lfloor \frac{988}{k}\rfloor +1</math>
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<math>\emph{Proof:}</math> Lets consider the members numbered <math>1,2,3,\ldots 989</math>. If these members belong to <math>1\leq k\leq 6</math> countries then, by the Pigeonhole principle, there must exist a country to which a minimum of <math>\left \lfloor \frac{988}{k}\right \rfloor +1</math> members belong. Let the numbers assigned to these members be <math>a_1,a_2,\ldots, a_{\left \lfloor \frac{988}{k}\right \rfloor +1}</math>, where we assume WLOG that
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<cmath>1\leq a_1<a_2<a_3<\cdots<a_{\left \lfloor \frac{988}{k}\right \rfloor +1}\leq 989.</cmath>
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Assuming that these numbers satisfy the conditions of the problem, then there exists no <math>(x,y,z)</math> triple (not necessarily distinct), such that <math>a_x+a_y = a_z</math>. Since every pair of the above <math>a_i</math> correspond to a unique sum less than or equal to 1978, then there are minimum of <math>\dbinom{\left \lfloor \frac{988}{k}\right \rfloor +1}{2}</math> numbers that must be assigned to members of the other countries. Note that this doesn't account for the case when <math>x=y</math>. Consequently, if we can show that
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<cmath>\dbinom{\left \lfloor \frac{988}{k}\right \rfloor +1}{2}>1978-\left (\left \lfloor \frac{988}{k}\right \rfloor +1\right ) (*)</cmath>
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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>

Revision as of 20:53, 5 August 2017

$\emph{Problem:}$ 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. $\blacksquare$

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