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BdMO Online Forum • View topic - IMO 1974/1

IMO 1974/1

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IMO 1974/1

Post Number:#1  Unread postby seemanta001 » Mon Jan 04, 2016 11:23 pm

Alice, Betty, and Carol took a series of examinations. There were one grade of $A$, one grade of $B$, and one grade of $C$ for each examination, where $A,B$ and $C$ are different positive integers. The final test scores were:

Alice = $20$
Betty = $10$
Carol = $9$

If Betty placed first in the arithmetic examination, who placed second in the spelling examination?
Last edited by Phlembac Adib Hasan on Sun Jan 10, 2016 5:10 pm, edited 1 time in total.
Reason: Fixed grammatical errors and unintended line-breaks
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Re: IMO 1974/1

Post Number:#2  Unread postby Phlembac Adib Hasan » Sun Jan 10, 2016 5:27 pm

seemanta001 wrote:Alice, Betty, and Carol took the same series of examinations. There were one grade of A, one grade of B, and one grade of C for each examination, where A;B and C are different positive integers. The final test scores were
Alice Betty Carol
$20$ $10$ $9$
If Betty placed first in the arithmetic examination, who placed second in
the spelling examination?

This version is quite ambiguous and not the official statement of IMO 1974-1. Let me add that here:
Three players $A,B$ and $C$ play a game with three cards and on each of these $3$ cards it is written a positive integer, all $3$ numbers are different. A game consists of shuffling the cards, giving each player a card and each player is attributed a number of points equal to the number written on the card and then they give the cards back. After a number $(\geq 2)$ of games we find out that A has $20$ points, $B$ has $10$ points and $C$ has $9$ points. We also know that in the last game B had the card with the biggest number. Who had in the first game the card with the second value (this means the middle card concerning its value).

@Kids from junior category, try it before reading further. ;)
Hint:
Add the numbers. Also note that, in each game (= exam), the total number of points (=grades) is fixed.
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Re: IMO 1974/1

Post Number:#3  Unread postby tanmoy » Wed Jan 13, 2016 10:10 pm

Phlembac Adib Hasan wrote:Three players $A,B$ and $C$ play a game with three cards and on each of these $3$ cards it is written a positive integer, all $3$ numbers are different. A game consists of shuffling the cards, giving each player a card and each player is attributed a number of points equal to the number written on the card and then they give the cards back. After a number $(\geq 2)$ of games we find out that A has $20$ points, $B$ has $10$ points and $C$ has $9$ points. We also know that in the last game B had the card with the biggest number. Who had in the first game the card with the second value (this means the middle card concerning its value).


A nice problem. :) Here is my solution:
Let $g$ be the number of games and $X,Y,Z$ be the numbers written on the three cards.So,$g(X+Y+Z)=20+10+9=39$.Now,given that $X,Y,Z$ are distinct positive integer,so,$X+Y+Z$ is at least $6$.$\therefore$ $g=3,X+Y+Z=13$.
WLOG,assume that $X> Y> Z$.If $X\leq 6$,then the total number of $A$ would be $\leq 17$,a contradiction.So,$X\geq 7$.As $B$ got $X$ points in the last game and got $10$ points in total,so,$X\leq 8 \Rightarrow Y\geq 4$.Therefore,$B$ got $Z$ points in both the other two games.$C$ got $9$ points and at least one $Z$.As $B$ got $Z$ points in two games,$C$ can not get $Z$ points in more than one game which implies that $C$ got one $Z$.So,he got either $X$ and $Y$ points in the other two games or $Y$ points in both the other two games.But in the first case,his total number would be $X+Y+Z=13$,a contradiction.So,he got $Y$ points in two games (on which $B$ scored $Z$).$\therefore$ $C$ had in the first game the card with the second value. :D
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