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).
Problem
Source: IMO ShortList 1974, USA 1, IMO 1974, Day 1, Problem 1
Tags: combinatorics, game, algebra, system of equations, IMO, IMO 1974
Peter
30.10.2005 18:59
sum = 39 points, so we have a sum of either 13 or 3. for 3 it is impossible to find 3 positive integers that sum to 3, so the sum is 13 and they played 3 games. The minimum number is at least 1, the maximum number is at most 10. (let's call highest number win, lowest loss, middle draw) Assume A lost once, then C can have lost at most twice, and we cannot find such numbers. So A has 2 wins and a draw. B and C both have a loss and we still have to split {win,loss,draw,draw} over B,C. Since B=C+1 and B won once we have: A = wwd B = llw C = ldd B won the last game, so lost the others and C drawed all the others, so C also drawed the first one.
hypogean
30.12.2014 06:08
*spends 2 hours on problem with friend *realizes you can add numbers So if A wrote down the value a, B wrote down the value b, and C wrote down the value c, and they played n rounds, then n(a+b+c) = 39 = 3*13 Since n is at least 2, n is either 39,13, or 3. If n is 39, then a+b+c = 1 which is impossible since a,b,c > 1. If n is 13 then a+b+c = 3. Since a,b,c > 0 this assumes a=b=c=1 which contradicts them being distinct. This means that the only possibility is n=3 and a+b+c = 13. WLOG let a>b>c. We know that a is at least 7 (because if a<7 then 3*a < 20). So we split this into cases Case 1: a=7 Then we know that B had to get 7 the last round. This means that the first 2 rounds B must have gotten 3 points in total. Since b and c are integers and 3 is not even, this means that b+c = 3. Since b>c and b,c>1 we see that b=2, c=1 But that means the greatest value A could have gotten is 7+7+2 = 16 < 20. This is bad Case 2: a=8 That means B got 8 the last round. That means in the first 2 rounds B got 2 points, which is only possible if the least value is 1. Thus in the first,second, and third rounds B got 1,1,and 8 points respectively. There are 2 possibilities from here, either A got 8 in 2 of the rounds or 8 in one of the rounds. If A got 8 in one of the rounds then b=6. But this is impossible as this assumes C got both 6 and 8 at least once, where 6+8>9 This means that the only possibility is A got 8 in 2 of the rounds, meaning b=4. That means that A got 8,8,4 in the first,second, and third rounds respectively. Also this means that C got 4,4,1 in the first, second, and third rounds respectively. Thus, in this case C got the middle number first Case 3: a=9 Then B got 9 once and got 1 in the first 2 rounds. This is obviously impossible. And obviously if a becomes larger this just gets more impossible. Thus the only possibility is the one listed above. Thus C drew middle the first round
quangminhltv99
30.08.2016 11:06
The sum of points is $39$. Hence there are $3$ shufflings and the sum of numbers on cards is $13$. Since $A$ has the highest point, hence the highets number, call $W$, must be $7$. On the ;ast shuffling, $B$ got $W$, hence $W$ is $8$, and the lowest, call $L$, have the number $2$ (if it is grater then $W$ max is $13-3-4=6$, a contradict. So we get the following possibilities ($W$, $D$ for middle number and $L$ respectively): \[ (8,4,1); (8,3,2); (7,4,2); (7,5,1) \]Since we get $20$ after $3$ shufflings, the triple must be $(8,4,1)$. Now, it is easy to see that person is C.
A-Thought-Of-God
24.11.2020 13:19
We claim that $C$ is the desired person. Notice that during each game, the sum $S$ of numbers is invariant. Let $x,y,z$ be the numbers written on three cards. Moreover, WLOG assume $x<y<z~(\spadesuit)$. Let there by $\ell$ games played. Let $a,b,c$ be the numbers on the cards picked by $B$ in three games respectively. Then we have $$S\ell=20+10+9=39=3\cdot 13$$which gives $\ell=3$ and $S=13,$ since otherwise we would have $1,1,1$ on three different cards, a contradiction. Clearly, $x+y+z=S=13$ and $z\ge 5$ using $(\spadesuit)$. We will proceed by dividing the value of $z$ in cases. If $z=c=5$ then only choice to have is $x=y=4$ but by $(\spadesuit),$ a contradiction. If $z=c=6$ then $x+y=7$ and $a+b=4$. Hence $(a,b)\equiv(1,3),(2,2)$. Checking $(a,b)\equiv(1,3)$ since $1,3,6$ are distinct hence we should have $1+3=7,$ a contradiction. Therefore, $(a,b)\equiv(2,2)$ which gives $(x,y,z)\equiv(2,5,6)$. But then attributing numbers to $A$ we have maximum sum of numbers for $A$ is $6+6+5<20,$ contradiction. If $z=c=7$ then $x+y=6$ and $a+b=3$ which gives $(a,b)\equiv(1,2)$. Getting aforementioned reasoning into the play, we get $1+2+7=13,$ a contradiction. If $z=c=8$ then $x+y=5$ and $a+b=2$ which gives $(a,b)\equiv(1,1)$. Therefore, $(x,y,z)\equiv(1,4,8)$. Consequently, we get following as our configuration of game played. $$\left[\begin{array}{ccc} 8&1&4 \\ 8&1&4 \\ 4&8&1 \\ \end{array}\right]$$It's easy to see that this indeed works and that $C$ gets the second number in first game. $\blacksquare$