Mods, if you want to post this, at the Stcky Topics discussion, you are welcome to!
Let's see if anyone here, can solve this, fun, challenging question...
Three players \(A,B\) and \(C\) play the following game: On each of three cards an integer is written. These three numbers \(p,q,r\) satisfy \(0 < p < q < r\). The three cards are shuffled and one is dealt to each player. Each then receives the number of counters indicated by the card he holds. Then the cards are shuffled again; the counters remain with the players.
This process (shuffling, dealing, giving out counters) takes place for at least two rounds. After the last round, \(A\) has 20 counters in all, \(B\) has 10 and \(C\) has 9. At the last round \(B\) received \(r\) counters. Who received \(q\) counters on the first round?
This is from the 1974 IMO...Please do not search this problem up, and copy the solution here...
If you guys like this, I can post a problem from IMO every day, and you guys can solve it.
After the last round, A has 20 counters, B has 10, and C has 9 . So...
(p + q + r) * the num of rounds = 20 + 10 + 9 = 39
p + q + r = 39 / the num of rounds
Since p + q + r must be an integer, the num of rounds must be 3, 13, or 39 .
If the num of rounds = 13 , p + q + r = 3 , but no 3 different positive integers add to 3 .
If the num of rounds = 39 , p + q + r = 1 , but no 3 different positive integers add to 1 .
So the num of rounds must be 3 , and p + q + r = 13
By trial and error, I think I found 3 different integers that work: 1, 4, and 8
sum of counters earned in 3 rounds for player A = 8 + 8 + 4 = 20
sum of counters earned in 3 rounds for player B = 1 + 1 + 8 = 10
sum of counters earned in 3 rounds for player C = 4 + 4 + 1 = 9
q = the middle number = 4
On the first round, player C earned 4 counters.
I just got lucky by guessing those numbers for p, q, and r.....Maybe there is a way to do it without guessing!