MathLabs

Problem 1

Three players AA, BB, and CC play the following game. On each of three cards an integer is written; the three numbers p,q,rp, q, r satisfy 0<p<q<r0 < p < q < r. The three cards are shuffled and one is dealt to each player, and each player receives as many counters as the number on the card they hold. The cards are shuffled and dealt again, and this is repeated for at least two rounds (counters from earlier rounds stay with the players). After the last round, AA has 2020 counters in all, BB has 1010, and CC has 99. In the last round, BB received rr counters. Who received qq counters on the first round?
Step 6 of 6: Reconstructing the first round
A={r,r,q}, B={p,p,r}, C={q,q,p} ⇒ round 1: A=r, B=p, C=qA=\{r,r,q\},\ B=\{p,p,r\},\ C=\{q,q,p\}\ \Rightarrow\ \text{round }1:\ A=r,\ B=p,\ C=q
Detailed analysis

With p=1,q=4,r=8p=1,q=4,r=8, the only way to write 20,10,920,10,9 as sums of three cards drawn from {1,4,8}\{1,4,8\} is 20=8+8+420=8+8+4, 10=1+1+810=1+1+8, 9=4+4+19=4+4+1; so over the three rounds AA held the cards {r,r,q}\{r,r,q\}, BB held {p,p,r}\{p,p,r\}, and CC held {q,q,p}\{q,q,p\}. Since BB's single rr-card is used in the last round (given), BB held pp in rounds 11 and 22. Because CC never holds rr at all, the rr-card in rounds 1,21,2 must go to AA, leaving qq for CC in those rounds. So round 11 was A=r, B=p, C=qA=r,\,B=p,\,C=q: the middle card qq went to CC.