Problem 1

Three players

A

B

, and

C

play the following game. On each of three cards an integer is written; the three numbers

p, q, r

satisfy

0 < 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,

A

has

20

counters in all,

B

has

10

, and

C

has

9

. In the last round,

B

received

r

counters. Who received

q

counters on the first round?

Step 5 of 6: Pinning down

p=1

q=4

r=8

Previous step Next step

r=8,\quad 10-8=2=2p \implies p=1,\quad q=13-1-8=4

Detailed analysis

So $r=8$ , and $B$ 's remaining $10-8=2$ counters over two rounds must equal one of $2p,\,p+q,\,2q$ ; since $2\le2p<p+q<2q$ , only $2p$ can equal the smallest possible value $2$ , so $p=1$ . Then $q=13-p-r=13-1-8=4$ .