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 2 of 6: Only

3

rounds are possible

Previous step Next step

p+q+r\ge 1+2+3=6,\quad n\ge2 \implies n=3,\ p+q+r=13

Detailed analysis

Since $p,q,r$ are distinct positive integers with $p<q<r$ , we have $p\ge1,q\ge2,r\ge3$ , so $p+q+r\ge6$ and $n=39/(p+q+r)\le6$ . Also $n$ must divide $39=3\times13$ , and the game lasts at least two rounds, so $n=1$ is excluded, while the bound $n\le6$ excludes $n=13$ and $n=39$ . Hence $n=3$ and $p+q+r=13$ .

Common mistake. It is tempting to test

n=13

n=39

directly; the bound

p+q+r\ge6

(from distinctness) rules both out immediately, leaving only

n=3