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 1 of 6: Total counters fix the number of rounds

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In plain words

Counting the same quantity — all the counters ever handed out — two different ways (round by round, and player by player) turns a messy dealing process into a single algebraic equation.

n(p+q+r)=20+10+9=39

Detailed analysis

Let $n$ be the number of rounds played. In every round the three players together receive exactly $p+q+r$ counters — one card each — so after $n$ rounds the grand total handed out is $n(p+q+r)$ . Adding the three final totals gives $20+10+9=39$ , so $n(p+q+r)=39$ .