A4. Twenty students go abseiling during a school trip. In each round, one student gets a turn to abseil, so after twenty rounds, everyone has safely descended. To determine who goes first in round 1, cards with numbers 1 to 20 are distributed to the students. The one who gets the 1 starts. In round 2, cards with numbers 1 to 19 are distributed to the remaining students; the student who gets the 1 is next. This continues for each round, until in round 20, only one student is left, who automatically gets a card with 1. By a bizarre coincidence, no student gets a card with the same number twice. Sara gets the card with 11 in the first round.

What result does Sara get if she adds up all the numbers on her cards?
B4. 66 For convenience, number the students in the order in which they take their turns. Student 20, who abseils last, has seen the numbers 1 through 20 once. Logically, he received the card with number 1 in the last round. In round 19, he must have received the card with number 2 (since there were only two cards). In round 18, he could only have received the card with number 3, and so on until round 1, where he received the card with number 20.

Now, let's look at student 19. In his last round, the 19th round, he received the card with number 1. In round 18, he must have received the card with number 2 (since the card with number 3 was already taken by student 20). In round 17, he must have received the card with number 3 (student 20 took the card with number 4, and numbers 1 and 2 were no longer available). Thus, we see that student 19 must have picked up the cards with numbers 1, 2, 3, ..., 19 in rounds 19, 18, 17, ..., 1, respectively.

If we continue this reasoning, we find that student number $n$ in rounds $n, n-1, n-2, \ldots, 1$ received the cards with numbers $1, 2, 3, \ldots, n$, in that order. Since Sara received the card with number 11 in round 1, she must be student number 11. If she adds up the numbers on her cards, she gets $11+10+9+\cdots+2+1=66$.