11. A kite pilot has had a good championship this year, finishing on the podium 16 times. In each race, the first-place finisher earns 10 points, the second 8, and the third 5, while no points are awarded from fourth place onwards. With how many different scores can the pilot have concluded the championship?
(A) 153
(B) 80
(C) 78
(D) 75
(E) None of the above.
11. The answer is $(\mathbf{D})$. We need to calculate all possible sums that can be obtained with 16 addends chosen from 5, 8, and 10. Let's write $n=5a+8b+10c$, where $a+b+c$ are non-negative integers whose sum is 16. First, we derive $a=16-b-c$, so $n=80+3b+5c$, where now $b, c$ are non-negative integers with a sum of at most 16. Since we only need to count the number of different possible scores, the addend 80 does not matter, and we simply need to determine how many numbers can be written as $3b+5c$. Moreover, it is now clear that two pairs $(b, c)$ and $\left(b^{\prime}, c^{\prime}\right)$ give the same sum if and only if $b^{\prime}=b+5h, c^{\prime}=c-3h$ for some integer $h$.

Let's start counting the scores for small $c$: with $c=0$ we have 17 choices for $b$ (from 0 to 16), with $c=1$ we have 16, and then 15 with $c=2$, and we have not introduced any repetitions; if $c$ is 3, then we can replace the pair $(b, 3)$ with the pair $(b+5,0)$, unless $b+5>16$, so the sums we have not yet counted are only those with $b>11$ (and on the other hand $b+3 \leq 16$), thus only those corresponding to $(b, c)=(12,3)$ and $(13,3)$. Similarly, for $c \geq 4$, from a pair $(b, c)$ we can reduce to the pair $(b+5, c-3)$, which we have already considered, unless $b+c+2>16$: the only cases in which this happens, given that $b+c \leq 16$, are those in which $b+c=15$ or 16, that is exactly 2 pairs for each value of $c$, unless $c=16$, in which case there is only the possibility $b=0$. In the end, the number of required sums is $17+16+15+13 \cdot 2+1=75$.