, 2003

Solution:

The answer is (D). Indeed, suppose for the sake of contradiction that 114 is an admissible score. Let us denote by $N$ the number of rolls with which it was obtained; $N$ can be 10 or 11. The first score made is less than or equal to six, so at least $114-6=108$ was made with $N-1$ rolls; note that $108=12 \times 9$, so if $N=10$ were the case, the only possible sequence of scores would be given by a 6 followed by nine consecutive 12's, but then the last roll would have given 6 and there would also have been an eleventh throw.

Hence $N=11$. It follows that at least 108 was obtained with 10 rolls (those from the second to the eleventh). Note that the eleventh roll gives a score less than or equal to six, so at least 102 must have been obtained with 9 rolls. Suppose that the scores of these rolls are not all 12. Then there is a score less than or equal to 10 (11 cannot be obtained). Suppose this occurred at the tenth roll, which consequently must be a 6. This is impossible because at least $96=8 \times 12$ would have been obtained with 8 rolls (from the second to the ninth), which should all give 12, but then the tenth would give 12 and not 6. So the score less than or equal to 10 occurs before the tenth roll, and at the following roll the score is less than or equal to 6; subtracting these two scores, it follows that at least $102-16=86$ was obtained with only seven rolls, and this is impossible.

In conclusion the only admissible sequence is the one given by an initial 6, nine consecutive 12's plus the score $p$ of the eleventh throw, and the total score is $114+p$ which cannot coincide with 114 because $p$ is between $1$ and $6$. We have obtained a contradiction.