3. In a mathematics competition there are four problems, carrying $1,2,3$ and 4 marks respectively. For each question, full score is awarded if the answer is correct; otherwise 0 mark will be given. The total score obtained by a contestant is multiplied by a time bonus of $4,3,2$ or 1 according to the time taken to solve the problems, and a further bonus score of 20 will be added after multiplying by the time bonus if one gets all four problems correct. How many different final scores are possible?
(1 mark)
In a mathematics competition, there are four problems, worth 1, 2, 3, and 4 points respectively. For each question, full points are awarded if the answer is correct; otherwise, 0 points will be given. The total score obtained by a contestant is multiplied by a time bonus of 4, 3, 2, or 1 according to the time taken to solve the problems, and an additional bonus score of 20 will be added after multiplying by the time bonus if one gets all four problems correct. How many different final scores are possible?
3. 27
3. If one gets all four questions wrong, his score is 0; if one gets all four questions correct, his score may be $30,40,50$ or 60. These give 5 possible scores.
Now suppose one gets one, two or three questions correct. Then the score before the time bonus may be any integer from 1 to 9. With the time bonus of 1 to 4, there are 36 possible scores from 1 to 36. Of these, 30 is double-counted, $11,22,33,13,26,17,34,19,23,29,31$ contain a prime factor greater than 10 and have to be discarded, and 25,35 have their smallest prime factor greater than 4 and have to be discarded. The rest are possible scores. Hence there are $36-1-11-2=22$ possible scores in this case, and so the answer is $5+22=27$.
Remark. Another (perhaps easier) way of counting is to list the values of $x y$ ( $0 \leq x \leq 10$, $0 \leq y \leq 4$ ) and then count directly.