A national math contest consisted of 11 multiple choice questions, each having 11 possible choices, of which only 1 of the choices is correct. Suppose that 111 students actually wrote the exam, and no two students have more than one answer in common. The highest possible average mark for the students can be expressed as a/b where a,b are coprime positive integers. What is the value of a+b?
To find the highest possible average mark, we need to maximize the number of correct answers.
Since there are 11 questions and 11 possible choices for each question, we can assign each student a unique answer key by using the student's number as the answer choice for the corresponding question. For example, Student 1 gets Answer 1 for Question 1, Student 2 gets Answer 2 for Question 1, and so on.
This way, no two students have the same answer for any question, ensuring that all the answers are unique.
Since there are 111 students, we can assign answers for the first 11 students for each question. For the subsequent students, their answers will be a repeat of the first 11 students (in cyclic order), so that every student has a unique combination of answers.
By doing this, we can guarantee that each of the 11 answer choices is chosen the same number of times (11 times), which maximizes the number of correct answers.
Thus, the total number of correct answers is 11 x 11 = 121.
Since each question carries the same weight, the highest possible average mark is (121/111).
The value of a+b is therefore 121 + 111 = 232.