MCQ highest

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Thanic Nur Samin
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MCQ highest

Unread post by Thanic Nur Samin » Mon Dec 09, 2013 11:46 am

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 $a+b$?

Nayeemul Islam Swad
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Joined:Sat Dec 14, 2013 3:28 pm

Re: MCQ highest

Unread post by Nayeemul Islam Swad » Tue Jan 21, 2014 1:00 pm

Yeeeeees!!!!!!!!! Got it!!! Highest possible average is $121/111$. So, $a + b = 121 + 111 = 232$ :D

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Labib
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Location:Dhaka, Bangladesh.

Re: MCQ highest

Unread post by Labib » Wed Jan 22, 2014 8:33 pm

It's a problem from Brilliant.org. The problem name is Mathematics Exam (Login required to see the problem/solution).
I'm posting a solution posted by a brilliant member named Patrick Corn.
Patrick Corn wrote:
Each question can be answered correctly on at most $11$ papers. This is because any two students who answer it correctly cannot have the same answers for any other given question, and that other question can have at most $11$ different answers. So the maximum total number of correct answers on all the papers is $11.11$, which would give an average score of $\frac{11.11}{111}=\frac{121}{111}$.

To show that this maximum average is attained, we exhibit a collection of papers that satisfies the condition. Label each student with an ordered pair $(a,b)$ in ${F_{11}}^{2}$, where ${F_{11}}$ is the field with $11$ elements (the integers mod $11$). Label the questions $0,1,....,10$ and the answers $0,1,....,10$. Now we mandate that student $(a,b)$ answers question $x$ with the answer $ax+b\in F_{11}$. Note that if $a_1x+b_1=a_2x+b_2$, then $x$ is determined uniquely as $x = \frac{(b_2-b_1)}{a_2-a_1}$, and this expression makes sense unless $a_1 = a_2$, but in that case $b_1=b_2$ and they are the same student. So the point is that each distinct pair of students have at most one answer in common.

Let's further stipulate that the $111$ students are made up of the $110$ ordered pairs $(a,b)$ with $a \neq 0$, plus the ordered pair $(0,0)$. And let's also stipulate that the correct answer to every question is $0$. Then the first students all have exactly one correct answer: student $(a,b)$ gets question $\frac{-b}a$ right. And student $(0,0)$ gets a perfect score. That adds up to $110+11=121$ correct answers, which we already saw is the maximum.
So the answer is $121+111=\boxed{232}$.
$\Rightarrow$ PS1 - If you saw this on brilliant, this is kind of cheating. Please do not abuse this opportunity. :mrgreen:
$\Rightarrow$ PS2 -
Nayeemul Islam Swad wrote:Yeeeeees!!!!!!!!! Got it!!! Highest possible average is $121/111$. So, $a + b = 121 + 111 = 232$ :D
@Nayeemul Islam Swad, please post your full solution. People actually post, because they do not know how to approach or solve the full problem.
If they just needed an answer, they would just use a computer. So kindly practice posting full solutions.
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Tahmid Hasan
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Re: MCQ highest

Unread post by Tahmid Hasan » Wed Jan 22, 2014 9:16 pm

Labib wrote: I'm posting a solution posted by a brilliant member named Patrick Corn.
Lord pco :lol:
বড় ভালবাসি তোমায়,মা

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