A group of n friends wrote a math contest consisting of eight short-answer problem $S_1, S_2, S_3, S_4, S_5, S_6, S_7, S_8$, and four full-solution problems $F_1, F_2, F_3, F_4$. Each person in the group correctly solved exactly 11 of the 12 problems. We create an 8 x 4 table. Inside the square located in the $i$th row and $j$th column, we write down the number of people who correctly solved both problem $S_i$ and $F_j$. If the 32 entries in the table sum to 256, what is the value of n?
Problem
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Tags: calculus, integration
10000th User
04.11.2012 18:09
Table diagram: \[\begin{array}{r|c|c|c|c|} \phantom{}&F_1&F_2&F_3&F_4\\ \hline S_1&\phantom{}&\phantom{}&\phantom{}&\phantom{}\\ \hline S_2&\phantom{}&\phantom{}&\phantom{}&\phantom{}\\ \hline S_3&\phantom{}&\phantom{}&\phantom{}&\phantom{}\\ \hline S_4&\phantom{}&\phantom{}&\phantom{}&\phantom{}\\ \hline S_5&\phantom{}&\phantom{}&\phantom{}&\phantom{}\\ \hline S_6&\phantom{}&\phantom{}&\phantom{}&\phantom{}\\ \hline S_7&\phantom{}&\phantom{}&\phantom{}&\phantom{}\\ \hline S_8&\phantom{}&\phantom{}&\phantom{}&\phantom{}\\ \hline \end{array}\]
Let $a$ be the number of people who missed a $S_i$ and $b$ the number of people who missed a $F_j$. Set up two equations: $a+b=n$, $(32-4)a+(32-8)b=256$. 2nd eqn leads to $7a+6b=64$. $a$ is even so let $a=2c$ and get $7c+3b=32$. This has only one positive integral solution by inspection: $(c,b)=(2,6)$. Thus, $n=a+b=4+6=10$.