Suppose there is a test with $n\geq1$ questions. For each question, there are no descriptions for either the questions or the choices for the answer, just $k\geq1$ choices to choose from, in which exactly one of them is correct. (This condition is to make sure that the answers can only be guessed.) For each submission of the test, any number of questions can be answered, and the score for that submission would be announced, but not which answers are correct or wrong. What are the optimal strategy and the maximum number of submissions in the worst case needed to get a full mark for any $n$ and $k$?
A confusing test with $n$ questions with $k$ choices that can only be guessed
combinatorics
Best Answer
For $k=3$, I think this gives $(m+3)×3^{m-1}$ questions using $2×3^m$ practice exams. By tripling the number of practice exams, I can triple the number of questions, and sneak in one more question for each block of six practice tests. It follows the idea in @MikeEarnest reference.
Let the three answers be $a,b,c$.
If $T$ is a practice test, let $S(T)$ be the test with all a's replaced by b's, all b's replaced by c's and all c's replaced by a's. Let $P(T)$ be test with a's replaced by c's, b's by a's and c's by b's.
Let w(T) be the number of correct answers in T, called the weight of $T$. Let $|T|$ be the number of questions in T. Then $$w(T)+w(S(T))+w(P(T))=|T|$$ So, if we know two of the weights of $T, S(T)$ and $P(T)$, we know the third.
Let A,B,C be three sets of $M$ questions each. We want the weights of 2N tests for A, namely $A_i, S(A_i),i=1..N$, and similar for B and C.
Let $D_i$ be a single question. There will be a total of $N$ $D_i$.
Do the following six tests on $3M+1$ questions:
$$W_{1,1}=w(A_i+B_i+C_i+D_i)\\ W_{1,2}=w(S(A_i)+S(B_i)+S(C_i)+S(D_i))\\ W_{2,1}=w(A_i+S(B_i)+P(C_i))\\ W_{2,2}=w(S(A_i)+P(B_i)+C_i))\\ W_{3,1}=w(A_i+P(B_i)+S(C_i))\\ W_{3,2}=w(S(A_i)+B_i+P(C_i))$$ Now $$W_{1,1}+W_{2,1}+W_{3,1}=3w(A_i)+|B|+|C|+w(D_i)\\ W_{1,2}+W_{2,2}+W_{3,2}=3w(S(A_i))+|B|+|C|+w(S(D_i))$$ so from $\pmod3$ we know the answer to $D_i$ and can remove it from $W_{1,1}$ and $W_{1,2}$. Then we know $w(A_i)$ and $w(S(A_i))$.
The weights for $B_i$ and $C_i$ can be found similarly because we know $w(B_i)+w(S(B_i))+w(P(B_i))=|B|$.
So, from $M$ questions solved in $2N$ tests, we get $3M+N$ questions solved in $6N$ tests. Starting with one question solved in two tests, that gives $(m+3)3^{m-1}$ questions solved in $2×3^m$ tests.
This works for prime $k$.
If $k\ge5$ then we can put in a second extra question, add $E_i$ to $W_{1,1}$ and $W_{2,1}$, and $S(E_i)$ to $W_{1,2}$ and $W_{2,2}$, $SS(E_i)$ to $W_{1,3}$ and $W_{2,3}$ and so on.
Working $\pmod k$, we get $$ \text{$0$ if $D_i$ and $E_i$ are wrong,}\\ \text{ $1$ if $D_i$ is right and $E_i$ wrong,}\\ \text{ $2$ if $E_i$ is right and $D_i$ is wrong, and}\\ \text{ $3$ if both $D_i$ and $E_i$ are right.}$$
In general, this gives $(\lfloor\log_2 k\rfloor m+k)k^{m-1}$ questions from $(k-1)k^m$ tests.