The mysterious coffer

i interpret the problem as follow:

1. the code can be a single digit 0~9.

2. you cannot input same number twice. as a consequence, you can input at most 10 times. (0~9)

3. every wrong attempt, the code cycles from 0~9 and back to 0.

under worse case scenario, is there a guaranteed way to enter the correct code?

the answer is no. here is the reason.

1. notice that in worse case scenario, we need input at least 10 times. hence, we focus on eliminating the possible code at last (10th) attempt.

2. during 1 st attempt, say we input 3 and its wrong, then last attempt the code is not 3 - 1 = 2 . we know this by tracking incorrect input 3 cycling through 4~9 and 0~2.

3. during 2 nd attempt, say we input 0 and its wrong, then last attempt the code is not 0 - 2 == 8 (mod 10) by same reasoning.

4. in general, during n -th attempt, we input a and its wrong, then last attempt the code is not (a-n) mod 10 .

5. let $a_n$ be the input during n-th attempt, let $b_n$ be the eliminated code on last attempt. so the examples in step 2 and 3 translate into: $a_1 = 3 , \ b_1 = a_1 - 1 = 3 - 1 = 2 \\ a_2 = 0 , \ b_2 = a_2 - 2 = 0 - 2 \equiv 8$

6. during last attempt, we wish to "cover" all cases. the goal is to find $a_n = 0 \sim 9$ and $b_n = 0 \sim 9$ such that $a_n - n \equiv b_n \ (\text{mod } 10)$ for $n = 1 \sim 10$

7. assume this can be done, we sum up $b_n$ $\begin{aligned} \text{on one hand, } \sum b_n &= 0+1+2+3+4+5+6+7+8+9 \\ &= 45 \\ &\equiv 5 \ (\text{mod } 10) \\ \ \\ \text{on the other hand, } \sum b_n &= \sum (a_n - n) \\ &= \sum a_n - \sum n \\ &= 45 - 55 \\ &= -10 \\ &\equiv 0 \ (\text{mod } 10) \\ \therefore \sum b_n \equiv 5 \equiv 0 \ (\text{mod } 10) \end{aligned}$

we arrive a contradiction, hence there is no guarantee way to enter the correct code.

Every time you try you are, in essence, trying to deduce the starting point while either leaving the option to choose the starting point or to select the starting point directly. Each time you try, you eliminate a possible starting state (ST). There is no way of knowing which state it started at until either you win the game by chance, or you have tried 9 times. After you have finished 9 times, do note that the code would be 1 less than the true ST, which you may have tried previously (after all, you had no way to guarantee that the true ST was the true ST).

Say we begin by guessing 1. We are incorrect and the code moves up one. We then guess it was originally 2(and now 3). Getting it wrong, we would guess: Guess 1:1 Guess 2:3 Guess3:5 Guess 4:7 Guess 5:9 Guess 6: At this point we would guess one but instead we would skip this and assume it was 7 Guess 6:2 Guess 7:4 Guess 8:6 If all these are incorrect then we have entered 7 incorrect answers and the original code was 6. So we would be forced to guess 3(correct, but blows up coffer) Luckily, with this method we have an 8/9 chance of getting it but still... Oh and to work out the three in the last answer I calculated(7+6)mod10=13mod10.