Ten students are seated around a (circular) table. Each student selects his or her own secret number and tells the person on his or her right side the number and the person his or her left side the number (without disclosing it to anyone else). Each student, upon hearing two numbers, then calculates the average and announces it aloud. In order, going around the table, the announced averages are 1,2,3,4,5,6,7,8,9 and 10. What was the secret number chosen by the person who announced a 6?
[Math] Secret Number Problem
problem solving
Related Solutions
English clock idioms: old clocks strike or chime the hour (sometimes also half-hour and quarter hour); a tick is the sound of the escapement mechanism, typically once per second.
Let's abstract away the clock: Chiming events occur every 30 minutes. $2015 = 67(30) + 5$. That $+5$ minutes means it doesn't matter whether the 2015 minutes is an open or closed interval (if it were $2010=67(30)$ minutes, we'd have to decide whether chimes at the start and end of the interval were in the interval).
How many chiming events can we fit in $2015$ minutes? In a $5$ minute interval, we can fit $0$ events or $1$ event, depending on when the interval starts. Each additional $30$ minutes is another event, so $2015$ minutes can give us $67$ or $68$ events. Since we're trying to maximise the total number of chimes, choose $68$ events. That gives $34$ on-the-hour events and $34$ on-the-half-hour events. The number of half-hour events is fixed, allowing us to put aside the half-hour events (subtotal 34 chimes) and focus on the hourly chimes.
What's the largest total of 34 consecutive terms from the repeating sequence $1,2,\cdots,10,11,12,1,2,3,\cdots$? Every 12 consecutive terms sums to $\dfrac{12(12+1)}{2} = 78$ regardless of where we start. $34 = 2(12) + 10$, so the problem reduces to finding the largest total of 10 consecutive terms in $1,2,\cdots,11,12,1,2,\cdots$. This will be $3+4+\cdots+11+12=75$.
So the maximum number of chimes is $34 \text{ (half-hour events) } + 78 + 78 + 73 \text{ (hour events) } = 265$ chimes. It could occur in, say, the interval starting 2:55 ending 12:40, or the interval starting 2:25 ending 12:10.
Once you have fixed $M$ and $S$, the task can succeed if and only if $N$ is at most $(M+1)^{S-1}$. In particular there is no $M$ and $S$ that will work for all $N$.
To see that the condition $N\le (M+1)^{S-1}$ is necessary, observe that after step $k$ nobody can possibly have a list with more than $(M+1)^k$ numbers on it. (This can be proved easily by induction on $k$). However, if any copying at all happens in the last step, the employes being copied from in that last step must have reached a full list in the next-to-last step at the latest. So those employees will end with at most $(M+1)^{S-1}$ numbers.
On the other hand, to prove that $N\le(M+1)^{S-1}$ is sufficient, we need to show a concrete strategy that works in that case. Here's one:
- To begin with, everyone stands up.
- Do the following step $S-1$ times:
- All employees who are still standing congregate in groups of size $M+1$.
- Each group elects a foreperson.
- Each foreperson copies the lists of the other $M$ persons in his/her group.
- Everyone who was not a foreperson sits down.
- (This invariant holds after $k$ steps: There are $(M+1)^{S-1-k}$ people still standing, and every room number appears on one of their lists).
- After $S-1$ steps there is a single person left standing; they have a complete list. Everyone else copies it off them.
This works as written if there are exactly $(M+1)^{S-1}$ employees. If there are fewer, we need to allow for one of the groups in each step to be smaller than $M+1$ employees. How to arrange that in a hall that is so large that nobody even knows in advance how many people there are in it, I will leave unsaid -- in other words, I assume the the original author of the problem did not intend for that to be a sticking point ...
Best Answer
Let us denote secret numbers as $x_i$ , where $i$ is announced number ,then we have following system of equations :
$\begin{cases} x_1+x_3=4 \\ x_2+x_4=6 \\ x_3+x_5=8 \\ x_4+x_6=10 \\ x_5+x_7=12 \\ x_6+x_8=14 \\ x_7+x_9=16 \\ x_8+x_{10}=18 \\ x_9+x_1=20 \\ x_{10}+x_2=2 \end{cases}$
According to Maple : $x_6=1$ , so requested secret number is $1$ .