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Ten girls are sitting around a table. Each of them picks a real number and whispers it to the two neighbors immediately to the left and to the right. (Hence: each girl communicates one number, and receives two numbers.) Each girl then loudly announces the average of the two numbers she received. The announced numbers in order around the circle are:
1, 2, 3, 4, 5, 6, 7, 8, 9, 10
Question: What was the number picked by the girl who announced the average number 6?
1
Consider 10 real numbers $a$ through $j$ and have each girl's whispered number be $2a, 2b, 2c$ etc (these do not have to be even numbers; $a$ through $j$ are not necessarily integers.)
It follows that
$$\dfrac{2a + 2c}2 = 1 \quad\text{or}\quad a + c = 1\\[10pt]\dfrac{2b + 2d}2 = 2\quad\text{or}\quad b + d = 2$$
etc
The girl who says 6 is sitting at g. Rearranging the
$$e + g = 5$$
equation into
$$g = 5-e$$
and then substituting around the system of equations, we get:
$$\begin{align}g &= 5-e \\&= 5 - (3-c) \\&= 2 + c \\&= 2 + (1-a) \\&= 3 - a \\&= 3 - (9-i) \\&= -6 + i \\&= -6 + (7-g) \\&= 1 - g\end{align}$$
therefore
$$2g = 1$$
which is what she says to her neighbours.
EDIT: After the fact, realize this is the same proof as Kate's. I will leave it here because I find it might be a bit clearer, but hers was first and also complete, so upvote that one.
~~~
Lets the number that the girls say be $S_i$ and the number the girls think be $T_i$. We then know $S_i=i$ and:
$$S_1=\frac{T_{10}+T_2}{2}=1 \implies T_{10}+T_2=2$$
We can repeat this for the rest and get:
$$S_2=T_1+T_3=4, S_3=T_2+T_4=6, S_4=T_3+T_5=8$$ $$S_5=T_4+T_6=10,S_6=T_5+T_7=12, S_7=T_6+T_8=14$$ $$S_8=T_7+T_9=16, S_9=T_8+T_{10}=18,S_{10}=T_9+T_1=20$$
We have 10 unknowns and 10 equations. In fact, we have 2 sets of 5 equations with 5 unknowns since the even subscripts are never in an equation with odd subscripts.
Lets solve the even subscript set first since $S_6=6$ and we are trying to solve $T_6$. $$T_{10}+T_2=2 \implies T_2=2-T_{10}$$ $$T_2+T_4=6 \implies 2-T_{10}+T_4=6 \implies T_4=4+T_{10}$$ $$T_4+T_6=10 \implies 4+T_{10}+T_6=10 \implies T_6=6-T_{10}$$ $$T_6+T_8=14 \implies 6-T_{10}+T_8=14 \implies T_8=8+T_{10}$$ $$T_8+T_{10}=18 \implies 8+T_{10}+T_{10}=18 \implies T_{10}=5$$
Thus, $T_6=6-T_{10}=6-5=1$ and $S_6=6$, so the girl who said 6 thought 1.
Less rigorous and mathematical, but perhaps more intuitive approach:
(a) note that for each girl,
there is no relationship at all between the number whispered and the number said; they are parts of different cycles.
(b)
in the group of girls we care about, the thought-of numbers averaged to $[1, 3, 5, 7, 9]$.
(c)
Note that this is symmetrical, so the number that was a component of both 1 and 9 must be 5. From there, the rest follows:
$5 + a = 9\times2 \therefore a = 13$
$13 + b = 7\times2 \therefore b = 1$
(d) to check the work, go the other way around the circle:
$5 + d = 1\times2 \therefore d = -3$
$-3 + c = 3\times2 \therefore c = 9$
$9 + b = 5\times2 \therefore b = 1.$
(The other 5 girls each thought of the number 1 higher than the girl sitting to one side of her).
The answer is
1
I have used simple matrix multiplication while solving the problem;
Let say who said 1 is A, 2 is B, 3 is C, 4 is B ... J said 10;
$\frac{J+B}{2}=1 \ for \ A$
an so on;
\begin{bmatrix} & & 1& 2& & & \\ & & A& B& & & \\ 10& J& & & C& 3& \\ 9& I& & & D& 4& \\ 8& H& & & E& 5& \\ & & G& F& & & \\ & & 7& 6& & & \end{bmatrix}
So If we put every equation as a matrix form;
$\begin{bmatrix} 1& 0& 1& 0& 0& 0& 0& 0& 0& 0\\ 0& 1& 0& 1& 0& 0& 0& 0& 0& 0\\ 0& 0& 1& 0& 1& 0& 0& 0& 0& 0\\ 0& 0& 0& 1& 0& 1& 0& 0& 0& 0\\ 0& 0& 0& 0& 1& 0& 1& 0& 0& 0\\ 0& 0& 0& 0& 0& 1& 0& 1& 0& 0\\ 0& 0& 0& 0& 0& 0& 1& 0& 1& 0\\ 0& 0& 0& 0& 0& 0& 0& 1& 0& 1\\ 1& 0& 0& 0& 0& 0& 0& 0& 1& 0\\ 0& 1& 0& 0& 0& 0& 0& 0& 0& 1 \end{bmatrix}\cdot \begin{bmatrix} J\\ A\\ B\\ C\\ D\\ E\\ F\\ G\\ H\\ I \end{bmatrix}=\begin{bmatrix} 2\\ 4\\ 6\\ 8\\ 10\\ 12\\ 14\\ 16\\ 18\\ 20 \end{bmatrix}$
Then we need to take inverse of the first matrix and multiply with the right hand side;
$\begin{bmatrix} J (10)\\ A (1)\\ B (2)\\ C (3)\\ D (4)\\ E (5)\\ F (6)\\ G (7)\\ H (8)\\ I (9) \end{bmatrix}=\begin{bmatrix} 5\\ 6\\ -3\\ -2\\ 9\\ 10\\ 1\\ 2\\ 13\\ 14 \end{bmatrix}$
The one who announced 6 was F then the answer is 1. You may find the other numbers as above.
Let's write down the sums:
$a_1 +a_3 = 2$
$a_2 +a_4 = 4$
...
$a_9 +a_1 = 18$
$a_{10} +a_2 = 20$
Then we can obtain two arithmetic series where the terms increase by 4:
$a_3,a_7,a_1,a_5,a_9$
$a_4,a_8,a_2,a_6,a_{10}$
The one who said 6 must have picked $a_7$, so we can find it using this and $a_7+a_9=14$ (or $a_7+a_5=10$), which means $a_7=1$.
Let Girl<#> be the girl who announced #. For example, Girl6 is the girl who announced 6.
Similarly, let Value<#> be the number chosen by, and whispered by, Girl<#>. For example, Girl6 chose Value6, and whispered Value6 to Girl5 and Girl7.
Let x = Value6.
As shown below, x = Value6 = 1. The girl who announced 6 chose 1.
Value4 = 10 - x. This allows (Value4 + Value6)/2 = ((10 - x) + x)/2 = 10/2 = 5.
Value2 = x - 4. This allows (Value2 + Value4)/2 = ((x - 4) + (10 - x))/2 = 6/2 = 3.
Value8 = 14 - x. This allows (Value6 + Value8)/2 = (x + (14 - x))/2 = 14/2 = 7.
Value10 = x + 4. This allows (Value8 + Value10)/2 = ((14 - x) + (x + 4))/2 = 18/2 = 9.
(Value2 + Value10)/2 = 1. Thus, ((x - 4) + (x + 4))/2 = 1, or 2x/2 = 1, or x = 1.
x = Value6 = 1. The girl who announced 6 chose 1.
Check by substitution.
Girl.. Choice Announced
Girl1... __...... 1
Girl2.... -3...... 2
Girl3... __...... 3
Girl4..... 9...... 4
Girl5... __...... 5
Girl6.... 1...... 6
Girl7... __...... 7
Girl8... 13...... 8
Girl9... __...... 9
Girl10... 5.... 10
Let $g_n$ be the number the $n$th girl picked, so that $(g_{n-1}+g_{n+1})/2=n$ for $1\le n\le 10$ (where indices wrap around the circle, so $g_0=g_{10})$. Then $$ \begin{align} g_6 &= (g_6-g_4)/2 + (g_6+g_4)/2\\ &= (g_6+(g_8-g_8)+(-g_{10}+g_{10})+(g_2-g_2)-g_4)/2 + (g_6+g_4)/2\\ &= (g_6+g_8)/2 -(g_8+g_{10})/2+(g_{10}+g_2)/2-(g_2+g_4)/2+(g_6+g_4)/2\\ &= 7-9+1-3+5=\boxed{1} \end{align} $$
"2" whispers v to .
` .
. "3" = (v+w)/2
.' . . `.
"4" whispers w to '. . . :
`. :
. "5" = (w+X)/2 :
.' . . :
"6" whispers X to '. . . + --> 3+9 = (v+w+y+z)/2 = 12
`. : .
. "7" = (X+y)/2 : .
.' . . : .
"8" whispers y to '. . . : .
`. .' .
. "9" = (y+z)/2 .
.' . . .
"10" whispers z to '. . . .
`. .
. "1" = (z+v)/2 .
.' . . .
"2" whispers v to ' . . .
. . .
3+5+7+9+1 = v+w+X+y+z = 25 .
: .
v+w+y+z = 24
:
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\:$
:
$\qquad\qquad\qquad\qquad\quad~~\!\;$"6" picked X = 25 - 24 = 1
We are given that the number $b_i$ announced by girl $i$ is the average of the numbers $x_{i^+}$ and $x_{i^-}$ chosen by the neighboring girls $i^+$ and $i^-$:
$$2 b_i = x_{i^+} + x_{i^-},$$
where
$$\begin{array}{rcl} i^+ & = & i + 1 - \left\lfloor \frac{i}{N} \right\rfloor N, \\ i^- & = & i - 1 + \left\lfloor \frac{N + 1 - i}{N} \right\rfloor N \end{array}$$
and $N$ is the number of girls (that is, $i^+$ and $i^-$ loop around the interval $[1, N]$). This forms a linear equation system, which can be expressed on matrix form as $A\mathbf{x} = \mathbf{b}$ where the number at row $i$, column $j$ of the matrix $A$ is
$$a_{i,j} = \begin{cases} \frac{1}{2}, \ i-j \equiv_N 1,\\ \frac{1}{2}, \ j-i \equiv_N 1,\\ 0 \ \ \mathrm{otherwise.} \end{cases}$$
For $N = 5$, $2A$ looks like:
$$\begin{array}{ccccc} &0 &1 &0 &0 &1 \\ &1 &0 &1 &0 &0 \\ &0 &1 &0 &1 &0 \\ &0 &0 &1 &0 &1 \\ &1 &0 &0 &1 &0 \end{array}$$
Solving this equation system for $\mathbf{x}$ with $N = 10$ and $\mathbf{b} = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]$ gives
$x_6 = 1$.
6. For her to announce an average of 6, the two numbers she receives must add up to 12. This happens to be the average of the two averages announced to either side of her. This pattern holds around the circle, which makes sense if the announced average is also the number picked.
1
Let $g_n$ be the number picked by the girl who said $n$. The average of $g_6$ and $g_4$ is two more than the average of $g_2$ and $g_4$, so $g_6$ must be four more than $g_2$. Symmetrically $g_6$ four less than $g_{10}$. $g_6$ is therefore the average of $g_2$ and $g_{10}$, but we already have already been told what this is.
I know there are plenty of correct answers, but here is a super-simple one.
Let's note $g_n$ the n'th girl's secret number and $a_n$ the average she gave aloud.
$g_2 + g_4 + g_6 + g_8 + g_{10} = a_1 + a_3 + a_5 + a_7 + a_9$
because each girl contributes in $a_n$ half or her two neighbor's values.
On the other side, rewriting some averages you have
$g_2 + g_4 = 2 a_3$ and $g_8 + g_{10} = 2 a_9$
If you subtract these from the 1st equation you get
$g_6 = a_1 - a_3 + a_5 + a_7 - a_9 = 1 - 3 + 5 + 7 - 9 = 1$
