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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?

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    $\begingroup$ by the way, why girls? :) $\endgroup$
    – Oray
    Commented Feb 20, 2016 at 10:12
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    $\begingroup$ @Oray because, click bait. ;) $\endgroup$
    – NVZ
    Commented Feb 20, 2016 at 12:34
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    $\begingroup$ @NVZ 10 girls, one table, one real conversation? $\endgroup$
    – Yakk
    Commented Feb 20, 2016 at 17:51

13 Answers 13

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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.

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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.

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    $\begingroup$ You've a typo in the last cluster of equations... T2+T4=6 ==> should be 2-T10+T4=6. (Too minor a change for a suggested edit.) $\endgroup$ Commented Feb 18, 2016 at 20:35
  • $\begingroup$ This was essentially the method I used. Glad to see I was doing it right, even if a bit more verbose than necessary. $\endgroup$
    – corsiKa
    Commented Feb 18, 2016 at 20:57
  • $\begingroup$ Seeing as you wanted $T_6$, I'm a little surprised that you solved for $T_{10}$ first. $\endgroup$
    – Neil
    Commented Feb 19, 2016 at 15:20
  • $\begingroup$ @Neil Meh... I just went around the circle. Seemed more natural to start at the beginning, but yes, that would have saved a step. $\endgroup$
    – Trenin
    Commented Feb 19, 2016 at 15:25
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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).

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    $\begingroup$ "Note that this is symmetrical": I don't find your argument clear here. $\endgroup$
    – cfh
    Commented Feb 19, 2016 at 12:43
  • $\begingroup$ @cfh The average of the five numbers is 5, you can see that by adding them all together twice and pairing them to get each of the averages in (1, 3, 5, 7, 9) counted once. The average of the five numbers plus the x that participates in both the 1 and the 9 is also 5: you can add all of those together and pair them to get each of the averages in (1, 5, 9) counted once. If adding x to the list of numbers has no effect on the average, then x must equal the average. $\endgroup$
    – hvd
    Commented Feb 20, 2016 at 18:13
  • $\begingroup$ What I meant by symmetrical was that they are symmetrically distributed about the average; one number is the average, another is two higher, another is two lower, another is four higher, another is four lower -- and they are arranged such that the two-above and two-below are equidistant -- physically, around the circle -- from the average. The unknown numbers to the left and right of the average response must be equidifferent, and since the next reponses are equidifferent (as each other, from the average), the next unknowns must also be equidifferent, etc. $\endgroup$
    – Vynce
    Commented Feb 24, 2016 at 20:07
  • $\begingroup$ @cfh ... which means that the number between 1 and nine (a) must be equidifferent from 1 and 9, ergo 5 and (b) must be exactly as much above 5 as it is below 5, ergo 5. As I said, intuitive, but I'm afraid not mathematically rigorously explained. $\endgroup$
    – Vynce
    Commented Feb 24, 2016 at 20:15
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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.

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    $\begingroup$ This method is overkill for this question. $\endgroup$
    – f''
    Commented Feb 19, 2016 at 21:51
  • $\begingroup$ @f'' :) i agree $\endgroup$
    – Oray
    Commented Feb 19, 2016 at 22:36
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    $\begingroup$ I didnt like the qustion, but I like this answer. thumb up $\endgroup$
    – Sadegh
    Commented Feb 20, 2016 at 20:55
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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 $a_7=1$.

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    $\begingroup$ I was with you until right after "next term increases by 4". Could you clarify what happens after that point? $\endgroup$ Commented Feb 18, 2016 at 17:54
  • $\begingroup$ $a_7$ can be inferred from $a_7+12=a_9$ and $a_7+a_9=14$ or $a_7+8=a_5$ and $a_5+a_7=10$. $\endgroup$
    – Nautilus
    Commented Feb 18, 2016 at 19:21
  • $\begingroup$ This answer is very straightforward and makes sense. Nice job. $\endgroup$
    – Timtech
    Commented Feb 18, 2016 at 22:06
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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

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  • $\begingroup$ @Neil -- Explanations of math problems should be pronounceable. Variable names can be short phrases, not just one letter with a subscript. The set-up of a problem should clearly describe all of the relevant variables. $\endgroup$
    – Jasper
    Commented Feb 19, 2016 at 14:55
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    $\begingroup$ @f'' -- Thank you for rejecting edits that deviate from the original intent of the post. $\endgroup$
    – Jasper
    Commented Feb 19, 2016 at 14:56
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    $\begingroup$ @Daedric -- Please reject edits that make posts worse. $\endgroup$
    – Jasper
    Commented Feb 19, 2016 at 14:57
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    $\begingroup$ @Gamow -- Please reject edits that make posts worse. $\endgroup$
    – Jasper
    Commented Feb 19, 2016 at 14:57
  • $\begingroup$ Apart from Neil, these pings don't notify anyone. Also, I thought this post was improved with Neil's edit. $\endgroup$
    – grg
    Commented Feb 20, 2016 at 14:35
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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 $$ \begin{align} g_6 &= (g_6-g_4)/2 + (g_6+g_4)/2\\ &= (g_6+(g_8-g_8)+(-g_{10}+g_{10})+(g_2-g_2)-g_4)/2 + (g_6+g_4)/2\\ &= (g_6+g_8)/2 -(g_8+g_{10})/2+(g_{10}+g_2)/2-(g_2+g_4)/2+(g_6+g_4)/2\\ &= 7-9+1-3+5=\boxed{1} \end{align} $$

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  "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

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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

$x_6 = 1$.

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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.

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  • $\begingroup$ Very elegantly thought and stated. Best answer. $\endgroup$
    – Vynce
    Commented Aug 30, 2017 at 2:20
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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.

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    $\begingroup$ Wait, I forgot to take into account that this is a circle. :( will edit answer once I reconsider $\endgroup$
    – Irishpanda
    Commented Feb 18, 2016 at 13:08
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    $\begingroup$ haha, I had this same thought process initially $\endgroup$ Commented Feb 18, 2016 at 16:03
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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
$g_6 = a_1 - a_3 + a_5 + a_7 - a_9 = 1 - 3 + 5 + 7 - 9 = 1$

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Going for an even simpler answer with the minimum amount of work needed.

Number the girls such that the $n^{th}$ girl announced $n$.

Assume the 2nd girl chose $x$. Then:

  1. The 4th girl chose $6-x$ (as $\frac{x+6-x}{2}=3$).
  2. The 6th girl chose $4+x$ (as $\frac{6-x+4+x}{2}=5$).
  3. The 8th girl chose $10-x$.
  4. The 10th girl chose $8+x$.

The average of the numbers chosen by the 2nd and 10th girls must be $1$, so:

$\frac{x+8+x}{2}=1\dashrightarrow x=-3$

Therefore the 6th girl must have chosen

$4-3=1$.

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