# Challenge Math Challenge - April 2019

#### fresh_42

Mentor
2018 Award
As (almost) always: have a look on previous challenge threads, too. E.g. in https://www.physicsforums.com/threads/math-challenge-march-2019.967174/ are still problems to solve, and some of them easy, which I find, and in any case useful to know or at least useful to have seen.

As a general note due to a given cause: Please do not edit your posts after they have been answered!
This is in general - not only here - not acceptable, as it makes the threads unreadable and worst case change the entire meaning and thus disrespects the ones who answered! At most an additional commentary marked as "edit (1):" etc. could be added, but nothing within the text, except spell or comma mistakes.

Questions

$1.$ (solved by @SpinFlop ; see also here. ) There are $12$ prisoners who are given the chance to be freed if they pass this test: the warden will put a hat on the head of each prisoner which will be either black or white. The prisoners will get informed that there will be at least one hat of each color. They will be able to see everyone else's hat except their own. Also, there will be no sort of communication among prisoners. The prisoners will be lined up every five minutes starting at $10:00 \space\space a.m.$ and ending at $10:50 \space\space a.m.$ In order to pass the test, all the prisoners with a black hat and only them, will have to step forward during the same line-up. If they succeed, all prisoners will be freed otherwise they will all be executed. Is there a way for the prisoners to pass the test?

$2.$ (solved by @lpetrich ) For which real numbers $\alpha$ does the improper integral $\int_0^1\int_{-\infty}^\infty\frac{1}{(x^2+y^2)^{\alpha}} dx dy$ converge.

$3.$ Let $a\in\mathbb{C}$ and define $s=t^3+3t^2+at\in\mathbb{C}(t)$. For which $a$ is $\mathbb{C}(t)/\mathbb{C}(s)$ a Galois extension?

$4.$ (solved by @wrobel ) Let $f$ be a continuous element of $L^1((0,\infty))$. Suppose that $x=x(t)$ solves the differential equation $\frac{dx}{dt}=-px+f(t)$. Prove that $\lim_{t\to\infty}x(t)=0$.

$5.$ Let $f:S^5\to S^3\times S^2$ be a smooth [edit: and surjective] function. Show that $f$ takes critical values in any set of the form $S^3\times\{p\}$ for $p\in S^2$.

$6.$ (solved by @lpetrich ) Find the area $A$ enclosed by the asteroid $(x,y)=(\cos^3 t,\sin^3 t)$ for $0\leq t \leq 2\pi\,.$

$7.$ (solved by @lpetrich ) Two surface ships on maneuvers are trying to determine a submarine's course and speed to prepare for an aircraft intercept. Ship $A$ is located at $(4, 0, 0)$, whereas ship $B$ is located at $(0, 5, 0)$. All coordinates are given in thousands of feet. Ship $A$ locates the submarine in the direction of the vector $2\mathbf{i} + 3\mathbf{j} - (1/3)\mathbf{k}$, and ship $B$ locates it in the direction of the vector $18\mathbf{i} - 6\mathbf{j} - \mathbf{k}$. Four minutes ago, the submarine was located at $A=(2, -1, -1/3)$. The aircraft is due in $20$ minutes. Assuming that the submarine moves in a straight line at a constant speed, to what position should the surface ships direct the aircraft?

$8.$ Calculate the following:
$a.)$ (solved by @fbs7 ) $\displaystyle{\int} \dfrac{\sqrt{(x^2-1)^3}}{x}\,dx$
$b.)$ (solved by @lpetrich ) The arc length $L$ of $y=-\dfrac{x^2}{8}+\log x$ for $1\leq x \leq 2$

$9.$ (solved by @lpetrich ) Find the similarity transformations to diagonalize the following matrices:
$a.)$ $A=\begin{pmatrix}1&-\sqrt{2}&1\\ \sqrt{2}&0&-\sqrt{2}\\ 1&\sqrt{2}&1\end{pmatrix}$
$b.)$ $B=\begin{pmatrix}\cos \varphi &-\sin \varphi \\ \sin \varphi& \cos \varphi \end{pmatrix}$

$10.$ (solved by @Math_QED ) Suppose that $\mathbb{F}$ is a finite field with say $|\mathbb{F}|= p^m = q$ and that $V$ is a vector space of finite dimension $n$ over $\mathbb{F}\,.$ Find the order of $\operatorname{GL}(V)\,.$

$1$. The king in chess can move to any neighboring square in three different ways: horizontally, vertically or diagonally. Now, we assume that we have an infinite chessboard and the king starts on some square.
$a.)$ (solved by @fbs7 ) In how many different squares can it be after $n$ moves?
$b.)$ (solved by @Master1022 ) Provide an optimal approach i.e. the minimum number of moves required to reach a destination.
$c.)$ (solved by @Master1022 ) Answer $a.)$ if there is a constraint of no diagonal moves.

$2$. The squares of an $8 \times 8$ chessboard are mistakenly colored in blocks of two colors, such that beginning from top left corner of the chessboard and moving in rows of $2 \times 2$ squares, always from left to right, the colors of $2 \times 2$ squares are $W - B - W - B$ for the odd-numbered rows and $B - W - B - W$ for the even numbered rows - we begin counting of rows with number $1$, where $W$ denotes White and $B$ denotes Black. This board needs to be cut in such a way so that a standard chessboard can be reassembled from the pieces obtained. What is the minimum number of pieces that you need to cut the board and in which way should they be reassembled?

$3$. An algorithm works like this: it starts with a single equilateral triangle and on each (subsequent) iteration, it adds new equilateral triangles all around the outside. When the algorithm finishes its $n$-th iteration how many small triangles will be there?

$4.$ (solved by @YoungPhysicist ) Can the numbers $1, 2, 3, \ldots, 16$ be arranged in a row so that each two adjacent numbers add up to a square number?
Example: $2, 7, 9, 16, \ldots$ would be a possibility for the first four numbers $2 + 7 = 9, 7 + 9 = 16, 9 + 16 = 25$; but then we are stuck.

$5.$ (solved by @ShreyJ ) We are looking for a ten-digit number $N$, where the first digit indicates how many zeros occur in $N$, the second digit, how many ones appear in $N$, the third digit, how many doubles occur in $N$, ... and the tenth digit, how many nines appear in $N$.

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Homework Helper
Gold Member
2018 Award
9b is a puzzle. I think it should be simple, but the solution eludes me.

#### fbs7

Why these are not solutions to Yoda level No 1?

(a) On 10:00 am every prisoner looks at the color of the hat to his left (left-most prisoner looks at the hat on the right-most prisoner), and then puts his own hat on the head of the prisoner to the right (right-most prisoner passes the hat to left-most prisoner). Then on 10:01 am every prisoner knows the color of his own hat.

(b) On 10:00 am every prisoner looks at the color of the hat to the left (left-most prisoner looks at the hat on the right-most prisoner), and then steps back if that is white. Then each prisoner just looks to his right and see if the other prisoner stepped back and will know the color of his own hat.

I guess a better question is which actions are allowed for the prisoners?

#### QuantumQuest

Gold Member
Why these are not solutions to Yoda level No 1?

(a) On 10:00 am every prisoner looks at the color of the hat to his left (left-most prisoner looks at the hat on the right-most prisoner), and then puts his own hat on the head of the prisoner to the right (right-most prisoner passes the hat to left-most prisoner). Then on 10:01 am every prisoner knows the color of his own hat.
How a prisoner can put his own hat onto another prisoner? The hats are put on only by the warden and then every prisoner can just see the hats of every other except his own hat. If a prisoner was allowed to put his hat off and put it on to another prisoner then he would immediately see its color.

(b) On 10:00 am every prisoner looks at the color of the hat to the left (left-most prisoner looks at the hat on the right-most prisoner), and then steps back if that is white. Then each prisoner just looks to his right and see if the other prisoner stepped back and will know the color of his own hat.
The allowable movements of the prisoners are also implied in the wording of the puzzle. They can neither step back nor communicate to each other.

I guess a better question is which actions are allowed for the prisoners?
I think that this is clear in the wording of the puzzle.

(The prisoners) will be able to see everyone else's hat except their own. Also, there will be no sort of communication among prisoners. The prisoners will be lined up every five minutes starting at $10:00 \space\space a.m.$ and ending at $10:50 \space\space a.m.$ In order to pass the test, all the prisoners with a black hat and only them, will have to step forward during the same line-up.

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#### fbs7

Hmm.. I think this means the only action the prisoner can do it to step forward or not.

But, how can a prisoner step forward based on what he sees, and that not be a communication to other prisoners? Like, say that a prisoner steps forward if he sees a prime number of white hats.

Then, the other prisoners see that one moving forward, and they know that that guy sees a prime number of white hats... that's implicit communication...

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#### fbs7

Trying my hand at Padawan Level 1(a) (ie High School 1)!

Call $δ_{xn}, δ_{yn}$ two variables, $δ_{xn}, δ_{yn} ∈ \{-n,...,-1,0,1,...,+n\}$.

If the king is initially at position $(x_o,y_o)$, then it can move on move n=1 to positions $(x_1,y_1)= (x_o+δ_{x1},y_o+δ_{y1})$, except to $δ_{x1} = δ_{y1} = 0$ as the king must move from $(x_o,y_o)$. Therefore, on move n=1 in particular the location $(x_o,y_o)$ is forbidden

On move n=2 it can move to $(x_2,y_2)=(x_1+δ_{x1},y_1+δ_{y1})=(x_0+δ_{x2},y_0+δ_{y2})$. That includes being back at $(x_o,y_o)$ for example by moving $(x_o,y_o)=>(x_o+1,y_o)=>(x_o,y_o)$, so on $n=2$ any such location is valid and there are no exceptions.

Induction: if at move $n$ then king can be in any location $(x_n,y_n)=(x_0+δ_{xn},y_0+δ_{yn})$, then at move $n+1$ the king can be at any location $(x_{n+1},y_{n+1})=(x_0+δ_{xn+1},y_0+δ_{yn+1})$. That's because the king can move from $(x_0+δ_{xn},y_0+δ_{yn})$ to $(x_0+δ_{xn}+δ_{x1},y_0+δ_{yn}+δ_{y1}) = (x_0+δ_{xn+1},y_0+δ_{yn+1})$. As the induction is true for n=2 and if true for n then it's true for n+1, then it is true for any n>=2.

As the set $\{-n,...,-1,0,1,...,+n\}$ has $2n+1$ members, then:

At n=0 the king can be at $1=(2n+1)^2$ positions
At n=1 the king can be at $(2n+1)^2-1$ positions
At n>=2 the king can be at $(2n+1)^2$ positions

#### QuantumQuest

Gold Member
Hmm.. I think this means the only action the prisoner can do it to step forward or not.

But, how can a prisoner step forward based on what he sees, and that not be a communication to other prisoners? Like, say that a prisoner steps forward if he sees a prime number of white hats.

Then, the other prisoners see that one moving forward, and they know that that guy sees a prime number of white hats... that's implicit communication...
I recommend reading the wording carefully. Only when in line can prisoners make a move forward, if they think that they have a black hat and this will be based on what they see and so, on what they logically deduce. Communication, in the context of this particular puzzle, means direct communication of any kind e.g. speaking, making any sort of gesture, making some sound / noise and in general what constitutes direct communication. On the other hand, making a decision from seeing is a logical deduction which does not constitute such communication.

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#### Math_QED

Homework Helper
Solution attempt for 10:

$GL(V)$ are invertible operators $V \to V$. The matrix associated with such an operator (fix a basis for this) is an $n \times n$-matrix, and because there is a bijective correspondence between the invertible linear maps and the associated (invertible) matrices, it suffices to count these matrices, which are precisely the ones with linearly independent rows.

For the first row, we can put any vector in $\mathbb{F}^n$ there, except the null vector. So, there are $q^n-1$ possibilities.

For the second row, all vectors will do except for multiples of the first row, so there are $q^n- q$ possibilities.

For the third row, we cannot have linear combinations of the first two rows, so there are $q^n-q^2$ possibilites.

Repeat the process to end with $q^n - q^{n-1}$ possibilities for the last row.

Hence, the total amount of such matrices is $(q^n-1)(q^n-q)(q^n-q^2) \dots (q^n-q^{n-1})$

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#### fbs7

Trying Yoda Level 8(a):

$I = \int \frac{\sqrt{(x^2-1)^3}}{x} dx$

using $v^2 = x^2-1$

$v.dv = x.dx, \frac {v.dv}{x^2} = \frac {v.dv}{1+v^2} = \frac{dx}{x}$

$I = \int \sqrt{(v^2)^3} \frac{dx}{x} = \int (v^3).\frac{v}{1+v^2}.dv$
$I = \int \frac{v^4}{1+v^2}.dv = \int \frac{1+(v^2-1).(v^2+1)}{v^2+1}.dv$
$I = \int \frac{1}{1+v^2}.dv + \int (v^2-1).dv$
$I= arctg(v)+v^3/3-v+C$

translating back the variables...

$v=\sqrt{x^2-1}$
$I = arctg(\sqrt{x^2-1})+\frac {\sqrt{(x^2-1)^3}}{3}-\sqrt{x^2-1}+C$

proof:

$v=\sqrt{x^2-1}$
$\frac{dI}{dx} = \frac{dI}{dv} . \frac{dv}{dx} = \frac{d}{dv} ( arctg(v)+v^3/3-v ).\frac{d}{dx} (\sqrt{x^2-1})$
$= ( \frac{1}{1+v^2} + v^2 - 1 ).(1/2).(x^2-1)^{-1/2}.2x$
$= \frac{v^4}{1+v^2} .(x^2-1)^{-1/2}.x$
$= \frac{(x^2-1)^2.(x^2-1)^{-1/2}}{x}$
$= \frac{\sqrt{(x^2-1)^3}}{x}$

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#### lpetrich

I will attempt to solve problem 2:
It is to find values of $\alpha$ where integral
$$I(\alpha) = \int_0^1 \int_{-\infty}^{\infty} \frac{1}{(x^2+y^2)^\alpha} dx dy$$
converges.

I will first consider convergence for the y integral, and then for the x integral.

For the y integral, I need the possible values of x, and those are between 0 and 1 inclusive. This means that x is nonzero at all but one point, so I will treat it as nonzero in this part. This means that the integrand is finite for $y \ll x$, so we turn our attention to $y \gg x$. In that limit, the y integral becomes
$$\int \left(\frac{1}{y^{2\alpha}} + O\left(\frac{1}{y^{2(\alpha+1)}}\right)\right) dy$$
That integral will converge whenever $2\alpha > 1$ or $\alpha > (1/2)$.

The integral can be evaluated with the help of a change of variables: $y = x u$ where u ranges between $-\infty$ and $\infty$. This gives us
$$I(\alpha) = \int_0^1 \int_{-\infty}^{\infty} \frac{1}{(x^2+x^2 u^2)^\alpha} x\, dx du = \int_0^1 \frac{dx}{x^{2\alpha-1}} \int_{-\infty}^{\infty} \frac{du}{(1 + u^2)^\alpha}$$
The integral for x will converge whenever $2\alpha - 1 < 1$, or $\alpha < 1$, and doing that integral gives us
$$I(\alpha) = \frac{1}{2(1-\alpha)} \int_{-\infty}^{\infty} \frac{du}{(1 + u^2)^\alpha}$$
Collecting the two bounds for $\alpha$ gives us $(1/2) < \alpha < 1$.

#### lpetrich

I will now do Problem 6:
It is to find the area inside a unit-sized astroid, a curve given by $(x,y) = (cos^3 t, sin^3 t)$ for $0 \le t \le 2\pi$.

It is possible to find this area by finding the bounds of y as functions of x, integrating over y, and then integrating over x. But these bounds are given by $y = \pm (1 - |x|^{2/3})^{3/2}$, and that does not look very easy to work with, and finding such expressions will sometimes be impractical or impossible in closed form.

So I seek a formula for the area that uses the curve's rectangular-coordinate statement while integrating over the curve's parameter. This can be taken as the continuous limit of the area of a polygon with vertices given in rectangular coordinates. There is a surprisingly simple formula for that area, the shoelace formula:
$$A = \frac12 \sum_i ( x_i y_{i+1} - x_{i+1} y_i ) = \sum_i (x_i (y_{i+1} - y_i) - y_i (x_{i+1} - x_i))$$
In the continuous limit, it becomes, for parameter $t$:
$$A = \frac12 \oint \left( x \frac{dy}{dt} - y \frac{dx}{dt} \right) dt$$
Substituting in the expressions for the problem:
$$A = \frac12 \int_0^{2\pi} \left( (\cos^3 t) (3 \sin^2 t \cos t) - (\sin^3 t) (- 3 \cos^2 t \sin t) \right) dt \\ = \frac12 \int_0^{2\pi} (3 \cos^4 t \sin^2 t + 3 \sin^4 t \cos^2 t) dt \\ = \frac32 \int_0^{2\pi} \sin^2 t \cos^2 t \, dt \\ = \frac{3\pi}{8}$$
So the area inside a unit-sized astroid is $3\pi/8$.

#### lpetrich

I will solve Problem 8b.
It is to find the length L given by the curve $y = - \frac{x^2}{8} + \log x$ for $1 \le x \le 2$.

For this purpose, I need the formula for arc length for a curve given in rectangular coordinates (x,y) as a function of some parameter t. It is
$$L = \oint \sqrt{ \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 } \, dt$$
For this problem, the parameter t is x, and that simplifies the formula to
$$L = \int \sqrt{ \left( \frac{dy}{dx} \right)^2 + 1} \, dx$$
Finding $dy/dx = - \frac{x}{4} + \frac{1}{x}$ for the problem condition, and plugging in both it and the bounds of integration,
$$L = \int_1^2 \sqrt{ \left(- \frac{x}{4} + \frac{1}{x}\right)^2 + 1 } \, dx$$
The expression inside the square root can be simplified:
$$\left(- \frac{x}{4} + \frac{1}{x}\right)^2 + 1 = \frac{x^2}{16} - 1/2 + \frac{1}{x^2} + 1 = \frac{x^2}{16} + 1/2 + \frac{1}{x^2} + 1 = \left(\frac{x}{4} + \frac{1}{x}\right)^2$$
It is easy to take the square root of the resulting expression, and it is always positive over the domain of integration, so we have
$$L = \int_1^2 \left(\frac{x}{4} + \frac{1}{x}\right) dx = \left. \frac{x^2}{8} + \log x \right|_1^2 = \frac38 + \log 2$$
Thus, $L = \frac38 + \log 2$.

#### fbs7

a.)a.)a.) (solved by @fbs7 ) ∫√(x2−1)3xdx∫(x2−1)3xdx\displaystyle{\int} \dfrac{\sqrt{(x^2-1)^3}}{x}\,dx
Ahhh--hahahaha!! I solved one at Top-Master-Jedi-Level! Hooray!!!

#### Infrared

Gold Member
I will attempt to solve problem 2:
The integral can be evaluated with the help of a change of variables: $y = x u$ where u ranges between $-\infty$ and $\infty$. This gives us
$$I(\alpha) = \int_0^1 \int_{-\infty}^{\infty} \frac{1}{(x^2+x^2 u^2)^\alpha} x\, dx du = \int_0^1 \frac{dx}{x^{2\alpha-1}} \int_{-\infty}^{\infty} \frac{du}{(1 + u^2)^\alpha}$$
I think this bit is enough (you don't need the earlier work) since in the last product, the first integral converges iff $\alpha<1$ and the second does if $\alpha>1/2.$

You also seem to have confused which bound went with which variable of integration but as the integrand is symmetric in $x,y$, this is okay!

Nice work.

#### lpetrich

I will solve Problem 7.
We want the position of the submarine at the time of the aircraft's arrival, at time +20, using minutes as the time unit. We have observations of the submarine's directions relative to two ships at time 0, and the submarine's position at time -4.

So we must first find the position of the submarine at time 0, then extrapolate from its positions at times 0 and -4 to time +20.

For the first part, we have the ships' positions and the directions of the submarine relative to each ship. The positions and directions form lines, and where they intersect is where the submarine is. For distances d1 and d2 along observation lines l1 and l2, we get for the positions along those lines
$$l_1 = (4,0,0) + (2,3,-1/3) d_1 ,\ l_2 = (0,5,0) + (18,-6,-1) d_2$$
For intersection, we set $l_1 = l_2$, and that gives us
$$4 + 2d_1 = 18d_2 ,\ 3d_1 = 5 - 6d_2 ,\ -1/3 d_1 = -d_2$$
The third equation gives us $d_1 = 3d_2$, and inserting it into the above equation gives us
$$4 +6d_2 = 18d_2 ,\ 9d_2 = 5 - 6d_2$$
or $4 = 12d_2$ and $5 = 15d_2$, giving $d_2 = 1/3$, $d_1 = 1$, and the intersection point as $(6,3,-1/3)$.

I now turn to the extrapolation part. For points $\mathbf{x_0}$ and $\mathbf{x_1}$ at times $t_0$ and $t_1$, $\mathbf{x}(t)$ is, for time t, is
$$\mathbf{x}(t) = \mathbf{x_0} + (\mathbf{x_1} - \mathbf{x_0}) \frac{t - t_0}{t_1 - t_0}$$
and likewise with indices 0 <-> 1. Plugging in the values for our problem, $t_0 = 0$, $\mathbf{x_0} = (6,3,-1/3)$, $t_1 = -4$, $\mathbf{x_1} = (2,-1,-1/3)$, and $t = 20$. That gives position $\mathbf{x}$ as
$$\mathbf{x} = (6,3,-1/3) + ((2,-1,-1/3) - (6,3,-1/3)) \frac{20 - 0}{-4 - 0} = (6,3,-1/3) - 5 (-4,-4,0) = (26,23,-1/3)$$
So the aircraft must be dispatched to point (26,23) and the submarine will be at depth (-1/3).

#### YoungPhysicist

High school 4:
You stuck because 2,7,9,16 is the end of the sequence, rather than beginning. Good misdirection though.

Brute force success on first go:

8,1,15,10,6,3,13,12,4,5,11,14,2,7,9,16

Their sum follows the pattern of 9-16-25-16.

#### SpinFlop

Questions

$1.$ There are $12$ prisoners who are given the chance to be freed if they pass this test: the warden will put a hat on the head of each prisoner which will be either black or white. The prisoners will get informed that there will be at least one hat of each color. They will be able to see everyone else's hat except their own. Also, there will be no sort of communication among prisoners. The prisoners will be lined up every five minutes starting at $10:00 \space\space a.m.$ and ending at $10:50 \space\space a.m.$ In order to pass the test, all the prisoners with a black hat and only them, will have to step forward during the same line-up. If they succeed, all prisoners will be freed otherwise they will all be executed. Is there a way for the prisoners to pass the test?

Solution to problem 1

Since there is no communication, we Just need to consider the perspective and thought process of a single prisoner with a black hat. Call him BHP.

Suppose only 1 hat is black. BHP will see all white hats and be able to deduce that his hat must be black.
This prisoner steps forward immediately, ie: 10:00

Suppose 2 prisoners wear black hats. BHP sees only one black hat in the crowd. If BHP's hat is white, then the person with the black hat on will step forward at 10:00. If he does not step forward at 10:00, then that must mean BHP's hat was black. By symmetry they both can now deduce there were two black hats and so they step forward at 10:05.

Suppose 3 prisoners wear black hats. BHP can use the same reasoning as above. At 10:05 if the other two black hats don't step forward, then he knows his was black as well. By symmetry all 3 can deduce there were 3 black hats total and step forward at 10:10.

Suppose N prisoners wear black hats. At 10:(N-2)*5 if the N-1 black hats that BHP sees don't step forward, then he knows his was black as well. By symmetry all N black hats can deduce there were N black hats total and step forward at 10:(N-1)*5

So the rule is as follows:
If a prisoner sees N black hats, then he should step forward at 10:(N)*5 because his hat is black. Thus, if 11 hats are black (the max number possible), BHP would see 10 hats and step forward at 10:50. Just in time!

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#### QuantumQuest

Gold Member
Suppose 2 prisoners wear black hats. BHP sees only one black hat in the crowd. If his was white, then the person with the black hat on will step forward at 10:05.
If I understand it correctly you mean that initially BHP, as you call it, sees only one black hat. If his hat was white then wouldn't this mean that there would be just one person with a black hat? Then, I see a discrepancy between the time you write and your above analysis for one black hat.

#### SpinFlop

Ahhhh, there were some number typos in my original post. I have fixed them. Let me know if that clears it up.

#### QuantumQuest

Gold Member
Of course, if all 12 were black, then none would step forward.
This cannot be the case as it violates the data given in the wording of the problem:

The prisoners will get informed that there will be at least one hat of each color.

"Math Challenge - April 2019"

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