Challenge Riddles and Puzzles: Extend the following to a valid equation

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77. Which ##n## times ##m## puzzles exist, that have as many boundary parts as interior parts?

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I will try #77, interpreting as "jigsaw puzzles".
Imagine a rectangle formed out of same-size squares, m on one side and n on the other. The problem states that the number of edge squares equals the number of interior squares. Since the total number is the number of both edge and interior squares, this means that the total number is twice the interior number.

In effect, ## mn = 2 (m-2)(n-2) ##.

Solving for m gives
$$ m = \frac{4(n-2)}{n-4} $$
with an interchange of m and n giving n as a function of m.

Since this problem is symmetric in m and n, we can impose the condition that m >= n without loss of generality. The next step is to try the possible values of positive-integer n and find which ones give positive-integer m. To get an idea of how feasible that is, I consider how m varies as a function of n. So I calculate
$$ \frac{dm}{dn} = - \frac{8}{(n-4)^2} $$
This means that increasing n makes m decrease.

Since m must be greater than 0, n must either be 1 or greater than 4. Since n must be at least 3 to allow both the interior and the edges to have a positive number of squares, this means n > 4.
  • n = 5 -> m = 12
  • n = 6 -> m = 8
  • n = 7 -> m = 20/3 = 6.6666...
That last one violates m >= n, so we stop there.

That means that the only two solutions are
  • (5,12) -- # squares = 60, # edge, interior squares = 30
  • (6,8) -- # squares = 48, # edge, interior squares = 24
 

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78. When a cyclist had covered two-thirds of his way, a tire burst. For the rest of the way he needed on foot twice as much time as for the previous ride on the bike.
How many times faster was he cycling than he was walking?

D94
 

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79. Find all solutions of the following addition:
\begin{align*}
&\quad &{}xx \\
&\quad &{}yy \\
+&\quad &{}zz \\
\hline
& \quad & xyz
\end{align*}
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#79
Apart from the trivial 00 + 00 + 00 = 000, 11 + 99 + 88 = 198 is the only solution in base 10.
 
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#79 proof
  1. The last digit of the solution ## = z \implies x + y = 10 ##
  2. From the addition of the most significant digits (plus the carry from the least significant) we have
  3. ## x + y + z + 1 = 10x + y \implies 9x = z + 1 ## so we must have ## x =1, z = 8 ##
  4. From 1 and 3 we have ## 1 + y = 10 \implies y = 9 ##
 
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#78
We want to find v(cycling)/v(walking). Velocity = distance / time, and I must calculate the appropriate distances and times.

Distance cycling = (2/3) (total distance), meaning that distance walking = (1/3) (total distance), and (distance cycling) = 2 (distance walking).

Time walking = 2 (time cycling), yielding time cycling = (1/2) (time walking).

Thus, velocity cycling = (distance cycling) / (time cycling) = (2)/(1/2) (distance walking) / (time walking) = 4 (velocity walking).

The cyclist had biked four times faster than he walked.
#79
Let us break down the sums by digits, and let us use ones and tens carries a and b. Thus,
$$ x + y + z = z + 10 a \\ x + y + z + a = y + 10 b \\ b = x $$
The third equation gives us the second carry, ## b = x ##, and the first one gives us ## x + y = 10 a ##. This only nonzero multiple of 10 that any digits add up to is 10, and thus ## a = 1 ## and ## x + y = 10 ##. Inserting into the second equation gives ## 10 + z + 1 = y + 10 x ## or ## 11 + z = 10 + 9 x ## or ## 1 + z = 9 x ##. The only possible value of x that can solve that equation is 1, and that solution gives y = 9 and z = 8.

The solution: ## x = 1, y = 9, z = 8 ##, giving us
$$
\begin{align*}
&\quad &{}11 \\
&\quad &{}99 \\
+&\quad &{}88 \\
\hline
& \quad & 198
\end{align*}
$$
 

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80. What is the ratio between the (sum of all) colored areas and the (area of the) entire square?
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76. ##n^3+ n^2 u +n^2 v+n^2w+nuv+nuw+nvw+uvw = 27,673,509,091## with ##u<v<w##.
What is ##u\cdot n^2\,?## (All numbers are non negative integers and ##n## maximal among all solutions.)
Brute force appears to yield ## n = 131, u = 0, v = 2,000, w = 99,000 \implies u \cdot n^2 = 0 ## as the solution with maximal ## n ##, but finding this has not given me any greater insight!
 

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81. Bart and Lisa are sitting in front of a huge heap of skittles. Since both of them want to eat as many as possible, they decide to play a game. Bart has to write two (different) numbers (positive integers) on two pieces of paper. Then Lisa turns around one of the papers and guesses, whether the other number is higher or lower. If she is right, then she will get 10 skittles, otherwise Bart will get them.

Is there a strategy for Lisa which improves her chances in comparison to a 50:50 guess? And can Bart counter this strategy?

D95
 
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jbriggs444

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81. Bart and Lisa are sitting in front of a huge heap of skittles. Since both of them want to eat as many as possible, they decide to play a game. Bart has to write two (different) numbers (positive integers) on two pieces of paper. Then Lisa turns around one of the papers and guesses, whether the other number is higher or lower. If she is right, then she will get 10 skittles, otherwise Bart will get them.

Is there a strategy for Lisa which improve her chances in comparison to a 50:50 guess? And can Bart counter this strategy?

D95
Suppose that Bart has chosen numbers i and j. One is chosen at random. 0.5 probability for i, 0.5 probability for j. Without loss of generality, assume i > j.

Lisa decides on a strictly monotone increasing function f from the positive integers to the reals between 0 and 1. After seeing the chosen number, she uses this function to determine her probability of staying with the number that has been revealed.

Lisa's probability of seeing i and then winning is ##0.5 \times f(i)##
Lisa's probability of seeing j and then winning is ##0.5 \times (1 - f(j))##
Total ##0.5 + 0.5\times (f(i) - f(j))##

Since f is monotone increasing and i > j, f(i) > f(j) and Lisa's winning probability exceeds 0.5

Bart can counter this strategy by choosing very large numbers. Lisa's function is bounded and monotone increasing. It must therefore have a least upper bound that is approached arbitrarily closely for large n. (i.e. ##\lim_{n \to \infty}f(n) = U##. It follow that for any epsilon, Bart can choose a delta large enough that that f(i)-f(j) < epsilon for all i, j > delta.

Edit: You'd only asked about the existence of a counter-strategy for Bart rather than for an actual implementation. The following mixed strategy should perform adequately.

Suppose that Bart wishes to obtain a winning probability within epsilon of 0.5. Bart arbitrarily picks a positive integer n that is larger than 1/epsilon. He then picks uniformly and at random an integer i between 1 and n. He writes i on one piece of paper and writes i+1 on the other.

Lisa's chance of winning will depend on her choice of f, of course. But averaged over all the n possible picks by Bart, her winning probability exceeds 0.5 by a total of:
$$\frac{1}{n}\sum_{i=1}^n 0.5 \times(f(i+1)-f(i))$$
Since f(1) and f(n+1) are bounded by 0 below and 1 above, this sum is no more than 0.5/n. This is less than epsilon.
 
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fresh_42

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I think you got the basic method, but I don't follow your reasoning, since ##f(i)=0## for any ##i##, considering the amount of numbers available. I don't think it can easily be put into a "proof", since the probabilities involved are very hard to determine appropriately without additional assumptions.

So I was looking for an answer to: What must Lisa do to improve her chances?

Very large numbers is a practical counter strategy based on assumptions of human behaviour, but there is a counter strategy which always work.
 

jbriggs444

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I think you got the basic method, but I don't follow your reasoning, since ##f(i)=0## for any ##i##, considering the amount of numbers available.
Huh? There is no guarantee that f(i) = 0 for any i. ##f(i)=0.75 - \frac{1}{2i}## works just fine and is bounded below by 0.25.
I don't think it can easily be put into a "proof", since the probabilities involved are very hard to determine appropriately without additional assumptions.
Again, what? Nothing hard about any of it.
Very large numbers is a practical counter strategy based on assumptions of human behaviour, but there is a counter strategy which always work.
Yup. Found one and had already edited it in.
 

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Maybe I didn't understand your solution in detail, esp. the definition of ##f##.
Anyway, I said I think you have it. Here's my wording:

Lisa chooses a third random number ##s## secretly and adds ##0.5##. If the displayed number is less than ##s##, then she answers "greater" and vice versa. This does not affect her chances if ##s## is smaller than the smaller or greater than the bigger of the two numbers. But if ##s## is in the interval of the two, she will always win. So all these instances increase her chances.

The counter strategy for Bart would be to choose intervals of length one.
 

jbriggs444

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Maybe I didn't understand your solution in detail, esp. the definition of ##f##.
Anyway, I said I think you have it. Here's my wording:

Lisa chooses a third random number ##s## secretly and adds ##0.5##. If the displayed number is less than ##s##, then she answers "greater" and vice versa. This does not affect her chances if ##s## is smaller than the smaller or greater than the bigger of the two numbers. But if ##s## is in the interval of the two, she will always win. So all these instances increase her chances.

The counter strategy for Bart would be to choose intervals of length one.
So, roughly speaking, my monotone increasing f is the inverse of the cumulative distribution of your random variable s. Sure, that works.

Intervals of length one is a good start. But it is inadequate, by itself, to assure a particular expected value against all instances of the Lisa strategy we have discussed.
 

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82. Fill in all numbers ##\{\,1,2,\ldots,15\,\}## such that the sum of all numbers in every ring is the same.
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D96
 

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83. A dealer shuffles a standard deck of ##52## cards. The player has ##\$\,100## at the beginning. He can bet any non negative amount on the color (red and black) of each card the dealer turns around. E.g. he could wait for ##51## cards to be drawn and put the entire ##\$\,100## on the missing color. In case he is right, he will receive twice the amount he has set, and nothing if he loses.

What is the optimal strategy?

[optimal: guarantees ##\$\,y## at the end of the game and no other strategy can guarantee ##\$\,z > y##. No statistical assumptions must be made and must work even if the dealer recognizes the strategy and cheats.]

D98
 

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84. A red ant and seven black ants are placed on a five-meter-long, straight branch. Each ant chooses (randomly) to go to the left or to the right, and then runs in that direction at a speed of 1 meter per 10 seconds. Whenever two ants hit each other, both turn and then continue in the opposite direction. When an ant reaches the end of the branch, then it falls off the branch and out of the game.

How long can the red ant stay on the branch at most?

D99
 
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83. A dealer shuffles a standard deck of ##52## cards. The player has ##\$\,100## at the beginning. He can bet any non negative amount on the color (red and black) of each card the dealer turns around. E.g. he could wait for ##51## cards to be drawn and put the entire ##\$\,100## on the missing color. In case he is right, he will receive twice the amount he has set, and nothing if he loses.

What is the optimal strategy?
If we want to maximize the minimal money we walk away with then we must be indifferent about the outcome of every bet: If we would favor winning then we should bet less money to make losing less bad and vice versa.
Let f(r,b) be the factor we can multiply our money with if there are r red and b black cards left. Clearly f(0,0)=1 as we can't bet any more and f(r,0)=2r as we bet all on red for r rounds.
Let's assume r>=b, we bet on red. Our bet fraction p must be chosen such that winning and losing will lead to the same result: (1+p)f(r-1,b) = (1-p)f(r,b-1). Solve: p = (f(r,b-1)-f(r-1,b))/(f(r-1,b)+f(r,b-1)). While negative bets were not allowed bets with p between -1 and 0 are effectively a -p bet on red, so we can use this formula everywhere.
Fill out the table: f(26,26)=9.08132954942779. Multiply by the initial $100 and we get $908.13.

I also looked at a different question: What if you bet everything when there is an imbalance in the remaining cards? You get $100*226 with probability 26/51 * 25/49 * ... = 26!/51!! and zero otherwise.. A one-in-7 million chance to win 6.7 billion. The expectation value? 9.08132954942779.

This is not an accident, but I don't have a simple argument why the highest minimal payout for n red and black cards is (2n)!!/(2n-1)!!.
 

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85. The gardener has a clear mission: he must plant ten trees in a way that there will be five rows of four trees. Can he manage this?

D100
 

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Not sure whether this one will be too easy or too hard. (revised)

86. I am a three digit even natural number which can be written as a sum of two primes in six different ways. None of them is in my factor decomposition, but if I sum up all products of those six pairs, then it is a five digit number and again a product of two primes.

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87. There are four six sided dice with an unusual numbering:
Die 1: 0-0-4-4-4-4
Die 2: 1-1-1-5-5-5
Die 3: 2-2-2-2-6-6
Die 4: 3-3-3-3-3-3
John chooses first which one he will play with, Jane second. Then they roll their dice once and the higher number will win. Who has the better chances to win, and with which strategy?

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Hey, 87 looks familiar...
86. I am a three digit even natural number which can be written as a sum of two primes in six different ways. None of them is in my factor decomposition, but if I sum up all products of those six pairs, then it is a five digit number and again a product of two primes.
I had tried the first three examples when it said eight different ways but then didn't check more.
With six different ways the first numbers are 60, 72, 100, 106, 110. The first three lead to a sum of products that is too small. 106 is twice a prime. 110 leads to a sum of products of 10334=2*5167. I didn't check if there are other numbers, but that is a solution.
84. A red ant and seven black ants are placed on a five-meter-long, straight branch. Each ant chooses (randomly) to go to the left or to the right, and then runs in that direction at a speed of 1 meter per 10 seconds. Whenever two ants hit each other, both turn and then continue in the opposite direction. When an ant reaches the end of the branch, then it falls off the branch and out of the game.

How long can the red ant stay on the branch at most?
I have seen this before, but as no one answered:
It doesn't matter which ant is red, for any arrangement of ants we can simply label the one red that will stay on the branch the longest. This also means we can replace the collision process with the ants passing each other without changing the survival times. Now the problem became trivial: 10 seconds per meter at 5 m gives 50 seconds. This maximum will always be reached if one ant starts at the end, facing towards the rest of the stick. Which ant will reach it depends on the details of the arrangement.
 

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88. Two shepherds meet. Says the first shepherd to the second: "Give me 8 sheep of yours, then I have twice as many as you." Then shepherd two says to number one: "If you give me 8 sheep of yours, then we have the same number of sheep." How many sheep does each shepherd have?

D102
 

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