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Can You Solve The Horse Grazing Area? Sunday Puzzle

A horse is tied to an outside corner of a 20 foot by 10 foot rectangular barn. What is the maximum area the horse can graze outside, if the rope has length L? Solve for L = 5 feet and L = 25 feet, and find the exact answer. If you are up for a challenge, solve for L = 50 feet. You can solve for the answer to 3 decimal places.

Can you figure it out? Watch the video for a solution.

Can You Solve The Horse Grazing Puzzle?

Or keep reading.

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Answer To The Horse Grazing Puzzle

If the rope has L = 5 feet, then the horse can graze the area of 3/4 of a circle with radius 5.

The horse can graze an area (3/4)π(52) = 75π/4 ≈ 58.905 square feet.

If the rope has L = 25 feet, then the horse can graze the area of 3/4 of a circle with radius 25. But along the 10 foot wall, there is also 15 feet of rope left over. The horse can then graze an extra 1/4 circle with a radius of 15 feet. Similarly, along the 20 foot wall, the horse can also graze an extra 1/4 circle with a radius of 5 feet.

The horse can graze an area equal to the sum of these shapes. The horse can therefore graze an area (1/4)π[3(252) + (152) + (52)] = 2125π/4 ≈ 1668.971 square feet.

The case of L = 50 feet is more complicated. The horse can graze 3/4 of a circle with a radius of 50. Along the 10 foot wall, the rope has an extra 40 feet, so the horse can reach an extra 1/4 circle with a radius of 40. And along the 20 foot wall, the rope has an extra 30 feet, so the horse can reach an extra 1/4 circle with a radius of 30. But the extra 1/4 circles overlap which makes the problem much harder to solve.

I will present two ways to solve for the area. One method is to use calculus to solve for the overlapping area directly. The other method is to use trigonometry to dissect the area into non-overlapping shapes.

Calculus approach to L = 50

If we add up the areas of the 1/4 circles, we would be double-counting the area of the region where the circles overlap. If we can solve for the area of the overlap, we could subtract that out.

grazing area = (area of 3/4 circle radius 50) + (area of 1/4 circle radius 40) + (area of 1/4 circle radius 30) – (overlap of 1/4 circles)

What is the area of the overlapping regions?

We can put a coordinate system and then set up integrals. The origin is placed at the left corner where the barn intersects the circle with a radius of 40. We then have one circle with a radius of 40 centered at the origin, and we have another circle with a radius of 30 centered at (20, -10). We can write equations for the y values of these circles (we only care about the semi-circles above the centers of the circles, so we can disregard the negative square roots):

x2 + y2 = 402
y = √(1600 – x2)

(x – 20)2 + (y + 10)2 = 302
y = -10 + √(900 – (x – 20)2)

We can set the equations for these curves equal to each other, and we can solve that they intersect at x = 24 + 4√11.

We now focus on only the overlapping region to illustrate the overlap is the sum of two definite integrals.

We have solved the overlapping region is approximately 341.77937.

We can now go back and solve for the total grazing area:

grazing area = (area of 3/4 circle radius 50) + (area of 1/4 circle radius 40) + (area of 1/4 circle radius 30) – (overlap of 1/4 circles)

grazing area = (π/4)[3(50)2 + 402 + 302] – (341.77937)

grazing area = 2500π – (341.77937) ≈ 7512.202

Trigonometry approach to L = 50

The problem can also be solved without calculus, although there are more steps. The key insight is to connect the point of intersection between the circles to opposite corners of the barn.

This creates a kite-like shaped quadrilateral. Above the kite-like shaped quadrilateral is a circular sector from the circle with radius 40. And below the quadrilateral is another circular sector from the circle with radius 30. We can solve for the angles of these circular sectors and then solve for their areas.

In other words, we can solve for the total area by adding up the areas of 3 non-overlapping shapes.

First we solve for the quadrilateral area. Let’s add 1/2 of the rectangular barn to make a large triangle. The diagonal of the rectangular barn is 10√5 because it is the hypotenuse of a right triangle with legs 10 and 20. The large triangle has sides of 40 and 30 corresponding to the radii of the circles. The area of the large triangle can then be found using Heron’s formula. And the area of 1/2 the barn is also readily calculated.

We have found the quadrilateral has an area of 100√11 – 100.

Now we need to find the angles for the circular sectors. We can use trigonometry.

We can find the angles of 1/2 of the barn readily using the inverse cosine of an adjacent leg over the hypotenuse. Then we calculate the angles of the large triangle using the law of cosines.

The angle of a circular sector with 40 is found by subtracting the upper angle from the right triangle and the angle from the large triangle opposite the side equal to 30. The angle of a circular sector with 30 is found by subtracting the lower right angle from the right triangle and the angle from the large triangle opposite the side equal to 40.

We can use those angles to calculate the area of the circular sectors, which then leads to the total area of the region.

This region has an area of approximately 1621.716. We add that to the 3/4 circle with a radius of 50 to get the total grazing area.

grazing area = (3/4)π(50)2 + 1621.716 ≈ 7512.202

We again get to the same answer without using calculus.

Sources
Math StackExchange
http://math.stackexchange.com/questions/1942222/a-goat-tied-to-a-corner-of-a-rectangle

Sketch of trigonometry approach
http://mathcentral.uregina.ca/QQ/database/QQ.09.08/h/misty1.html

Another sketch of the trigonometry approach
http://mathforum.org/dr.math/faq/faq.grazing.html

See the original post:
Can You Solve The Horse Grazing Area? Sunday Puzzle

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Originally posted here:
Puzzle Games – Free Online Puzzles for Kids and Adults …

Can You Solve The Apples And Oranges Riddle? The Mislabeled Boxes Interview Question. Sunday Puzzle

In front of you are 3 boxes. One box contains only apples, another box contains only oranges, and the last contains both apples and oranges.

Currently the first box has the label “apples,” the second “oranges,” and the third “apples and oranges.” Unfortunately all of the labels are wrong. Your job is to fix the labels.

You are not allowed to peek inside any of the boxes. But you can ask for a sample from any box. You point to a box, and you get a fruit from that box.

What is the minimum number of samples you need to label all of the boxes correctly?

Watch the video for a solution.

Can You Solve The Apples And Oranges Riddle?

Or keep reading.

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Answer To The Apples And Oranges Riddle

This problem is sometimes asked as an interview brain teaser. Many people think you will need to sample at least a few boxes. After all, when you get an apple, you may not be sure it is from a box of apples or from the box of apples and oranges.

But in fact, you can do much better. You only need one sample to figure out how to label all the boxes. The solution depends on the key detail that all of the boxes are labeled incorrectly.

Here is what you do. Ask for a sample from the box labeled “apples and oranges.” Suppose you get an apple. Since this box is labeled incorrectly, it must contain only one fruit (it cannot have both apples and oranges), so you know this box must contain only apples. Next consider the box labeled “oranges.” The choices left are to contain only oranges or to contain both apples and oranges. Since the label is wrong, it must contain both apples and oranges. This means the box labeled “apples” has to contain only oranges.

Suppose instead you got an orange from the box labeled “apples and oranges.” By similar logic as above, this box must contain only oranges. Next consider the box labeled “apples.” The choices left are to contain only apples or to contain both apples and oranges. Since the label is wrong, it must contain both apples and oranges. This means the box labeled “oranges” has to contain only apples.

Once you get the sample from the box labeled “apples and oranges” you can readily figure out all of the labels!

Explanation with permutations

Suppose box 1 is labeled “apples,” box 2 is labeled “oranges,” and box 3 is labeled “apples and oranges.” How many different ways are there to label 3 boxes?

There are 3 choices for the first box, 2 choices for the second, and 1 choice for the last. In total there are 3(2)(1) = 6 total ways to label the boxes. These can be enumerated as:

box 1
(“apples”)
box 2
(“oranges”)
box 3
(“apples and oranges”)
applesorangesapples and oranges
applesapples and orangesoranges
orangesapplesapples and oranges
orangesapples and orangesapples
apples and orangesapplesoranges
apples and orangesorangesapples

Of these 6 possible ways, only the rows (oranges, apples and oranges, apples) and (apples and oranges, apples, oranges) correspond to all of the labels being wrong. A permutation where all the elements are in the wrong spot is known as a derangement. We have encountered those before and calculated the formula for the number of derangements in a previous puzzle.

If we pick from box 1 or box 2, we could be picking from a single fruit box or a box with both fruits.

If we pick from box 3, it must have a single fruit. The sample provides exactly enough information to distinguish between the two cases and label all of the boxes correctly.

Sources
Math by Matt
https://matthewhr.wordpress.com/2013/04/07/permutations-and-mislabeled-jars/

Apple interview question
https://www.glassdoor.com/Interview/There-are-three-boxes-one-contains-only-apples-one-contains-only-oranges-and-one-contains-both-apples-and-oranges-The-b-QTN_114922.htm

Cut The Knot
http://www.cut-the-knot.org/SimpleGames/PickLabel.shtml#solution

Continued here:
Can You Solve The Apples And Oranges Riddle? The Mislabeled Boxes Interview Question. Sunday Puzzle

Can You Solve It? Use A Coin To Simulate Any Probability. Sunday Puzzle

You have a fair coin that shows heads 50 percent of the time and tails 50 percent of the time.

You and a friend play a coin flipping game. Can you create a rule classifying the coin flips so that you win with probability exactly equal to p and your friend wins with probability exactly equal to 1 – p? The probability can be any rational or irrational number between 0 and 1. The game has to end in a finite number of coin flips with probability 1.

Extension: Now you have a coin that shows heads with probability q, where 0 q q, and it could be rational or irrational! Can you now figure out a game where you win with probability exactly equal to p? The value p is known to you. The same restrictions apply: the game has to end in a finite number of flips with probability 1.

Watch the video for a solution.

Can You Solve It? Use A Coin To Simulate Any Probability

Or keep reading.

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Answer To Use A Coin To Simulate Any Probability

There is a straightforward method if the probability is a fraction with a power of 2; a fraction that is of the form x/2n. Here is a game that works. Flip a coin n times and record the result of each toss. As each toss is equally likely to be heads or tails, there are a total of 2n equally likely sequences of tosses. Define x of those sequences to be a win for you and let every other sequence be a loss for you. This rule has you winning with probability x/2n.

What about other denominators, like if you need to win with a probability of 1/3? The game can be modified as follows. Flip a coin 2 times and record the sequence of tosses, which can be any of the 4 possible outcomes:

HH
HT
TH
TT

Now define the one sequence as a win for you, two sequences as loss for you, and the last sequence as a re-toss.

HH – win
HT – loss
TH – loss
TT – start over

Occasionally you will have to flip again. But eventually you will toss one of the other sequences, and there is a 1/3 probability of winning.

The game can be modified for any rational probability x/y. Flip a coin n times where 2n – 1y ≤ 2n. From a list of equally likely coin flipping sequences, define x sequences as a win, yx sequences as a loss, and the remaining sequences as requiring to start over.

Sometimes you will have to flip again. But eventually the game ends, and you win with probability x/y.

But what if you want to estimate an irrational probability, like 1/π?

There are rational numbers arbitrarily close to 1/π, so you could design a procedure that gets arbitrarily close to 1/π. But you cannot win with probability exactly 1/π in any of these methods. And better approximations require more coin flips, so you cannot find a solution in finite coin flips either.

To solve the problem we have to ditch this method and create another rule.

The solution: use binary numbers!

Here is a game that works for any probability p, whether it is rational or irrational. First, write p as a decimal in binary:

p = b1/2 + b2/22 + b3/23 + …

= 0.b1b2b3

Flip the coin until it shows heads. Let n be the number of tosses.

You win if bn = 1 and you lose otherwise.

The probability you win the game is:

Pr(you win) = Pr(toss 1 is first H)Pr(b1 = 1) + Pr(toss 2 is first H)Pr(b2 = 1) + …

= b1/2 + b2/22 + …

= binary expansion of p

= p

You win this game with probability exactly equal to p. Furthermore, if you toss a coin, it will eventually show heads, so this procedure ends in a finite number of flips with probability 1.

(Technical point: for numbers with two binary expansions, use the representation that ends with an infinite string of 0s rather than an infinite string of 1s. This is similar to how 0.1 = 0.0999…. in decimal numbers.)

Alternate procedure

There is another way to use binary numbers that has a nice geometric interpretation.

First write p in binary. Flip the coin and record the results, with heads = 1 and tails = 0. Flip the coin, and continue to flip, if the kth flip matches bk. Stop flipping if the result does not match.

If the final toss was heads, then you lose. If the final toss was tails, you win.

The game can be interpreted as follows. Your coin tosses determine a random number z between 0 and 1. You win the game if the number z ends up between 0 and p. You lose the game if z is larger than p.

Here is why the rule works. The sequence of tosses is like a binary search. The first toss puts you in the intervals tails = [0, 0.5] or heads = [0.5, 1]. The second toss puts you in an interval tails/tails = [0, 0.25], tails/heads = [0.25, 0.5], heads/tails = [0.5, 0.75], heads/heads = [0.75, 1]. The third toss divides each of those intervals in half, and so on.

At some point your toss will be different from the binary representation of p. If you tossed a tails, then that means your interval maximum is less than p and you will win for sure (you converge to a value less than p). If you tossed a heads, then your interval minimum is greater than p and you will lose for sure (you converge to a value larger than p).

In other words, the coin flips determine a random number (uniform distribution) between 0 and 1, and you win if the number is less than or equal to p. Thus the probability you win is p.

Formally, recall p = 0.b1b2b3

When do you win the game? This happens if you flip until one of the bits in p is 1 and you flip a tails (0). For each non-zero bit bk, you would need to match for k – 1 flips, and then differ on the next toss. The probability of this is 1/2k. The overall probability is then the sum over all the non-zero bits. This is an infinite sequence of 1/2j where j is an index for each non-zero bit. The sum of this series is again the binary expansion of p, so the probability of winning is p.

Extension: coin with unknown bias

Now this seems like it’s impossible. You don’t know the bias of the coin, and yet you have to use it to simulate any probability.

Amazingly, there is a solution! The insight is that you can make a fair coin toss out of any biased coin, even if you do not know the bias. Here is how to do it. Flip the coin two times.

HH
HT
TH
TT

The probability of heads is q and tails is 1 – q, so here are the associated probabilities of the tosses.

Pr(HH) = q(q)
Pr(HT) = q(1 – q)
Pr(TH) = (1 – q)q
Pr(TT) = (1 – q)(1 – q)

Notice the tosses HT and TH have exactly the same probability of occurring. So our procedure is the following. If HH or TT occurs, then disregard the flips and start over. If HT occurs, call it heads; and if TH occurs, call it tails.

HH – start over
HT – “Heads” q(1 – q)
TH – “Tails” (1 – q)q
TT – start over

This procedure eventually terminates in “Heads” or “Tails” with equal probability.

In other words, we have used a coin with an unknown bias to create a fair coin toss.

So we essentially have a fair coin at our disposal. We can then use one of the rules above to generate a game where we win with probability exactly equal to p.

We have used a coin with an unknown bias to simulate any probability we want!

Thanks to all patrons! Special thanks to:

Kyle
Shrihari Puranik

You can support me and this site on Patreon.

Sources

1989 Putnam exam, problem A-4
http://www.math.hawaii.edu/~dale/putnam/1989.pdf

Sorta Insightful
http://www.alexirpan.com/2015/08/23/simulating-a-biased-coin-with-a-fair-one.html

Not really blogging
https://amakelov.wordpress.com/2013/10/10/arbitrarily-biasing-a-coin-in-2-expected-tosses/

Janko Gravner’s course Math 189 problem set 11
https://www.math.ucdavis.edu/~gravner/MAT189/materials/ps11.pdf

Original post:
Can You Solve It? Use A Coin To Simulate Any Probability. Sunday Puzzle

Can You Solve The Integer Polynomial Puzzle?

The points (1, 1) and (2, 2) can be connected by line y = x.

The points (1, 1) and (2, 7) can be connected by parabola y = 2x2 – 1.

These are examples of connecting points using polynomials that have integer coefficients.

Can the points (1, 3) and (3, 2) be connected by a polynomial with integer coefficients? The equation should be of the form

y = cnxn + cn – 1xn – 1 + … + c1x + c0

Additionally, all of the coefficients ck need to be integers (positive or negative whole numbers).

If it is possible, then find a polynomial. If it is impossible, then prove why it is impossible.

Watch the video for a solution.

Can You Solve The Integer Polynomial Puzzle?

Or keep reading.

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Can You Connect The Dots? An Integer Polynomial Puzzle

You might think that some polynomial has to work because there are only 2 points and the degree of the polynomial can be any arbitrary value. But somewhat amazingly, it is impossible to connect the points (1, 3) and (3, 2) with a polynomial consisting of integer coefficients!

The reason is parity. For a polynomial with integer coefficients, p(x), both values p(1) and p(3) have to be odd or both have to be even. This is because p(1) is the sum of the coefficients and p(3) has the same parity as the sum of the coefficients, which can be seen by considering the polynomial modulo 2 and noting 3k ≡ 1 (mod 2).

Therefore, it is impossible for p(1) to be odd and p(3) to be even. No such polynomial with integer coefficients exists!

Longer explanation

For the polynomial to go through (1, 3), we have:

3 = cn(1)n + cn – 1(1)n – 1 + … + c1(1) + c0

3 = cn + cn – 1 + … + c1 + c0

This equation implies the sum of the integer coefficients is 3, which is an odd number.

For the polynomial to go through (3, 2), we would need:

2 = cn(3)n + cn – 1(3)n – 1 + … + c1(3) + c0

Next we observe that 3 raised to any whole number power is always an odd number. This is because 3 is an odd number and an odd number times another odd number is odd.

3, 9, 81, 243, etc.
3 ≡ 1 (mod 2)
3n ≡ 1n ≡ 1 (mod 2)

For each term ck(3)k, the oddness or evenness only depends on the value ck. If ck is odd, it is multiplied by an odd term, so the product is odd; if ck is even, the product is even because an even number times another number is always even.

ck(3)kck (mod 2)

We can simplify term by term to get:

2 = cn(3)n + cn – 1(3)n – 1 + … + c1(3) + c0

2 ≡ cn + cn – 1 + … + c1 + c0 (mod 2)

This implies the sum of the coefficients is an even number, and that contradicts the previous result that the sum of the coefficients has to be an odd number.

Therefore no such polynomial exists.

Source

I learned about this problem from Aha! Solutions by Martin Erickson. Link to book on Amazon: http://amzn.to/1sOwMnN

I also saw it on Reddit Math
https://www.reddit.com/r/math/comments/5fppvp/polynomials_with_integer_coefficients_can_not/

Continued here:
Can You Solve The Integer Polynomial Puzzle?

Can You Solve The Diluted Wine Puzzle? Famous 16th Century Math Problem

A servant has a method to steal wine. He removes 3 cups from a barrel of wine and replaces it with 3 cups of water. The next day he wants more wine, so he does the same thing: he removes 3 cups from the same barrel (now with diluted wine) and replaces it with 3 cups of water. The following day he repeats this one more time, so he has drawn 3 times from the same barrel and has poured back 9 cups of water.

At this point the barrel is 50% wine and 50% water.

How many cups of wine were originally in the barrel?

Watch the video for a solution.

Can You Solve The Diluted Wine Puzzle?

Or keep reading for a solution.

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Answer To The Diluted Wine Puzzle

Suppose the barrel started with x cups of wine. We will keep track of how much wine is in the barrel and the concentration of wine in the barrel. Initially the barrel has x cups of wine and it is 100% wine (a concentration of 1).

Wine Amount (starting), Wine Concentration (starting)
x, 1

The servant first takes 3 cups of wine and replaces them with water. The amount of wine left is x – 3, and the concentration is the amount of wine, x – 3, divided by the total volume of liquid, which is x.

Wine Amount (step 1), Wine Concentration (step 1)
x – 3, (x – 3)/x

x – 3, 1 – 3/x

The servant then removes 3 cups from the barrel, and each cup contains a concentration 1 – 3/x of wine. The amount of wine left is x – 3 – 3(1 – 3/x)/x, and the concentration is the amount of wine, x – 3 – 3(1 – 3/x)/x, divided by the total volume of liquid, which is x.

Wine Amount (step 2), Wine Concentration (step 2)
x – 3 – 3(1 – 3/x), (x – 3 – 3(x – 3)/x)/x

x – 6 + 9/x, (x – 6 + 9/x)/x

x – 6 + 9/x, (1 – 3/x)2

The servant finally removes 3 more cups from the barrel, and each cup contains a concentration (1 – 3/x)2 of wine. The amount of wine left is x – 6 + 9/x – 3(1 – 3/x)2, and the concentration is the amount of wine divided by the total volume of liquid, which is x.

Wine Amount (step 3), Wine Concentration (step 3)
x – 6 + 9/x – 3(1 – 3/x)2, (x – 6 + 9/x – 3(1 – 3/x)2)/x

x – 9 + 27/x – 27/x2, (x – 9 + 27/x – 27/x2)/x

x – 9 + 27/x – 27/x2, (1 – 3/x)3

The final concentration should be equal to 50%, or 1/2.

(1 – 3/x)3 = 1/2

(1 – 3/x) = 1/21/3

21/3x – 3(21/3) = x

x(21/3 – 1) = 3(21/3)

x = 3(21/3)/(21/3 – 1) ≈ 14.54

Thus the original barrel contained approximately 14.5 cups of wine.

There is a shortcut to solving this problem! You can save many steps by noticing the concentration is 1 – 3/x after the first step.

Wine Amount (step 1), Wine Concentration (step 1)
x – 3, 1 – 3/x

The subsequent steps iterate the same process of removing 3 cups and then diluting the wine with water. Accordingly, the wine is diluted by the same percentage in each step. To find the new concentration, multiply by the factor 1 – 3/x.

Wine Concentration (step 2)
(1 – 3/x)2

Wine Amount (step 3), Wine Concentration (step 3)
(1 – 3/x)3

Now we can set the concentration equal to 1/2 and find the answer as before.

Sources

I read this problem in Famous Puzzles of Great Mathematicians by Miodrag S. Petkovic.

The puzzle appears in Niccolo Tartaglia’s work “General Trattato di Numeri” (1556).

StackExchange has the idea to multiply concentrations
http://puzzling.stackexchange.com/questions/28776/turning-wine-into-water/28781

See the article here:
Can You Solve The Diluted Wine Puzzle? Famous 16th Century Math Problem

Chess Puzzles – 365Chess.com

The best way to improve your chess is practicing! And the funniest way to practice is solving puzzles! Train your tactical abilities solving all kind of situations like the ones you’ll find over a chess board at a real game.

You can play in unrated mode, just for fun, without recording your solving history. Or you can have full tracking of your progress solving in Rated mode. If you play in this mode you’ll also obtain problems according to your current level.

Solve chess problems taken from real games. You should find the best continuation for you opponent’s moves.

Train your abilities facing chess studies from the most complete database of problems, created by the greatest composers.

Good news! Only a few minutes every day solving puzzles and you will boost your chess!

If you want to improve your game the fastest, easiest and most fun way, start solving puzzles right now.

One thing that is absolutely undisputed in chess training philosophy is: solve tactical chess puzzles regularly and you’ll get better and better everyday.

Train your abilities facing chess studies from the most complete database of problems, created by the greatest composers.

“My fascination for studies proved highly beneficial, it assisted the development of my aesthetic understanding of chess, and improved my endgame play” Vasily Smyslov

“I always urge players to study composed problems and endgames” Pal Benko.

Just a few samples of recently added games to our database.

Chess Composition

Chess Problem

Chess Composition

Go here to see the original:
Chess Puzzles – 365Chess.com

Can You Solve The Mind-Bending Self-Counting Sentence Riddle? Sunday Puzzle

In this sentence, the number of occurrences
of the digit 0 is __,
of the digit 1 is __,
of the digit 2 is __,
of the digit 3 is __,
of the digit 4 is __,
of the digit 5 is __,
of the digit 6 is __,
of the digit 7 is __,
of the digit 8 is __, and
of the digit 9 is __.

The challenge is to fill in the blanks with numerical digits (1, 2, etc.) so that the sentence is true.

There are 2 solutions. Can you find them?

Watch the video for a solution.

Can You Solve The Mind-Bending Self-Counting Sentence Riddle?

Or keep reading.

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Answer To The Mind-Bending Self-Counting Sentence Riddle

The answers can be found by experimentation. To start, each of the digits appears at least once, so fill in 1 for every line (except the digit 1 which is left blank at the moment).

TEST CASE 1
In this sentence, the number of occurrences
of the digit 0 is 1,
of the digit 1 is __,
of the digit 2 is 1,
of the digit 3 is 1,
of the digit 4 is 1,
of the digit 5 is 1,
of the digit 6 is 1,
of the digit 7 is 1,
of the digit 8 is 1, and
of the digit 9 is 1.

How many times does the digit 1 occur? The sentence already contains 10 occurrences of the digit 1, so let us try that value.

TEST CASE 2
In this sentence, the number of occurrences
of the digit 0 is 1,
of the digit 1 is 10,
of the digit 2 is 1,
of the digit 3 is 1,
of the digit 4 is 1,
of the digit 5 is 1,
of the digit 6 is 1,
of the digit 7 is 1,
of the digit 8 is 1, and
of the digit 9 is 1.

This sentence is false, however, because the digit 1 appears 11 times in total (including the “1” in the number “10.”). We can try to adjust the sentence to 11 occurrences of the digit 1.

TEST CASE 3
In this sentence, the number of occurrences
of the digit 0 is 1,
of the digit 1 is 11,
of the digit 2 is 1,
of the digit 3 is 1,
of the digit 4 is 1,
of the digit 5 is 1,
of the digit 6 is 1,
of the digit 7 is 1,
of the digit 8 is 1, and
of the digit 9 is 1.

But now this sentence is also false because it contains 12 occurrences of the digit 1 (the number “11” adds another “1” to the count). If we increase the count to 12, then the occurrences of the digit 2 would increase to 2, thereby reducing one of the occurrences of the digit 1. So let’s try keeping the occurrences of the digit 1 as 11, and increasing the occurrences of the digit 2 to be 2.

TEST CASE 4 = Solution One
In this sentence, the number of occurrences
of the digit 0 is 1,
of the digit 1 is 11,
of the digit 2 is 2,
of the digit 3 is 1,
of the digit 4 is 1,
of the digit 5 is 1,
of the digit 6 is 1,
of the digit 7 is 1,
of the digit 8 is 1, and
of the digit 9 is 1.

This sentence is true and is the first solution to the puzzle. The digit 1 occurs 11 times, the digit 2 occurs 2 times, and every other digit occurs 1 time.

But there’s another solution this problem too! Let us adjust the above solution by considering if the digit 1 only has 9 occurrences to avoid having a double-digit number.

TEST CASE A
In this sentence, the number of occurrences
of the digit 0 is 1,
of the digit 1 is 9,
of the digit 2 is 2,
of the digit 3 is 1,
of the digit 4 is 1,
of the digit 5 is 1,
of the digit 6 is 1,
of the digit 7 is 1,
of the digit 8 is 1, and
of the digit 9 is 1.

The sentence if false because there are 2 occurrences of the digit 9. If the digit 9 occurs 2 times, then that means the digit 2 has to occur 3 times total. And that also means the digit 3 occurs 2 times.

TEST CASE B
In this sentence, the number of occurrences
of the digit 0 is 1,
of the digit 1 is 9,
of the digit 2 is 3,
of the digit 3 is 2,
of the digit 4 is 1,
of the digit 5 is 1,
of the digit 6 is 1,
of the digit 7 is 1,
of the digit 8 is 1, and
of the digit 9 is 2.

This sentence is false; now there are only 7 occurrences of the digit 1. To fix the sentence, we can swap the occurrences of the digits 7 and 9. That is, write that the digit 7 occurs 2 times, and write that the digit 9 occurs 1 time.

TEST CASE C = Solution Two
In this sentence, the number of occurrences
of the digit 0 is 1,
of the digit 1 is 7,
of the digit 2 is 3,
of the digit 3 is 2,
of the digit 4 is 1,
of the digit 5 is 1,
of the digit 6 is 1,
of the digit 7 is 2,
of the digit 8 is 1, and
of the digit 9 is 1.

This sentence is true and it is the second solution. It is in fact the unique solution if all the digits occur less than 10 times.

And there you have it: two true sentences that count themselves!

Sources and further reading
Douglas Hofstadter Metamagical Themas on Google Books
https://books.google.com/books?id=o8jzWF7rD6oC&lpg=PA390&ots=jRCg7rLufs&dq=The%20number%20of%200s%20in%20this%20sentence%20Metamagical%20Themas&pg=PA390#v=onepage&q&f=false

Math Central December 2002 Problem. Proof there are only 2 solutions
http://mathcentral.uregina.ca/mp/archives/previous2002/dec02sol.html

Self-descriptive sentences (and two solutions described)
http://www.cut-the-knot.org/ctk/SelfDescriptive.shtml

Unique solution if all occurrences less than 10
http://math.stackexchange.com/questions/19061/puzzle-digit-x-appears-y-times-on-this-piece-of-paper?rq=1

Sketch of proof that there are only 2 solutions
https://www.reddit.com/r/math/comments/2oh9s9/heres_a_maths_puzzle_my_friend_posted_on_facebook/cmn8f26/

Solving by iterative process
http://web.archive.org/web/20120428023510/http://www.lboro.ac.uk/departments/ma/gallery/selfref/index.html

Variations for self-counting sentences
http://lkozma.net/blog/self-counting-sentences/
http://lkozma.net/blog/self-counting-sentences-ii/

Autogram (self-counting with letters)
https://en.wikipedia.org/wiki/Autogram

Variation with letters (autogram)
http://www.braingle.com/news/hallfame.php?path=language/english/sentences/self.ref/self.ref.letters.p&sol=1

More here:
Can You Solve The Mind-Bending Self-Counting Sentence Riddle? Sunday Puzzle

Chess Problems, Puzzles, & Compositions – Chessopolis

135 Studies And Problems A whole career of composed problems from J.T. Sanderse.

Archives des problmistes In French. Problem site from Quebec.

Beautiful Chess Problems Only a few at the moment, but more promised.

Brunos Chess Problem of the Day Just like it sounds.

BYCA Chess Puzzles In addition to the collection of puzzles, features a puzzles league where you can participate in a friendly solving competition.

Chess Composition Microweb Featuring various sections on composed chess problems & problem-solving.

Chess Online Problems Problems galore. More added frequently.

Chess Problems Online A decent collection, intelligently displayed, from Mate in One to Mate in Six.

Chess Problems From Around The World Problems from everywhere.

Chess Problems Of 1001 Years Ago From the Chess Variants Homepage really old problems mainly of historical interest.

Chess Problems Unlimited Not quite unlimited, there are but several.

Chesspros Chess World Problems and instructive games.

Chess Puzzles by GMs Taken from the games of the greats.

Chess Puzzles Collection A nice collection taken from real games. Includes an index of the puzzles by theme.

CHEST Source code (in ANSI C) for a problem solving program.

Classic Chess Problems 2 and 3 movers from renowed composers.

Endgame Of The Day An endgame study a day. Also available as a mailing list.

Exeter Chess Club: Studies and Problems Intro to the world of problems.

GorFo Problem Chess Pages Has a problem of the month composed by the author, plus an archive of past problems.

Key Moves A new chess combination (from recent games) every week for your instruction and amusement.

Loiodices Chess Problems Lots of training positions to download.

Manolis Stratakis Chess Problems Page Pretty good page with plenty of problems and puzzles.

MateMaster Shareware software for solving chess problems.

Mater Will find any mate thats there, eventually.

Mat Plus The best chess problems from magazine Mat Plus.

Practical Chess Endgame Solve the endgame of the week.

Problemiste Chess Problem software.

Public Domain Chess Composition Books 19th Century (mostly) chess problem books in PDF format so you can print them out yourself. Nice!

Retractor Software for creating Retrograde Analysis chess problems.

The Retrograde Analysis Corner Retro problems and resources for retro enthusiasts.

Sack The King! A chess problem a day, with special animated interface to display the solution.

Solving Chess Tracking the world of competitive chess problem solving.

Torsten Lin Problem Chess Pages Problems and articles from noted composer. Some of the text in German.

Vincents Chess Problems Page Dozens of problems to pour over.

See the article here:
Chess Problems, Puzzles, & Compositions – Chessopolis

Grade 12 Geometry Problem From Australia. The Marshmallow Chocolate Dessert Volume. Sunday Puzzle

A marshmallow has the shape of a cylinder with a diameter of 5 and a height of 3.

A dessert is made with 24 marshmallows. There are two stacked layers, and each layer consists of 3 rows of 4 marshmallows each. The gaps between the marshmallows are filled with chocolate, as shown in the diagram.

marshmallow-area-problem-from-grade-12-australia-test-diagram

What is the volume of chocolate in the dessert? This problem was asked to grade 12 students in an Australian test. Give it a try, and watch the video for a solution.

Can You Solve This Grade 12 Geometry Problem From Australia? The Marshmallow Chocolate Puzzle

Or keep reading.

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Answer To The Marshmallow Chocolate Dessert Volume

The volume of chocolate is equal to the area of a cross section times the height.

Volume = (area of cross section) × (height)

The height of chocolate is equal to the height of 2 marshmallows, so the height is equal to 6.

Volume = (area of cross section) × 6

The area of a cross section is equal to the 6 gaps between the marshmallows, so the volume formula is directly related to the area of a single gap.

Volume = 6 × (area of a gap) × 6
Volume = 36 × (area of a gap)

What is the area of the gap? It is the area between four circles. As each marshmallow has a diameter of 5, each circle has a radius r = 5/2.

Now draw a square connecting the centers of the four circles. The side of the square, s, spans the radius of two adjacent circles, so the side has a length equal to the diameter, 5, of a circle.

The area of the gap is then the area of the square minus the area of 4 quarter-circles. But 4 quarter circles together equal a single circle. Therefore the area of the gap is the area of the square minus the area of a circle.

marshmallow-problem-area-of-gap

area of gap = s2 – πr2
area of gap = 25 – 25π/4

This can be substituted back into the volume equation to arrive at the answer.

Volume = 36 × (area of a gap)
Volume = 36 × (25 – 25 π/4)
Volume = 900 – 225 π ≈ 193.1

Sources
http://www.smh.com.au/nsw/hsc-2016-mixed-verdict-on-maths-exam-from-students-but-formula-reference-sheet-a-bonus-20161021-gs7inz.html

ATAR Notes
http://atarnotes.com/forum/index.php?topic=168134.0

The formula volume = (area of cross section) x height is an example of Cavalieri’s principle
https://en.wikipedia.org/wiki/Cavalieri%27s_principle

Continued here:
Grade 12 Geometry Problem From Australia. The Marshmallow Chocolate Dessert Volume. Sunday Puzzle


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