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100. We are looking for an equilateral dodecagon whose sides form right angles at all vertices.
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Riddles and Puzzles: Extend the following to a valid equation
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KnotTheorist
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My first guess at the problem was 107, but that can be permuted to 170 and 710, which are not prime.fresh_42 said:
I read the question as the smallest number that was both emirp and permutably prime.
KnotTheorist
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I do not have not the time nor the patience to write out the move set for completing this puzzle, even though it is one of my favorites. I own my own board and love to see people struggle with the challenge.
The penultimate move (of a fair and winning game) will always leave you in (5,10), (15, 16), (17, 18), or (23, 28). Thus you can either end the game with 1 peg in the center or on the respective edge.
The penultimate move (of a fair and winning game) will always leave you in (5,10), (15, 16), (17, 18), or (23, 28). Thus you can either end the game with 1 peg in the center or on the respective edge.
KnotTheorist
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A plus sign, just like the logo of the American red cross
KnotTheorist
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I believe the 4,294,967,295 -gon is the largest odd sided n-gon that can be constructed. It a Fermat number (2^2^5 -1), and it is the product of the primes 641 and 6700417.
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#99
Being constructible with ruler and compass translates into algebra as being constructible with a finite number of basic arithmetic operations and square roots starting with integers. Construction of a regular n-gon requires finding trigonometric-function values, and one may conveniently start with ##\cos (\pi/n)## and determine ##\sin (\pi/n)##, ##\cos (2\pi/n)##, and ##\cos (2\pi/n)## using basic arithmetic and square roots. One does so by solving ##T_n(x) = -1## where ##x = \cos (\pi/n)## and ##T_n(x)## is the Chebyshev polynomial of the first kind.
There is a theorem that this Chebyshev-polynomial equation can be solved with basic arithmetic and square roots only if the number of sides n has form ##n = 2^m \cdot p_1 \cdots p_k## where ##p_1 \cdots p_k## are distinct Fermat primes. They have the form ##F_n = 2^{2^n} + 1##, and only five of them are known, for n = 0, 1, 2, 3, and 4: 3, 5, 17, 257, and 65537. All the others, up to n = 32, are known to be composite, and it is not known whether or not there are any more prime ones.
Turning to the problem statement, n = 4,294,967,295, I find that it is 3 * 5 * 17 * 257 * 65,537. I used Mathematica's FactorInteger[] function to do the factoring. Since all five factors are distinct Fermat primes, it is thus possible to construct a polygon with that many sides with ruler and compass.
There is a theorem that this Chebyshev-polynomial equation can be solved with basic arithmetic and square roots only if the number of sides n has form ##n = 2^m \cdot p_1 \cdots p_k## where ##p_1 \cdots p_k## are distinct Fermat primes. They have the form ##F_n = 2^{2^n} + 1##, and only five of them are known, for n = 0, 1, 2, 3, and 4: 3, 5, 17, 257, and 65537. All the others, up to n = 32, are known to be composite, and it is not known whether or not there are any more prime ones.
Turning to the problem statement, n = 4,294,967,295, I find that it is 3 * 5 * 17 * 257 * 65,537. I used Mathematica's FactorInteger[] function to do the factoring. Since all five factors are distinct Fermat primes, it is thus possible to construct a polygon with that many sides with ruler and compass.
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#100
Make it in the shape of a plus sign:
Code:
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#
bluej
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#99 : I would simply draw a circle.
##\frac {2\pi} {4294967295 }## is an angle too small for the human eye to differentiate the polygon side from a smooth curve on a piece of paper.
Of course, using an abstract compass and straight edge and open intergalactic space - then the polygon side might not look like a smooth curve anymore, to the human eye, that is
##\frac {2\pi} {4294967295 }## is an angle too small for the human eye to differentiate the polygon side from a smooth curve on a piece of paper.
Of course, using an abstract compass and straight edge and open intergalactic space - then the polygon side might not look like a smooth curve anymore, to the human eye, that is
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101. At each station, tickets are sold to every other station. Then the route network will be expanded and 34 new tickets will be added. How many stations are there?
(99. was a joke.)
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(99. was a joke.)
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We go from n(n-1) tickets to (n+k)(n+k-1)=n(n-1) + 2kn + k2 - k tickets.
34 = 2kn + k2 - k
Solve for n: n = (34+k-k2)/(2k) implies k must be a factor of 34=2*17 and smaller than 7, that means k=1 or k=2.
k=1 leads to n=17: We add 17 tickets to the new station and 17 tickets coming from the new station.
k=2 leads to n=8. We add 8*2 tickets going to the new stations, 8*2 tickets from the new stations to old stations, and 2 tickets between the new stations.
8 or 17 existing stations.
34 = 2kn + k2 - k
Solve for n: n = (34+k-k2)/(2k) implies k must be a factor of 34=2*17 and smaller than 7, that means k=1 or k=2.
k=1 leads to n=17: We add 17 tickets to the new station and 17 tickets coming from the new station.
k=2 leads to n=8. We add 8*2 tickets going to the new stations, 8*2 tickets from the new stations to old stations, and 2 tickets between the new stations.
8 or 17 existing stations.
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#101
Each kind of ticket has a departure station and an arrival station, and they are different from each other. That means that the total number of tickets is ##N(n) = n(n-1)##. So we must find some old number of stations n1 and some new number n2 such that ##N(n_2) - N(n_1) = 34##. This equation becomes ##(n_2 - n_1)(n_1 + n_2 - 1) = 34##.
Since 34 is 1*34 = 2*17, I try each combination of factors. The two solutions are n1 = 17, n2 = 18 and n1 = 8, n2 = 10.
Since 34 is 1*34 = 2*17, I try each combination of factors. The two solutions are n1 = 17, n2 = 18 and n1 = 8, n2 = 10.
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102. A horse, a giraffe and an elephant have agreed to make a race. It's over ##1,000## meters, and only two animals compete against each other per run.
In the first race, the horse defeats the giraffe. The moment it crosses the finish line, the horse is ##100## meters ahead. Race two is won by the giraffe - ##200## meters ahead of the elephant.
Finally, the horse and the elephant compete against each other. How far will the horse be ahead when it crosses the finish line?
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In the first race, the horse defeats the giraffe. The moment it crosses the finish line, the horse is ##100## meters ahead. Race two is won by the giraffe - ##200## meters ahead of the elephant.
Finally, the horse and the elephant compete against each other. How far will the horse be ahead when it crosses the finish line?
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KnotTheorist
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Turning the problem into equations, we have
(1) 1000 = R_h * t_h
(2) 900 = R_g *t_h
(3) 1000 = R_g *t_g
(4) 800 = R_e * t_g
and we need to solve for x in these two equations
(5) 1000 = R_h * t_h
(6) 1000 - x = R_e * t_h
Dividing the 1st equation by the second and third by the forth, we get,
10/9 = R_h/R_g and 10/8 = R_g/R_e
Multiplying these together, we get 100/72 = R_h/R_e
Then dividing the the fifth equation by the sixth, we get,
1000/(1000-x) = R_h/R_e = 100/72 (from our previous ratio)
Thus 1000/(1000-x) = 1000/720
This x = 280 meters
(1) 1000 = R_h * t_h
(2) 900 = R_g *t_h
(3) 1000 = R_g *t_g
(4) 800 = R_e * t_g
and we need to solve for x in these two equations
(5) 1000 = R_h * t_h
(6) 1000 - x = R_e * t_h
Dividing the 1st equation by the second and third by the forth, we get,
10/9 = R_h/R_g and 10/8 = R_g/R_e
Multiplying these together, we get 100/72 = R_h/R_e
Then dividing the the fifth equation by the sixth, we get,
1000/(1000-x) = R_h/R_e = 100/72 (from our previous ratio)
Thus 1000/(1000-x) = 1000/720
This x = 280 meters
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103. Arrange ##1,1,2,2,3,3,4,4## in a way that between the ones is exactly one of the other numbers, between the twos two other etc.
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KnotTheorist
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104. How many different (connected) plane figures can be created with five squares?
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105. Why equal all sums of any diagonal which same number?
$$
\begin{bmatrix}
9&14&21&35&11&7\\
13&18&25&39&15&11\\
21&26&33&47&23&19\\
39&44&51&65&41&37\\
33&38&45&59&35&31\\
25&30&37&51&27&23
\end{bmatrix}
$$
(A diagonal of a matrix is a set where one matrix entry from each row and each column is chosen: ##a_{j \pi(j)}##.)
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$$
\begin{bmatrix}
9&14&21&35&11&7\\
13&18&25&39&15&11\\
21&26&33&47&23&19\\
39&44&51&65&41&37\\
33&38&45&59&35&31\\
25&30&37&51&27&23
\end{bmatrix}
$$
(A diagonal of a matrix is a set where one matrix entry from each row and each column is chosen: ##a_{j \pi(j)}##.)
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##a_{ij}=b_i+c_j## for tuples b,c. Any diagonal sums to ##\sum b_i + \sum c_j##.
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106. Someone has walked south for five kilometers, then five kilometers to the west, and finally five kilometers to the north, to return to their starting point. All the same, he was not at the North Pole.
Where else is this possible?
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Where else is this possible?
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SSequence
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South Pole? Like it seems that the second movement (the one in west) is sort of redundant (in the sense that it can be removed or the distance walked can be changed)? Or am I missing something too obvious?fresh_42 said:
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KnotTheorist
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He would be somewhere close to the south pole. There are actually a family of points that satisfy these conditions.
You start somewhere north of the south pole, travel south 5km, then walk around in a circle 5km such that you have moved an integer number of circumferences about your current great circle (you are back in the same spot after moving South) then you travel 5km North.
You start somewhere north of the south pole, travel south 5km, then walk around in a circle 5km such that you have moved an integer number of circumferences about your current great circle (you are back in the same spot after moving South) then you travel 5km North.
SSequence
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Nice point. Shouldn't there also be a possibility of other solutions if we are allowed to adjust the radius R (or the distance walked for that matter)?
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107. A quadrilateral piece of paper is cut into six pieces with two straight cuts. The paper is neither bent nor is it folded. In addition, the pieces of paper must not be rearranged or superimposed after the first cut.
How can that be?
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How can that be?
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fresh_42 said:
quadrilateral ABCD where AB and CD intersect in their interiors?
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A double triangle merged in one corner?jbriggs444 said:
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Yes. But I see that it can also be achieved with a V-shaped quadrilateral.fresh_42 said:
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There is also a solution with a proper quadrilateral.
KnotTheorist
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108. We are looking for the smallest eight digit natural number, where all digits are different, and the number is divisible by ##36##.
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OK I think my answer in post#440 is supposed to be wrong (because I was confusing how the directions are supposed to work). But there is still something that is bit ambiguous for me.
Suppose you were at a distance of 5km from southpole. You move 5km to south and reach the southpole. Now is the movement of "5km to the west" supposed to keep you at the same point? I would guess, so in a way since for all other points (except northpole) the movement towards west is making one move in a circle?
If the convention in previous paragraph is supposed to hold, then the distance of 5km from southpole should also be a solution. Right?
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Also, another small question (for my own understanding regarding directions). If I am at southpole then what does movement such as "5km towards south" would mean?
Suppose you were at a distance of 5km from southpole. You move 5km to south and reach the southpole. Now is the movement of "5km to the west" supposed to keep you at the same point? I would guess, so in a way since for all other points (except northpole) the movement towards west is making one move in a circle?
If the convention in previous paragraph is supposed to hold, then the distance of 5km from southpole should also be a solution. Right?
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Also, another small question (for my own understanding regarding directions). If I am at southpole then what does movement such as "5km towards south" would mean?
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