Francais
........... Enigms
The "pertuisanier"
225533 can be decomposed in primes numbers as following: 101 x 11 x 7 x 29The only reasonable repartition is then:
- 11 = half the age of the captain
- 7 = length of a "pertuisane" in feet
- 29 = last day of February of the bissextile year 1916
- 101 = the quarter of the difference between 1916 and the year of the "pertuisanier" death.
Thus this year was 1512 and there was in this year a famous battle in France: the battle of Ravennes, where were killed many "pertuisaniers" but also their captain whose name was Gaston de Foix, 22 years old, nephew of ...
The french wine bottle of 25 FF
Each of the three men has paid 9 FF and the 2FF on the table must be substracted not added to 27 FF. And we find again the price of the wine bottle: 25 FF.4 triangles with 6 matches
You must think to the third dimension: a tetraedre is composed of 4 triangles and needs 6 matches.6 triangles avec 6 allumettes
This time you must think to the roman writing:
\ / | /\
\ / | / \
\/ | /____\
8 triangles avec 6 allumettes
Still you need to add a dimension. It is obtained thanks to a mirror. Put on a mirror the tetraedre made with 6 matches and you will see 8 triangles.Still in the matches land
Counting the number of squares, don't forget the squares which contains other squares. Then, the answers are almost as simple as the figures below.
A hole through a sphere
The answer depends only at the fourth order of ratio of the hole diameter and the hole length. Intuitively, we can already see that the larger is the hole, the less longer it is. Taking a hole of diameter zero, that is a hole of 6cm through a sphere of 6cm diameter, we find the volume 4/3*pi*27 cm3.Some calculus to convince you:
If r is the radius of the hole, R the radius of the sphere and 2L the length of the hole, one can write: R*R = L*L + r*r
The volume of the sphere is V = 4/3 * pi * R*R*R and the volume of the hole is v = 2*L * pi*r*r, while the volume of the spherical hats removed by the creation of the cylindrical hole is vc = 2*pi*h*h*(R-h/3).
The searched volume is then V-v-vc = 4/3 * pi*L*L*L which is independent of r, so independent of the hole's diameter.
(Thank you to Nicolas Gufflet who corrected my mistake).
The mathematical painter
Mr Math will reach the roof, because the serie 1+1/2+1/3+... does not converge. The length of the pile is infinite. But Mr Math will be able to paint his cubes (in an infinite time!) because the total surface to paint is6 *(1 + (1/2)2 + (1/3)2 + ...) cm2.
that is pi2. The surface of the pile is finite. A small amount of paint is needed.
Logic of numbers
The logical raw of numbers: 1, 11, 21, 1211, 111221, ... can be very well understood if it is said instead of read. One 1 (thus 11), Two 1 (thus 21), One 2 One 1 (thus 1211), and so on...The following number of the raw is logically 312211.
The logic of letters
The logical raw of letters: O, T, T, F, F, S, S, ... , ... is of course the raw of the first letters of the corresponding numbers: One, Two, Three, Four, Five, and so on...The two following letters are thus E for Eight and N for Nine.
The three sisters
The product of the three sisters ages is 36. It lets many possibilities like (6,6,1) or (9,4,1) or (3,3,4) and so on...But the sums are all different except for (9,2,2) and (6,6,1) whose sum is 13. Since the factor can not find the answer after having seen the number of the front house, we deduce that this number was 13. The additional information "the elder is blond" allows the factor to deduce that the twins are the cadets and that the solution is (9,2,2). The daughters of Mr Dupont are 9 years, 2 years and 2 years old.
Age and age
If a=my present age and b=you present age, I can say that:a = 2(b+b-a)
a+(a-b) = 2b
The only solution is a=b=0. Of course, with such tortuous sentences...!
The fly and the spider
In 5 minutes, the spider can run 0.5 * 1000 * 5 / 60 = 500/12 = 41.65 meters.
If it goes on the deck, in the median plane of the box,
the spider must run 42 meters. But, unfold the box as shown on the figure,
the six faces on the floor. Then, you can see immediately the shortest
way that the spider can use. And Pythagore says to us that the length
of this way is the square root of
242 + 322, that is 40 meters.
This is shorter than the straight line first seen (42 meters) and this is
a way that the spider can travel in less than 5 minutes.
A single fly
The two trains run over 100 km at a speed of 50 km/h. Then, they meet after 2 hours. During this time, the fly has flied over 150 km at the speed of 74 km/h. You do not have to compute a serie.Money is not Gold
The solution is simple, as often in such type of puzzle. You just need to think about it! You take a coin in the first pack, 2 coins in the second pack, 3 coins in the third one, etc... And you weight the sum. Il there were no false coin, you would obtain the sum of an algebraix serie, that is: 500x501/2 times 10 grams. But you find this number minus n times 5 grams. And you can conclude that the pack number n contained false coins. Cqfd!The only constraint is to have less packs than coins per pack, which is a common case with avarious people!
If you have a lot of coins in each pack, there are many other solutions, but more complicated.
For instance, you can take 1 coin in the first pack, 2x1 coins in the second pack, 3x2x1 coins in the third one, etc...
Or you can take 1 coin in the first pack, 2 coins in the second pack, 4 coins in the third one, 8 coins in the fourth one, etc...
A train exchange
Let's note 1,2 and 3 the initial positions of wagon A, wagon B and locomotive.
The locomotive takes B and brings it to 3. Then it passes the tunnel and takes A.
It brings A to 3 and takes AB. It brings AB to 2. The locomotive brings then A back to 3,
brings B to 1 AND IS STOPPED BETWEEN TUNNEL AND POSITION 1)
and brings A to 2. Then, the locomotive comes back to its initial position 3.
If you find a solution, let me know...
A strange man
The man is simply too small to reach the 25th floor button of the lift.The logic of Protagoras
The teacher and the student are equally right. It is an undecidable assertion (thank you Mr Godel!). But you could give the preference to the student because Protagoras, great master of logic was not able the preview that he could be at the origin of the first suit of his student et go then into this greek paradoxe.The crossed laders
I could say only that 3/2 is greater than sqrt(2). For now, I only obtained a fourth order equation. The point O, intersection between the two laders belongs to the two straight linesy = sqrt(4-a*a)/a * x and y = -sqrt(9-a*a)/a * x + sqrt(9-a*a) from which we cqn deduce the coordinqtes of O:
xo = a / (1 + sqrt((4-a*a)/(9-a*a)))
yo = sqrt(4-a*a) / (1 + sqrt((4-a*a)/(9-a*a)))
But, the problem said that yo=1. After some calculus and taking X = 4 - a*a, we find:
X*X*X*X + 6*X*X*X - 5*X*X -50*X + 25 = 0
which has two solutions: X1 = 0.4912 and X2 = 2.48418.
The only solution which gives yo = 1 is X2 which gives a = 1.23118.
The logic of mathematics is crazy
When I will get the puzzle, I will give you the answer...Improbables probabilities
If F(b) = P(X<b) and 1-F(a) = P(X>=a).Then: P(a<=X<b) = F(b)*(1-F(a)).
Here is the mistake! In fact, the probabilities multiplication is valuable only for uncorrelated variables. This is not the case of F(b) and F(a) when a tends to b.
The speed of snails
After one second, the end of the elastic has runned over 1 km and the snail has runned over 1km+1mm (since he is on the elastic). Thus, every second the snail diminishes its late. After 1 million of seconds (about 11.6 days), the snail will have reached the elastic end.Wine and water
If a=water and b=wine, at the end of the operation you have in your glass: (10a+3b)x10/13and your friend has in his glass: 7b+(10a+3b)x3/13.
That is: 100a/13+30b/13 for you and 30a/13+100b/13 for him.
You amount of wine is equal to his amount of water.
A camel story
You come with your camel and you put it with the 11 other camels. Then, you have 12 camels and all is solved. The elder gets 12/2 = 6 camels, the cadet gets 12/4 = 3 camels and the youngest gets 12/6 = 2 camels. 6+3+2 = 11 camels is OK. There is still one camel, yours. You go back with it, smiling!Black and white hats
(thanks to Frederic Faure)Let's call W for white and B for black. If the last of the column, who can look the two others, could have answered, we could then have conclude that the two hats before him were black. This is not the case, so the available solutions are: WB, BW or WW.
But, the second man was not able to answer too. This is because of the ambiguity between the solutions BW and WW. Because the he can not see his own hat that can be B or W. This excludes the solution WB.
By this way, the first man of the column knows that his hat is white.
The two locomotives
At point A, the locomotive can never meet. At point B, the two locomotives are able the meet. Already after one rotation, they encouter if D=R. More generally, we find that the only solution is D=R. Then, locomotives encouter for all the couples of integers N and N' verifying 3*N'-2*N = 1.Au point A: Si tN et tN' sont les instants de passages des locomotives 1 et 2 au point A, et si V1 et V2 sont les vitesses des locomotives 1 et 2, on a:
tN = N * 2*pi/(V1/R)
tN' = N' * 2*pi/(V2/R) + 2*(pi-alpha)/(V2/R)
et il y a collision si et seulement si il existe N et N' tels que tN = tN'
c'est a dire: N*V2 - N'*V1 = V1 * (1-alpha/pi)
soit encore: 3*N - 2*N' = 2 * (1-alpha/pi)
Il faut donc que 1-alpha/pi soit un demi-entier k/2.
Mais on sait que 0 < alpha < pi/2 donc 1 < k < 2
Les seules solutions possibles sont k=1 (deux cercles superposes) ou k=2 (cercles tangents).
At point A: If tN and tN' are the instants of passage of locomotives 1 and 2 at point A, and if we note V1 and V2 the speed of the locomotives 1 and 2, we get:
tN = N * 2*pi/(V1/R)
tN' = N' * 2*pi/(V2/R) + 2*(pi-alpha)/(V2/R)
and there is collision if and only if there exists N and N' verifying tN = tN'
which means: N*V2 - N'*V1 = V1 * (1-alpha/pi)
that is: 3*N - 2*N' = 2 * (1-alpha/pi)
Then 1-alpha/pi must be an half integer k/2.
But we knoz that 0 < alpha < pi/2 then 1 < k < 2
The only possible solutions are k=1 (two circles superimposed) or k=2 (tangent circles).
Mice in flat land
(thanks to Jerome Damet)There are many complicated solutions, but the only simple one that does not need general relativity is that the third mouse is always lying.
Ah, ah!
The english cochonnet
(thanks to Jerome Damet)Do not search too complicated! This game can be played only after one has said 1,2,3! May be there is a third player, but we do not know what he does!
The square of the fool
To solve the puzzle, you must go out of the space in which your mind is closed. It was never said that the lines should stay within the square. Here is the solution:
. . * . * . *
. . .
* * *
. .
* * *
. .
.
The circle of the fool
You must use one more dimension than the common two dimensions of the paper space.
Take the paper and fold it on 1/4th of its length. Mark the center of the circle
just close the paper border if the folded part. Then draw what you want on the
folded part until you rech the border point were you will begin the draw
the circle on the unfolded part. Draw half of the circle, unfold the paper and
draw the other half of the circle.
Additions with letters
It asks for some time, but one can solve it, rejected the impossible solutions one after an other.
123416
+ 123416
+ 123416
+ 134 8324
+ 134 + 0913
---------- -------
370516 09237
A coins game
The magician question was: "How many results can you obtain?..." The number of results obtained is effectively 10 but the number of results you CAN obtain is 11.Dwarfs and Giants
The first problem is solved by the 3 following moves:
GpGpG --> G..pGpG --> GGpp..G --> ppGGG
-- -- --
The second problem is more difficult because among the previous 5
allowed moves, the GG and pp moves are not allowed.
If you have the answer, write to me!...
.. Thanks to C. Baro for the answer below which let place couples Gp
on non standard locations.
GpGpGp --> GpG..Gp --> G..GpGp --> GpGG..p --> GGGpp
-- -- -- --
The fool of the king
This puzzle is better to say than to write...The first sentence must be pronounced "the fool of the king is thirty fools". Then, you find that there was 16 "fool" and "seals": the thirty "fools" of the king, the seal of the king and the two last fools: the fool of the king and you. But since you found the puzzle, there is one less fool: that makes 15 "fools and seals"!
Some clever ratios
All is based on french words... Knowing that the SI simplify, that SOL FE RINO (SOL makes RINO) and that RINO CE ROS (RINO is ROS), you find:ROSSINI NI ------- = ---- = 1 since NI VO DO (NI equals DO). SOLSIDO DOIn the same way, noting beta the greek letter, knowing that OISEAU = beta L (animal with wings), we can simplify by L. We gets then CHEVA/beta. But CHEVA is like VACHE (cow) and VACHE = beta PI (animal with pie). Then, simplifying by beta, we find:
CHEVAL ------ = PI OISEAUFinally, knowing that we can simplify by R, that V is nothing (V n'est rien) and that who says UMI says T (qui dit UMI dit T), one obtains:
VERT E ------- = ---- = CASSOULET KROUMIR KROsince RO s'biffe (RO can be slashed), it gives K under the E (CASSOULET)
A nice logic
The fact that P => Q allows to induce that no Q => no P has nothing to do in fact with the sentences given as example, because those sentences say only possibilities.A logical mistake
That ((P=>Q) ou (Q=>P)) is right whatever P and Q is quite remarquable, but very logic. No hypothesis is made on the rightness of P or Q. Yet, the following example shows that our langage keeps an ambiguity with respect to the interpretation of (P=>Q) by (noP or Q).Example: (the caravane passes, thus the dog calls) or (the dog calls, thus the caravane passes) is always right except if the dog does not call when the caravane passes, that is when P is right and Q is false.
An other logical mistake
In fact, the third excluded principle does not tell us that ( noQ => Q ) is always false. It says only that Q is only right or false. If Q is right, (noQ =>Q) is right!Example: Q="every second passes one second"... OK, an other example!
Example; Q="The sun shines", then you can say wihtout mistake that ("the sun does not shines, thus "the sun shines"). It is not very common by it is as right as "the sun shines".
Logic and kindness
Simple logic taken at its basement!...Any triangle is isocel
The showed figure is a common triangle where the red lines are the bissector
of the angle (AB,AC) and the median line of BC.
Their intersection is the point O and the orthogonals projections of O
on AC and AB are the points H and K. Then, we can already write
OH=OK and AH=AK (bissector) and OB=OC (median line).
Then, from Pythagore theorem, we can conclude that HC=KB.
Until this point, everything is fine... Since AH=AK and HC=KB, we conclude that AH+HC = AK+KB, that is AC=AB. Any triangle is isocel. Here is the mistake! By summing AH+HC = AK+KB we based on the false figure. In fact, H is out of the triangle and the algebraic length HC is negative. We must add the algebraic lengthes AH and HC. The same holds for AK and KB. We can not conclude then that AC=AB.
The irreductible triangle
... since an enternity, it should have been completed...An other irreductible triangle
... idem ...A strange triangle
The upper triangle hypothenuse is in fact composed of two segments with an angle between them going into the triangle.
The lower triangle hypothenuse is in fact composed of two segments with an angle between them going out of the triangle. The surface difference is the surface of the apparently missing piece.
The U2 bridge
Mr 1mn and Mr 2mn cross the bridge together: 2 minutes
Mr 1mn comes back: 1 minute
Mr 10mn and Mr 5mn cross the bridge together: 10 minutes
Mr 2mn comes back: 2 minutes
Mr 1mn et Mr 2mn cross the bridge again: 2 minutes
Total is: 17 minutes.
One climb for 3 lights
Switch on a button for 5 minutes. Then switch it off. Switch on an other button and go up to the first floor. The light bulb that is on corresponds to the switch just put on, the light bulb that is warm corresponds to the first switch you put on for 5 minutes, and the light bulb that is off correspond to the switch you never touched.The self-refering sentences
Non significant: "What more do you want me to say!"The strange little i
Of course, sqrt(a) is different from - sqrt(a). A function as the square root does not behaves the same way with real numbers and complex numbers.The imaginary number "i" can also be written exp(i*(pi/2+n*2*pi)) (some definition by itself!), whose square root is: exp(i*(pi/2+n*2*pi)/2)= exp(i*(pi/4+n*pi)). We have then two solutions, according to wether n is even or odd. In the same way, i*i*i*i=1 can be written exp(i*(2*pi+n*2*pi)) whose square root is exp(i*(pi+n*pi)), that is 1 if n is odd and -1 if n is even.
We just forgot this other solution, which is the good one:
sqrt(i*i*i*i) = 1 different from i*i.
Fraction is running
The mistake consists in writing the fraction (2/3) / (2/5) in the fromat ((2/3)/2)/5. The value 1/15 is thus false. It is one of the most frequent calculus mistakes.The king Pepin
I tell you the story in french and every time I say a letter, you slashes it:The ROI PEPIN without O (water), without R (air), without PIN (bread), without l'I (the bed) and wihtout the PE (the few) that rests, G putted in a corner (G mis dans un coin and "G mis" is (cries)).
The ill monks
The solution can be found by a recurrent way: Let n be the number of ill monks and x the number of days needed by a monk to realize that he is ill and to go away.If n=1: The ill monk observes during the priest that all the others are sane. Since there is al least a ill monk in the community, he concludes that he is ill and he goes away immediately. Then x=0.
If n=2: Suppose you are one the two ill monks. During the priest, you see only one ill monk. And you say to yourself that, since he is ill, he will go away and the day after you will not see him at the priest. By the day after, you see him again! The only solution is that there an other ill monk and that this fact allowed the monk you saw to suppose that he is sane. You conclude that you are this other ill monk and you go away. The same conclusion is in the mind of the ill monk and he goes away. Then x=1.
And you find, then that for n ill monks, x=n-1 days will be needed for all the ill monks to go away from the monastery.
Didier Verkindt