wolfgang
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Posts posted by wolfgang
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Hint..... you can use a ribbon to make the circles.
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a three dimentional figure, two matches makes a cross and the third one crosses them through the centre perpendicularly,in this case we will have 12 right angles
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according to a rule which states that,In case of equal amounts, if the bases are equal then the powers are equal too!!
1^1= 1^2
Thus: 1=2
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It should be only ONE square, and you don`t need an extra line.
If I can use a vertical line with them, there are two squares I can make: 81 and 100,
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1080 degrees= 12 right angles....
so we can make a triangle having three crossed sides and forming 12 angles having a total of 1080 degrees...so are equal to 12 right angles. -
Can you make a square using two circles?
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Yes...thats right...Thanks
If the fuel is used for guidance and anti-explosibility measures only, and not for propulsion; then all five spaceships could be flung from the moon towards Mars starting at the same time. All the ships would be tethered in such a way that adjusting the course of one would alter the course of the fleet as a whole. For the first quarter of the trip, Spaceship #1 does all of the course correction. For the second quarter, Spaceship #2; and so on until landing on Mars. Once on Mars, Spaceship # 5 redistributes it's fuel equally between all of the spaceships to power their lasers and defeat the Martians.
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Dear Bushindo....you should notice that I have concidered the moon as our base station(!) and not the earth.So you should think a little bit (laterally),and try to reduce the number to 5 ships only!...note : all of them will reach mars
Well...9 ships are too much...try to reduce the number
I don't think the number of ships can be reduced below 9. Here is why
Let's say that the moon is at location (x = 0), and Mars is at (x=1). Each ship can travel a distance of (1/4) on a full tank.
The moon base has unlimited fuel, so if we only have 1 space-ship, the max range we can reach (and return safely) is (1/4)*(1/2) = 1/8.
We can imagine that each spaceship we use can extend the fuel capacity of the base by a distance up to (but not including) 1/8. For instance, at a lesser distance of (1/9), a single spaceship can take as many trips as required to shuttle back and forth to deposit fuel at location (x=1/9). Each trip the ship can deliver 1/36 of a tank, but we can take as many trip as necessary. Note that the single ship *cannot* extend the fuel-supply to the precise distance of (1/8), as it would be not able to deposit any fuel at (x=1/8) and return home safely.
So, if a single space-ship can extend the fuel-supply line by 0 < d < 1/8, then we would need minimum of 8 ships to extend the fuel-supply line past (x=7/8). We then need 1 more ship to go to Mars and return. That makes 8+1 = 9 ships. -
Well...9 ships are too much...try to reduce the number
I want the fewest number of ships required , and at least one of them should launch with enough energy keeping it working.
If you want the number of unique ships required, then
You would need 9 different ships. Let's label the ships 1-9. Let the distance to Mars be 1 unit. Let the base space station be located at (x=0) and Mars be at (x=1). The idea is as follows
1) Let ships 2-9 go to the location (x = 1/9) and stay there. Ship 1 would then shuttle back and forth between the starting point (x = 0) and (x = 1/9) to deliver gas to the ships 2-9 until all 8 ships are full of gas. Each trip ship 1 can deliver 1/36 of a tank.
2) Let ships 3-9 go to the point (x=2/9). Ship 2 now shuttles back and forth between (x=1/9) and (x=2/9) to deliver gas to ships 3-9 until all are full of gas. Each trip ship 2 will have to wait at (x=1/9) for ship 1 to deliver gas to it, and then it can deliver (1/36) of a tank at (x=2/9).
3) Ship 4-9 will then proceed to point 3/9, and then ship 3 shuttles back and forth as above. Eventually ship 9 will be able to reach Mars and go back. -
The thing about the space travel is that it doesn't require fuel to cover the distance. The fuel is required to break free of the gravity and to establish the direction and speed. Once you're far enough from any large celestial bodies you can cover arbitrarily large distances without any fuel consumption. So the fuel consumption is not proportionate to the distance, but is proportionate to the masses of the celestial bodies you're launching from or you're in gravitational field of.
Anyway, assuming that the fuel is consumed at a constant rate for the distance travelled I have a couple of questions:
1) Does the spaceship that reaches Mars also have to return to the station?
2) Can the spaceships that return to the base be used again, and if so, then do we count the number of spaceships used or do we count the number of spaceship launches?
k-man makes a good point. For the sake of the puzzle, let's assume that the fuel spent is proportional to distance travelled. Let's also assume that the ship that gets to Mars will need to return home.
It's not clear whether the OP wants the minimum unique number of ships required or the minimum unique number of ship launches.
Let's assume that we want the minimum unique number of ships (a ship can launch more than once). I'm getting an answer of 9 ships required if we allow a ship to turn off its engine and wait for fuel; see below for an outline of the solution. This will take a *long* time, though.
Let's assume that we want the minimum number of ship launches (also the one that takes the least time to get to Mars). Here are where things get interesting.
Let the distance to Mars be 12 unit. On a full tank a ship can travel 3 units. We define a function s(d) as the number of unique launches required to put 1/3 gas tank at distance d (and have all the ships return home to base). It is easy to see that
s(1) = 1
s(2) = 3
s(3) = 9
And in general,
s(N) = 2 * ( s( 1) + s( 2 ) + … s( N - 1 ) ) + 1 = 3(N - 1 )
The above is derived by assuming for distance N, there is a special ship designated to reach distance N. At every integer distance less than N on the trip out and on the trip back, it gets (1/3) of a tank from a supporting network on other ships.
So, if we want to put (1/3) gas tank on Mars, then we would need 311 launches. Since we don't want to put any gas on Mars, we would only need 2*310.
I want the fewest number of ships required , and at least one of them should launch with enough energy keeping it working.
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The thing about the space travel is that it doesn't require fuel to cover the distance. The fuel is required to break free of the gravity and to establish the direction and speed. Once you're far enough from any large celestial bodies you can cover arbitrarily large distances without any fuel consumption. So the fuel consumption is not proportionate to the distance, but is proportionate to the masses of the celestial bodies you're launching from or you're in gravitational field of.
Anyway, assuming that the fuel is consumed at a constant rate for the distance travelled I have a couple of questions:
1) Does the spaceship that reaches Mars also have to return to the station?
2) Can the spaceships that return to the base be used again, and if so, then do we count the number of spaceships used or do we count the number of spaceship launches?
Thanks...I know that the space travel doesn`t need fuel,but I wanted to be a kind of energy used by the machine to keep it working.
for your 1st question: The spaceship or ships reaching mars should have an amount of energy keeping them doing their mission.
2- Yes, the spaceship that returns can be sent again and again, so I need number of ships used.
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Suppose that we have on the moon a space station which is the only source of fuel and on it there are unlimited number of identical spaceships, We want to send one of them to Mars, each spaceship has a fuel capacity to allow it to fly exactly 1/4 way to Mars (without fuel the spaceship will miss its direction and may explode!).
Each spaceship has the ability to refuel by a second spaceship through a specific connection between them without loss of speed . What is the fewest number of spaceships necessary to accomplish this mission without losing any one of them?Note:1- Each spaceship must have enough fuel to return safe to the base space station .
2- The time and fuel consumption of refueling can be ignored. (so we can also assume that one spaceship can refuel more than one spaceship at the same time).
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238 (2x=476, 4x=952)
It was surprising how small the space of answers was. I was afraid this was going to be an Excel search through a large number of numbers, but logic could rule out a lot of them. Neat puzzle, Wolfgang!
Thanks my friend
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Take the numbers 2 to 9 to make three numbers each of three digits, so that the first number is twice the 2nd number , and the 2nd is twice the 3rd number.
Each number should come only once except for one number should come twice.
Zero is not included.
what are the three numbers?
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<1st7 <2nd4 <3rd2 <4th8 <5th1 <6th6 <7th5 <8th3 row order and the number on their backsTable____________________________________________________[1] [2] [3] [4] [5] [6] [7] [8] numbered square orderr8b3c4 row order the nearest to numbered square the earliest to put mini cuber7b5c6r6b6c5r5b1c2r4b8c7r3b2c1r2b4c3r1b7c8___________________________________________________________The 8th in row must place the cube as to hint his row order and the back number he see (nearest to number squared in front of square 5).The 7th in row will know his back number and cube number from previous cube position...6th, 5th, 4th,3rd & 2nd in row will know too.The 1st in row will know too and since square [3] has nothing in front, it is the 8th in row's back number and his cube number must be 4..All info complete..he place those mini cubes on numbered squares.Note: back1 cube2, back2 cube1, back8cube7, back7cube8, back3cube4, back4cube3, back5cube6, back6cube5
Very nice trial....I had another approach, but yours is ok and clever
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The way I'm reading the problem it seems that row order is random, i.e. the last person does not need to have 8 on their back, so...
Due to the constraints of the problem (each of 1 to 8 must be used once and exactly once on the cubes and each back number must have +/- 1 the cube number), there is only one possible one the cubes can be allocated to the numbers, as TSLF specified previously. So if you know the person's back number, you know the person's cube number.
So the plan is this:
Calling the dimension parallel to the numbered squares X, and the dimension perpendicular Y, the nth person who enters places
their cube at Y=n/8 the distance from the row of numbered squares to the edge of the table, at X=the number he sees on the person in front of him or the number he knows they must have on their cube (they agree on which before hand). This will 'graph' the numbers, and the last person will be able to figure out where to place each box fairly easily by 'reading' it, and figure out which number the last person had by process of elimination.
The OP should begin with the last one,who can see the number infront of him,but still don`t know his own number.
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If the grid form may not be used due to a lack of table space
If the orientation of the cube is known 24 states can be represented.
6 sides x 4 easily assigned orientations = 24.
This can be achieved with a dot in one corner of one of the faces on the cube or similar marker (such as noting were the hinge is).
Whilst this on it's own cannot represent the 56 states required to establish a complete map there are needs to still be enough room on the table for 7 cubes outside the 8 square receptacles.
These can be assigned the numbers 1-7 from left to right and top to bottom.
If there is less than 7 non receptacle spaces, the cubes need to be stacked.
This brings the total number of states that can be represented to 168 (much more for stacked version), more than enough without having to worry about grids with correct spacing.
The cubes are not allowed to be marked.
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o o o......o o o
o....o......o....o
o o o......o o o
o....o......o....o
o o o......o o o
Left........Right
I gave to each circle( from left to right) a letter, so the most upper left is A,then the next is B,then C........D E F
like this:
A B C.....D E F
G....H.....I......J
K L M.....N O P
Q....R....S......T
U V W...X Y Z
If I want to write 1 using the left side,it will be: AGKQU
and by using the right side it will be: DINSX
to write 2 with the left side,it will be: ABCHMLKQUVW
and so on...
Thanks
Nice puzzle, left & right using 13 each (complete alphabet) . How you say 3 ,4 & 8?
Thanks
...to write 3 to the left: ABCHKLMRWVU...and 4 to the left:AGKLCHMRW......8 to the left:HCBAGKLMRWVUQ... -
Eight persons having numbers on their backs( 1 to 8) were standing randomely behind each other in a straight raw( so that the first person can see nothing,while the others,each one can see only the number of the person standing infront of him), each one has a mini cube in his hand containing one number,from 1 to 8 at random, no one of them know neither the number on his back nor the number inside his cube, but the number inside the cube for each person is either lager than or smaller than the number on his back by 1.
They have to enter another room,starting from the last one,,, in the other room there is a rectangular table drawn on it 8 squares in a straight raw, in each square there is a number, 1 to 8.
Each person when enters the room ,he should put his cube on the table( according to previous plan ),and get out through another door, and only the last person is allowed to put all the cubes inside the squares so that they should match each other,i.e. the 1st square should have number 1, 2nd square number 2...and so on.
Any kind of comunications between them is forbidden.
They are allowed to set a plan before starting the mission.
What is the PLAN ?
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o o o......o o o
o....o......o....o
o o o......o o o
o....o......o....o
o o o......o o o
Left........Right
I gave to each circle( from left to right) a letter, so the most upper left is A,then the next is B,then C........D E F
like this:
A B C.....D E F
G....H.....I......J
K L M.....N O P
Q....R....S......T
U V W...X Y Z
If I want to write 1 using the left side,it will be: AGKQU
and by using the right side it will be: DINSX
to write 2 with the left side,it will be: ABCHMLKQUVW
and so on...
Thanks
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1 to the left=AGKQU
1 to the right= DINSX
2 to the left= ABCHMLKQUVW
2 to the right= DEFJPONSXYZ
7 to the left= ABCHMRW
7 to the right= DEFJPTZ -
If...
AGKQU DEFJNOPTZYX = 13
AGKLCHMRW DEFJPONSXYZ=42
CBAGKLMRWVU DINSXYZTPJFE=50
CBAGKQUVWRML NIDEFJPTZ=67
..
..
..
Find...89
Note: no mathematics are needed
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Capture the fish inside an open, buoyant container (something like a large pot with some inherent buoyancy), so that the fish is contained within the pot which is now floating in the aquarium. Measure the water level both inside and outside the pot. Release the fish, and fill the pot with aquarium water until it sits as deeply in the water as before (the "outside" water level is the same). Note the water level within the pot again, and how this differs from the interior water level as measured with the fish. You can determine the weight of the additional water by calculating its volume (current interior water depth minus interior water depth with fish included, multiplied by the cross-sectional area of the pot...) and multiplying by the known density of water. This is the weight of the fish.
A very good idea...
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Let it inside a plastic bag, tie a string on the opening (let no air),
fish in the bag of water is heavier than the bag of water without the fish.
Tie the end of string to a weigh hook with the fish bag fully submerged.
Yes...thats was my idea...thanks
clever Timmy
in New Logic/Math Puzzles
Posted · Edited by wolfgang