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I'm not sure if I understand what you're saying here keith.
4/3 x pi x (h/2)^3 = pi x h^3 / 6
So your initial answer was definitely correct.
I'm not sure if I understand what you're saying here keith.
4/3 x pi x (h/2)^3 = pi x h^3 / 6
So your initial answer was definitely correct.
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Yes, I'm confused too. Both formulae are the same
Sorry - meant to type 4/3 x pi x r ^ 3. In my 1st example r did equal h/2.
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E & 9. The other two cannot possibly violate the stated rule.
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As it could be an optical illusion, the easiest way to check is to look at the image type. It is a JPG, so the images aren't moving at all. It would need to be a GIF or some other non-static image type to actually have the wheels moving.
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Correct.
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It is an optical illusion ... most people look at "A" and notice that "B" is rotating and vice versa. Peripheral vision causes this phenomenon. It is widely accepted by most psychologists that if both wheels are spinning together you are clinically insane.
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Firstly treat the board as a 33 X 33 groups of 3 X 3. (i.e. 1089 groups of 9 squares)
(i) If there are no non-empty groups place a checker in the last available square of a group, otherwise place the checker in the central square of an empty 3 X 3 group. A further group has now been eliminated from play leaving an even number of playable groups.
Whenever the opponent responds by placing a checker on a central square within another empty group, repeat step (i).
(ii) Whenever the opponent places a checker on anything other than the central square within one of the vacant 3 X 3 groups, place a checker a chess knight's jump away ("L" shape) to the opponent's checker within that same group. This will leave exactly one more play within that group ensuring that an even number of groups remain in play.
As there are 1089 groups of 3 X 3, strict subscription to the aforementioned rules will result in the commencing player always having the last available move.
(i) If there are no non-empty groups place a checker in the last available square of a group, otherwise place the checker in the central square of an empty 3 X 3 group. A further group has now been eliminated from play leaving an even number of playable groups.
Whenever the opponent responds by placing a checker on a central square within another empty group, repeat step (i).
(ii) Whenever the opponent places a checker on anything other than the central square within one of the vacant 3 X 3 groups, place a checker a chess knight's jump away ("L" shape) to the opponent's checker within that same group. This will leave exactly one more play within that group ensuring that an even number of groups remain in play.
As there are 1089 groups of 3 X 3, strict subscription to the aforementioned rules will result in the commencing player always having the last available move.
I treated the board as two identical halves - either 2 triangles or 2 rectangles with a line of squares along the axis.
i) choose a line of symmetry (either horz, vert or a diagonal)
ii) place the 1st checker in the centre on the board.
iii) whenever the opponent places a checker NOT on your chosen line of symmetry, you place yours in the symmetrically opposite position.
iv) whenever the opponent places a checker on the line of symmetry, you place yours also on the line of symmetry, but the same distance away from the centre as the opponents.
TL;DR;
The board must always have reflective symmetry. After taking up the only square (the centre) that cannot copied, you simply copy the opponent in the opposite half of the board.
I guess you could also use rotational symmetry.
i) choose a line of symmetry (either horz, vert or a diagonal)
ii) place the 1st checker in the centre on the board.
iii) whenever the opponent places a checker NOT on your chosen line of symmetry, you place yours in the symmetrically opposite position.
iv) whenever the opponent places a checker on the line of symmetry, you place yours also on the line of symmetry, but the same distance away from the centre as the opponents.
TL;DR;
The board must always have reflective symmetry. After taking up the only square (the centre) that cannot copied, you simply copy the opponent in the opposite half of the board.
I guess you could also use rotational symmetry.
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I haven't worked through Cynic's detailed answer yet, but Keith's answer is the one I know that will ensure a win.
You must place the first draught (should I be calling them checkers?) in the centre square and then you place all remaining draughts symmetrically opposite what your opponent played. I worked on the basis of diagonally opposite (e.g. if your opponent placed a dart 6 up and 3 across to the right from the centre piece, you place yours 6 down and 3 to the left from the centre piece). You could also, as Keith suggested, place the draughts horizontally or vertically opposite, with the adjustment he suggested for draughts placed on the line of symmetry. Doing it diagonally opposite does that automatically without the need to state that special case.
Once you place the centre draught, you are effectively then imitating on one side of the board what your opponent has done on the other side. And since after you place your piece, one side of the board will be a reflection of the other, then by definition, if your opponent is able to place a piece on the board that abides by the rules, you will be able to do so as well as the same symmetrically opposite position will also be available to you. Your opponent will always be the first to encounter a situation where he cannot place a piece, so you will always win.
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I've just noticed a flaw in my method which would definitely require additional rules (adding to its complexity) in order to ensure success.
Keith's answer is ,of course, a far less complex and more sound approach to this problem.
(Congratulations again Keith!)
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They chose the numbers 6 & 7.
On day 1 (when nobody leaves) they realise that neither could have chosen 11-13 because they are all greater than 10 meaning that the other could be determined by subtraction from 13.
After day 2 they know that nobody could have chosen 0-2 as these can only contribute to a sum of 10 now that 11-13 have been eliminated.
After day 3 they know that nobody has chosen 8-10 as this can now only contribute to the sum of 13 since 0-2 have been eliminated.
After day 4 they know that nobody has chosen 3-5 because these can now only contribute to a sum of 10 since 8-10 have been eliminated.
This leaves only the combination 6 & 7.
On day 1 (when nobody leaves) they realise that neither could have chosen 11-13 because they are all greater than 10 meaning that the other could be determined by subtraction from 13.
After day 2 they know that nobody could have chosen 0-2 as these can only contribute to a sum of 10 now that 11-13 have been eliminated.
After day 3 they know that nobody has chosen 8-10 as this can now only contribute to the sum of 13 since 0-2 have been eliminated.
After day 4 they know that nobody has chosen 3-5 because these can now only contribute to a sum of 10 since 8-10 have been eliminated.
This leaves only the combination 6 & 7.
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Congratulations! Perfect answer.
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2,3,4 & 5.
There cannot be more than 2 yellow marbles without the sum exceeding 17. So it is known that there are either 1 or 2 yellow marbles.
The fact that Mary asked whether there was more than 1 yellow indicates that Mary has noticed that the house number can be produced from valid configurations containing either number of yellow marbles.
Therefore the house number is 120 as this is the only product that occurs for valid configurations containing exactly 1 yellow and also for a valid configuration containing exactly 2.
The fact that Mary instantly knew from the Father's reply indicates that the number of yellow marbles must be 2 (had there been 1 then Mary would have needed further information in order to to decide between the 2 valid configurations containing 1 yellow that produce the number 120).
The only valid configuration for 2 yellow, which has a product corresponding to 120 and a sum less than 18, is the one containing 3 blue, 4 green and 5 red marbles.
There cannot be more than 2 yellow marbles without the sum exceeding 17. So it is known that there are either 1 or 2 yellow marbles.
The fact that Mary asked whether there was more than 1 yellow indicates that Mary has noticed that the house number can be produced from valid configurations containing either number of yellow marbles.
Therefore the house number is 120 as this is the only product that occurs for valid configurations containing exactly 1 yellow and also for a valid configuration containing exactly 2.
The fact that Mary instantly knew from the Father's reply indicates that the number of yellow marbles must be 2 (had there been 1 then Mary would have needed further information in order to to decide between the 2 valid configurations containing 1 yellow that produce the number 120).
The only valid configuration for 2 yellow, which has a product corresponding to 120 and a sum less than 18, is the one containing 3 blue, 4 green and 5 red marbles.
Congrats. I agree. The additional bit of info that Mary requires (1 or 2 yellows) is sufficient to distinguish between the duplicate 120s. All the others products are unique & don't require the additional info.
Code:
Y B G R Product Sum
1 2 3 4 24 10
1 2 3 5 30 11
1 2 3 6 36 12
1 2 3 7 42 13
1 2 3 8 48 14
.........
1 3 4 6 72 14
1 3 4 7 84 15
1 3 4 8 96 16
1 3 4 9 108 17
1 3 5 6 90 15
1 3 5 7 105 16
[B]1 3 5 8 120 17
1 4 5 6 120 16[/B]
1 4 5 7 140 17
[B]2 3 4 5 120 14[/B]
2 3 4 6 144 15
2 3 4 7 168 16
2 3 4 8 192 17
2 4 5 6 240 17- Joined
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Damn. I thought this one would survive the weekend, but only lasted overnight. Congratulations again Cynic on being first and Keithj on arriving at the same answer.
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An easy one that can be solved with no math, just some logic.
A plane flies in a straight line from Perth to Sydney, then back in a straight line from Sydney to Perth. It travels with a constant engine speed (constant air speed) and there is no wind. Will its travel time for the same round trip be greater, less, or the same if, throughout both flights, at the same engine speed, a constant wind blows from Perth to Sydney?
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Since air speed is constant, should be the same either way.
But if engine speed being constant mean the engine revs x per minute and that will speed the plane at say 800km/hr, then will be slower from Sydney as the headwind will slow it down. But assuming that engine will rev up/down to keep a constant air speed (it goes at 800km/h regardless), then same time.
But then if Bronwyn is at the gate and there's no front seat in first class, Perth to Sydney will be delayed a bit.
A plane flies in a straight line from Perth to Sydney, then back in a straight line from Sydney to Perth. It travels with a constant engine speed (constant air speed) and there is no wind. Will its travel time for the same round trip be greater, less, or the same if, throughout both flights, at the same engine speed, a constant wind blows from Perth to Sydney?
----
Since air speed is constant, should be the same either way.
But if engine speed being constant mean the engine revs x per minute and that will speed the plane at say 800km/hr, then will be slower from Sydney as the headwind will slow it down. But assuming that engine will rev up/down to keep a constant air speed (it goes at 800km/h regardless), then same time.
But then if Bronwyn is at the gate and there's no front seat in first class, Perth to Sydney will be delayed a bit.
Q1. His strategy will make no difference - the ratio will stay the same at 50%. Each row below represents one set of parents, and each column their successive offspring (stopping when a Girl is produced). The ratio of total number of offspring remains at 50%
B B B B
B B B G
B B G
B B G
B G
B G
B G
B G
G
G
G
G
G
G
G
G
Count of B = 15
Count of G = 15
Q2. still thinking...
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So that we are all on the same wavelength, a constant engine speed means a constant air speed, that is the speed of the plane relative to the air it is passing through. Ground speed is the speed of the plane relative to the ground, which will be the air speed + or - the wind speed, depending on whether there is a tail or head wind.
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Then going back to Perth will be slower as there's headwind.
That both way need to cover the same ground, going to Sydney with the wind meant no headwind (wind don't speed up the plane from behind, i don't think) while going back to Perth faces a drag so ground speed will be slower so will take longer over same distant.
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