harey

You forget the lift by air depression on the wings. When the space shuttle (which is basically a plane) takes off vertically, as long as the thrust is less than 10 [N/kg], it does not move. When the thrust exceeds this value, it moves up. [If you then cut the engines, it slows down, stops and falls like a stone.] That's what your equations describe. When a plane flies horizontally at a constant speed, the gravitation is compensated by depression on wings - if you cut the engines, the plane glides. You still can keep it at constant horizontal speed, it slowly loses potential energy, but does not vertically fall like a stone. This implies that on constant speed/height, you need to furnish less energy than by a vertical takeoff and therefore less thrust. Another approach: The energy of a cruising plane is constant, so the furnished energy must be equal to the energy lost. As the lost energy is partly transformed to lift on the wings...

And what about this:

Thinking it over and over again, I always finish with a problem I cannot solve. The holes are numbered 1, 2, 3.... Three hunters check: 1 2 3 3 4 5 5 6 7 .... On day n, they will have checked up to the hole 2 * n +1 The groundhog starts in the hole k and moves to the right. On day n, he will be in the hole k+n Question 1: Will the hunters catch the groundhog (and if so, when)? 2 * n +1 grows faster than k + 1, I already have the answer. Nevertheless: 2 * n + 1 = k + n n ⁼ k + 1 No matter how high the number of the starting hole the groundhog choses, it will be caught. Question 2: Is there a starting hole so that the groundhog is not discovered on day n? k + n > 2 * n + 1 k > n + 1 No matter how long the hunters hunt, it always is possible that the groundhog started in a hole leading him outside the checked area.

Never mind how big the area you chose, I always can say "Bad luck, the groundhog is somewhere about 17 holes outside the defined area." I agree that the size of the cleared area will -> inf. Just the size is expressed as a number. No matter how big it is, there always is a bigger number.

I think I can do better: But not sure it is the max.

Please delete.

Still confused.

Nice beginning... (If this were the full solution, first round would be enough.) Oh, I got it now!!! Great!!!

It is not because of the gender. the Claytons indicates that Mrs Clayton is still alive and Jasper's mom is deceased. Therefore, Jasper is not a Clayton.

In the same way, you exclude that Jasper is a Clayton: 2. In deference to an influential family member, the Claytons agreed that if they ever had a daughter they would name her Janice.

There are 8 boxes, not 4. And there is another possible distribution you have to exclude.

The way how the marbles were selected is not known, so you cannot do better then Bonanova. However, instead of grabbing the calculator:

A big step forward. Now that we found the result, it remains to find the way to find the result.

@rocdocmac Pas du tout. Mais c'est ma langue maternelle, je sais même écrire.

I see now: plainglazed's formula is the simplification of my formula I was so hard looking for. On paper, I got a kind of unreadable proof, so I fed them into my computer. Up to n=30, no difference with 6 decimals.

Nice try, but they do not share and they survive all. Hint:

Though they are similar, I see at least one huge difference. In the traffic jam puzzle, you cut ANY part. In the marble problem, you cut the REMAINING part. Enough for different formulae. Remains me of http://brainden.com/forum/topic/18168-squirrel/

A large colony of squirrels dug holes during the summer and in each hole, they put between 1 and 100 nuts (each quantity has the same probability). If each squirrel has to eat 100 nuts during the winter, how many holes must he find (in average)? Each hole contains 50.5 in average, so 2 should be enough, right?