[SFW] Riddles / Puzzles / Brain Teasers - Page 31

Ahh, this brings me back to my abstract algebra days/cyclic groups. All numbers that are relatively prime to 12 will do this, so 1,5,7, and 11. (to be relatively prime is to not share any divisors other than 1). The answer is the same for any M.

pf/
It follows easily that if they are relatively prime then it will go to every number. Here's the proof that it won't go to every number if they are not relatively prime.

You return to the original position in the clock whenever the number of moves you've made times the number you move by, n, is divisible by m, the total number of clock positions. Suppose m and n (everything here is a natural numbers s.t. m>n) have a common divisor, d. There exists d2 and d3 s.t. m=d(d2) and n=d(d3). Therefore, d2(n)=d(d2)(d3), which is divisible by m. Since d2<m, n returns to the original position in fewer than m moves meaning it couldn't have possibly gone to all m positions.

Any prime number that is also not a non-trivial divisor (non-trivial divisor being a number 'n' that divides 'm' without leaving a remainder and isn't equal to 1 or m) of 12. That is:
1, 5, 7, 11 work.
2, 3 are prime but are non-trivial divisors of 12, so they do not work.
4, 6, 8, 9, 10, 12 are not prime, so they do not work.

On second thought, if you want to consider numbers outside of the range 1-M, then take the number modulo M then apply the above rule. 13 and 17 will work, as they mod 12 are 1 and 5, respectively, which both work.

Same as above, but I don't have a proof. First take N mod M, then check to see if the remainder is a prime number that is not a non-trivial divisor of M

I assume that the answer is simply move it ahead 1 hour at a time. This will always guarantee that it will work.

EDIT: Oh, numbers, not number. XD

It's a trick question. My answer is it depends, but if the plane did take off, it would not be in a good way.

First, let's note that the mythbusters clip is not directly relevent. In their formulation, they move the treadmill at the plane's takeoff speed, not at the wheel speed. The plane's engine provides thrust pushing against the air, the thrust easily overcomes the friction of the wheels on the treadmill (which, while higher than the usual friction of wheels on stationary ground, is still orders of magnitude smaller than the thrust the engines provide), the plane begins to move forward (at which point, the wheel speed exceeds the treadmill's speed), and when the plane reaches the necessary takeoff speed through the air, it takes off (at that point, the wheel speed is roughly double the treadmill's speed).

So we're going to need a treadmill that can go faster - several orders of magnitude faster - if it's to match the plane's wheel speed. The constraint that the treadmill's speed match the plane's wheel speed means that, as long as the wheels do not skid, the plane must remain stationary relative to the ground. The plane's thrust remains the same, and the thrust is provided by pushing against the air. So the treadmill must move so fast that the rolling resistance is equal to the thrust.

But there are physical limits to what the wheels can take (ignoring the physical limits of the conveyer belt setup, since that's already magical and arbitary). Those wheels aren't designed to turn at thousands of times their usual takeoff speeds. And with a treadmill moving that fast, the airspeed (relative to the stationary ground) isn't going to remain zero either. Let's consider both these effects, starting with the wheels.

Now, if the wheels fall off, then while they are still spinning (and the treadmill still moving backwards at super speed), the plane will collapse onto it's now wheelless undercarridge onto the treadmill. You will effectively be dropping your plane onto a spinning grinder. Bad news for the plane, and bad news for anyone standing anywhere nearby, as bits go flying. The plane will take off, but in pieces (and going backwards).

How might we make our wheels survive? Ordinarily, we would try to minimise rolling friction to allow for higher speeds, but here, that makes things worse. The lower the friction, the faster the wheels have to spin to generate enough backwards pull on the plane to match the engine's thrust (which we have to do in order to keep the plane stationary relative to the ground, which was shown earlier to be necessary to satisfy the problem's requirements). So lowering friction just leads to an even faster treadmill.

No, the trick is to go the other way. If we apply the plane's wheel brakes (and they are sufficiently strong), then the plane's wheel speed will remain 0 (and thus, the treadmill will also remain stationary). The problem then comes down to this: does the plane have sufficient thrust to skid the wheels? If so, can it reach takeoff speed with the wheels locked? If so, it will take off (after burning through a lot of wheel rubber - don't try this at home folks!). Most likely, it will not.

What about air movement? Could that make the plane take off? Well, that depends on something not specified in the problem - how big si the treadmill? There could be a short piece of treadmill under each wheel, or a massive treadmill stretching for miles ahead of the plane. Under short treadmill conditions (as are depicted), the treadmill speed required for this would not be reached until well after the wheels fell off or caught fire. But what about a large treadmill? You'd probably still suffer wheel failure before the airspeed got high enough, but let's suppose the wheels somehow survived; what would happen?

Well, the plane would liftoff. But the plane is still stationary relative to the ground at the moment of takeoff, and all of its airspeed comes from the treadmill-induced wind. This windspeed falls off rapidly with height. A few meters further up, the air is barely moving - and so is the plane. No airspeed means no lift, and also no control over the plane. It would gently nose-down and dive back into the wind, and into the ground. The best you could hope for would be a bumpy landing (slightly further forward from your starting point). The worst would be.. worse.

A.
The given pictures have no recurring pattern/shape, thus we can assume that B, D and E can be ruled out.

But A or C? C can be ruled out because none of the shapes have only vertical/horizontal lines.

Did i do anything wrong?

If I understand the riddle correctly, they will do the following:
First one will say symbol of person before him, Second one will say his own symbol to save himself. This goes on in pairs to the end, so 50 can be saved for sure. The indians sacrificing themselves however have a 1/12 chance that their symbol is the same as the one before them in the queue, so after them in the queue.

So the expected amount that is saved would be 50+ 50/12, this last part can obviously range from 1-50

I think it might be D.

The way I see it is that you need to look at how many triangles you can identify in each figure: the first has 4, the second 2, and the third and fourth have zero triangles. The fifth one also has 4 triangles implying a restart of the pattern (4,2,0,0). If the above logic makes any sense, than the correct answer should be D, as it has 2 triangles in it.

I think the answer is B. The reason being that I think that you have to count the number of shapes in the picture in the first picture its 4 then 5 then 1 then 2 then 3 it would follow if you have a recouring pattern of 1 to 5 then the next number of shapes would be 4. B has four shapes. At least that's what i think :3

They can try something like the following, although i'm not sure it's the easiest way:
They number the zodiac from 1 to 12 (are they allowed to do this?), or use their normal order if they know it. They then add all the amounts from all zodiacs they can see in front of them. Naturally, a mod 12 is used since you can't count further than 12. The back indian will answer the total of all the others and is sacrificed. The next one in line can calculate the difference between the number answered and his own calculated number, so he can save himself. The next one will subtract that number from the first, allowing him to know the difference between the total he calculated and total he counted, which is his own zodiac. This continues until the end, for every next indian is saved. Better make sure all the indians can count well, though, one mistake may cost a lot of lives .