64 is easy it's knowing what other switches are on and why that is the main part of the question.
Odd Ball Interview Questions
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64 is easy it's knowing what other switches are on and why that is the main part of the question.
The latter part of the question isn't terribly well worded. What is it asking for? Which lights beyond 64 are on after all 100 people have walked past?
If so, the answer is (-1)^n where n is the number of factors of the given number excluding 1, and where a result of 1 denotes the light being on at the end of the exercise.
So for light 65, the non-1 factors are 5, 13 and 65, which gives (-1)^3 = -1, so the light will be off. Looking at it number by number, that's 1 on, 5 off, 13 on, 65 off. For 66, non-1 factors 2, 3, 11, 22, 33, 66, giving (-1)^6 = 1, so the light is on. 1 on, 2 off, 3 on, 11 off, 22 on, 33 off, 66 on.
Etc.
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The last part is asking which other lights are on.
No. See my 65 and 66 example above. Neither is prime, but one is on and one is off.
It's to do with the number of factors (excluding 1) that each number has. Even number of factors = on. Odd number of factors = off.
Actually your examples are wrong. 66 would be off as you forgot the factor 6. Your line of thinking about about the odd and even number of factors determing the light being on/off is correct though. As a hint 64 is the example used for a reason. It has a property that ensures it has an odd number of factors.
Hahahaha, I forget the 6, what a retard.
It's nothing to do with friction and/or energy used lifting one side v.s. the other.
The point is that a rolling cylinder has rotational kinetic energy as well as translational KE. In other words some of the energy converted from gravitational potential energy has to go towards spinning the cylinder, leaving less to go towards straight line speed.
That is not less than 3 steps?
step 1: split into 2 piles and keep the 4 that weigh the least
step 2: split into 2 piles of 2 and keep the 2 that weigh the least
You cant use a 3rd step because it says in less than 3 steps...
Curiosity is getting the better of me... What property is this? Even power of 2?
Even power of 2 is sufficient, but not necessary...
I originally thought you could weigh the elephant using laws of displacement too.
But even simpler, couldn't you just use a pivot with the elephant one end and a known weight the other, providing both are equal distance from the pivot?
But even simpler, couldn't you just use a pivot with the elephant one end and a known weight the other, providing both are equal distance from the pivot?