The hour and minute hands of a clock overlap exactly 11 times throughout the day. Keeping the clock in the center of the pizza, just make the cuts along the lines defined by the hands during these times.

The answer is THE LETTER N.

A lost it. Since there have been (10 + 15 + 17) / 2 = 21 games played in total, and each player never misses 2 games in a row, the only way for A to play just 10 games is if he plays the 2nd, 4th, 6th, etc. games, and every time loses.

No, you can not construct such a polyhedron. Assume the opposite and consider its face, which has the largest number of sides, say k. Then the polyhedron contains at least k more faces with different numbers of sides, all less than m. However, this is clearly impossible.

The correct position of the bishop is e3. Otherwise, no White’s piece could have captured the last Black’s piece, moving on black squares.

The answer is THE LETTER M.

Just 2 evaluations are enough. First, get P(1). This will give you an upper bound on the number of digits of the largest coefficient of the polynomial, let’s say that is N. Now evaluate the polynomial at the point M = 10 to the power of N. This will make all of the coefficients of P appear in the number P(M), separated by strings of zeros.

We assume the cells in the prison are arranged in a circle. The prisoners agree every day each of them to pass the key they receive to their left neighbor, except for their own key which they keep. It is easy to figure out which key is their own since they can easily calculate when they will receive it. For example, if prisoner 8 knows that his key is at prisoner 3 in the beginning, then he will get it on the 5th day. Therefore, within 10 days, all prisoners will have their own keys.

The answer is NO.
The question is equivalent to analyzing the intersection of a cube and a sphere which share a common center. Thus the question gets reduced to figuring out whether such intersection, which is a curve, can connect two opposite points on the sphere/cube.
Let the edge of the cube has length 1. If you pick the radius of the sphere equal to √2/2, the intersection will consist of 6 circles inscribed in the sides of the cube. Then the rover can just move along these circles from one point to its opposite and NASA won’t be able to figure out the exact shape.
Remark: It is not hard to see that 2:√2 is the only edge-radius ratio, for which NASA can’t figure out the shape.

Empty the first glass, fill around 1/4th of it with liquid from the second glass, and the rest 3/4ths with liquid from the third glass. Then, measure the first and fourth glasses simultaneously. If their total weight is:
– 500 grams -> the first glass is (was) the poisoned one
– between 500 and 505 grams -> the second glass is the poisoned one
– between 505 and 510 grams -> the third glass is the poisoned one
– 510 grams -> the fourth glass is the poisoned one