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The 73rd person will survives.
Consider the case when there are 2^n numbers in the circle. Each time the number reduces by half and the number 1 remains till the end.
In the given question,
1 kills 2,
3 kills 4 and so on till 71 kills 72.36 people have been killed till now.
64 people remain in the circle.
64 is a power of 2. So the first guy after 72 will be the new number 1 in a circle of 2^6.So 73 will survive.
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The place would be southern Hemisphere. There is a ring near the South Pole with a circumference of one mile.Now if we are standing on one mile north of the ring at any point. Then let’s say if we walk on the southern direction and cover one mile, we will be standing on the ring. Traveling one mile east will bring us on the circumference of the ring. Walking one mile north from there, we will be standing on the exact point where we have started. Now if we start counting, we will understand that while we walk 1 mile north in the end, we can reach an infinite number of points.
In such a case, the total number of possible points possible are 1 + infinite.
Now let’s consider the ring that is half a mile in circumference near the South Pole. If we walk a mile along the ring, we would circle twice but will reach the point where we started from. In such a case if we start from the point that is located one mile north of a half mile ring, it will also help us reach the starting point after traveling as per asked.
Now with every possible integer N, there is a circle with radius R = 1 / 2 (2*pi*n); which is centered at the South Pole.
If we walk along these rings, we will be circling N times and again returning to the point where we started. We must note that the possible values for N are infinite. Also, we could have infinite ways of selecting a starting point which is located one mile north of the rings which means that there are (infinite * infinite) possible points.
Concluding with our statements, the possible number of points are equal to 1 + infinite * infinite which is equal to infinite.
Thus there are infinite points possible.
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The ants can only avoid a collision if they all decide to move in the same direction (either clockwise or anticlockwise). If the ants do not pick the same direction, there will definitely be a collision. Each ant has the option to either move clockwise or anticlockwise. There is a one in two chance that an ant decides to pick a particular direction. Using simple probability calculations, we can determine the probability of no collision.
N(No collision) = N(All ants go in a clockwise direction) + N( All ants go in an anticlockwise direction) = 0.5 * 0.5 * 0.5 + 0.5 * 0.5 * 0.5 = 0.25
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The Answer is : Square
Square the first shape’s sides and the remaining shapes add up to this number.
E.g.
triangle pentagon square
>> 3×3 = 5 + 4
square hexagon hexagon square
>> 4×4 = 6 + 6 + 4Therefore:
hexagon octagon octagon octagon octagon square
>> 6×6 = 8 + 8 + 8 + 8 + 4regards..
ronret45 4462 views
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