Test wisdom questions

1 How many parts can n straight lines on a plane divide a plane?

Answer: one entry can be divided into two parts; two, four parts; three, seven parts; four, eleven parts; and N parts, 1 + 1 + 2 + 3 +... + N: Part 1 + N * (n + 1)/Part 2

2. A circle divides space into two parts. Two Circles divide space into four parts at most?

A: First, let's look at how many knots are added after adding a circle, and then add a circle to k circles to add 2 k knots, and add n knots to add n areas, therefore, you can add a circle to k circles to add 2 k areas. Therefore, a circle can be divided into 2 parts at most, two circles can be divided into 2 + 2 = 4 parts at most, and three circles can be divided into 4 + 4 = 8 parts at most, four circles can be divided into 8 + 6 = 14 parts at most, five circles can be divided into 14 + 8 = 22 parts at most, and six circles can be divided into 22 + 10 = 32 parts at most.

Spread to N circles. N circles divide the plane into 2 + 2 (1 + 2 + 3 +... + N-1) = 2 + N (n-1) = n ^ 2-N + 2.

How many parts can a space be divided?

A: The derivation method of this problem is still recursive, first look at how many parts can be added after adding a sphere, in the existing N-1 on the basis of a sphere, add a sphere, this sphere to multi-energy is divided by the N-1 sphere into (N-1) ^ 2-(N-1) + 2 (see N circles to divide the plane into a maximum of several parts? In the conclusion) block area, where each area will be in the original part of the space is divided into two, so in the existing N-1 on the basis of a spherical surface, the spherical surface at most increase (N-1) ^ 2-(N-1) + 2 space areas. Therefore, a sphere can be divided into 2 parts at most, two globes can be divided into 2 + 2 = 4 parts at most, and three globes can be divided into 4 + 4 = 8 parts at most, the four globes can be divided into 8 + 8 = 16 parts at most, the five globes can be divided into 16 + 14 = 30 parts at most, and the six globes can be divided into 30 + 22 = 52 parts at most. Spread to N globes. N globes divide space into 2 + (1 ^ 2-1 + 2) + (2 ^ 2-2 + 2) +... + (N-1) ^ 2-(N-1) + 2) = n (n ^ 2-3n + 8)/3 part.

4 What is the angle between the hour hand and the minute hand at 3:15:00?

A: At 03:15 = 13/4, at 03:15, the angle between the sub-needle and the 0-point is (15/60) * 360 = 90 degrees. The angle between the hour hand and the 0-point is [(13/4) /12] * 360 = 97.5 degrees, so at 03:15, the angle between the hour hand and the minute = 97.5-90 = 7.5 degrees

5. There are two unevenly distributed fragrances. The burning time is one hour. How can you determine a 15-minute period?

A: Two incense sticks at the same time. The first two ends are ignited, the second one is lit at one end, and the first one is lit for half an hour. Then, both ends of the second one are ignited, the second point is 15 minutes.

6 there are four people, A, B, C, and D, who want to go through a bridge at night. It takes 1, 2, 5, and 10 minutes for them to pass through the bridge. They only have one flashlight and at most two people can bridge the bridge together. How long is the shortest time for all four people crossing the bridge?

A: A and B go over the bridge for 2 minutes. A goes back to the flashlight for 1 minute. C and D go over the bridge for 10 minutes. B goes back to the flashlight, it takes 2 minutes to go, and a and B go over the bridge for 2 minutes. Therefore, the minimum time is 17 minutes.

Seven monkeys are surrounded by 100 bananas. The monkeys are 50 meters away from their homes. The monkeys move up to 50 bananas at a time, but the monkeys will eat a banana every time they walk, how many bananas can a monkey take home?

A: A monkey moves bananas in two stages. The first stage is to move bananas back and forth. When the number of bananas is greater than 50, the monkey needs to eat three bananas for each rice. Stage 2: The number of bananas <= 50. Move them back directly. Eat one tablet each day.
Take 50 root, put one rice, take one back, and then take the remaining 50 root, put it together with the original, eat a total of 3 root. Take 50 more and take one rice, put it down, take one and go back, and then take the remaining 47, put it together with the original, eat three more, and so on, eat three for each rice, eat 51 after 17 meters, 49 are left, take 49 to go home directly, don't come back, eat 33 for the remaining 33 meters, and finally get 16 at home.

8, 12 balls and a day round. Now I know that only one ball is slightly heavier than the others. How many times can I find the ball if I try again? What about 13?

A: The simplest method is the binary search strategy, but the number of times is not the least. The Binary Search Strategy reduces the number of balls by half each time, however, if you divide the ball into three equal parts (or nearly three equal parts), you can exclude nearly 2/3 of the balls each time.
For example, 12 balls are divided into 4, 4, and 4. For the first time, four balls are placed at both ends, and the heavy ball is at the heavy end. If the ball is flat, it is in the third heap. The second time: divide the heap with a duplicate ball into 1, 1, and 2 for the first time, and put one at each end. The duplicate ball is at the same end, that's in the third heap, two balls. The third time: two balls are found again.
This question satisfies the following equation: if the number of balls is n and the minimum number of times is X, then n <= 3 ^ x