Answer: 9 balls are divided into three groups: A, B, C, B and C, with A and B being weighed once and B and C being weighed once. If there is something wrong with the ball in A, then the weights of B and C are equal, and A is heavier or lighter than B. (If A is heavier than B, it means that the different balls are biased, because the equal weights of B and C have proved that B and C are normal, the same below).
If there is something wrong with the ball in C, then A and B are equal in weight, and C is heavier or lighter than B. ..
If there is something wrong with the ball in B, then B is both heavier/lighter than C and heavier/lighter than A. ..
After weighing twice in this way, you can be sure which of the three balls the different ball is, and you can know whether it is overweight or underweight.
Then weigh any two of the three balls on the balance. If the weight is equal, the remaining one is different; If the weights are not equal, we can judge which is the abnormal ball according to the weight of the abnormal ball judged before.
2. The family lives in T city and works in other places. Every day after work, they arrive at T City Station by train, and his wife drives to the station to meet him on time. One day, he got off work early, took the early train at 5: 30 to T City Railway Station, and then walked home. His wife drove up as usual and met him halfway, that is, to take him home. Now she finds it ten minutes earlier than usual. How long has he been gone?
Answer: The topic seems a bit ambiguous. I met him halfway, that is, I took him home. Now I found it ten minutes earlier than usual. Did I find it when I met him or when I went home?
If it is the former, it cannot be solved because the conditions are not available; In the latter case, the walking time is 10 minute.
3. Three people went to 30 yuan for a night, and each paid 10, making up 30 yuan for the boss. Later, the boss said that 25 yuan was enough for today's discount, so he took out the 5 yuan and asked the waiter to return it to them. The waiter secretly took 2 yuan, and then distributed the remaining 3 yuan money to three people, each of whom got 1 yuan. In this way, everyone paid 65430 yuan at the beginning. Now it's back to 1 yuan, that is, 10- 1=9, each person only spent 9 yuan money, three people each 9 yuan, 3*9=27 yuan+2 yuan given by the waiter =29 yuan, where did the one yuan go?
Answer: In fact, this topic only confuses people's thinking direction. Fundamentally speaking, there is no problem at all. The answer is simple. The boss has dollars. The formula is "boss 25+ waiter 2+ customer 3=30". A lot of things are like this. If you always think from that angle, sometimes you won't get it. It doesn't seem so difficult to change.
4. There are a pair of mice. He gives birth to a pair of mice every month. After the mice grow to the fourth month, they will also give birth to a pair of mice every month (starting from the fourth month). Excuse me, how many mice will there be in a * * in 100 years?
PS: Don't think about men and women and death.
Answer: (Sn stands for mouse logarithm after n)
S 1=S2=S3= 1
Sn=S(n- 1)+S(n-3)
Simple recursive formula, variant Fibonacci sequence, simple procedure.
5. A strange logical math problem.
The manager will promote one of his three assistants to assistant manager. Among them, one person went to ask the manager which one can't be the assistant manager, B or C. The manager told him that C couldn't be the assistant manager. What is the probability that A will become a manager at this time?
6. It is known that the mother is 2 1 year older than the child, and she is 5 times older than the child after 6 years.
Excuse me: where is dad now?
(Pay attention to the exam)
7. The topic starts with a game. . .
There are three people in the game: A, B and the referee.
1, A has a number A and B has a number B. They only know their own numbers, but they don't know each other's numbers.
2. Party A and Party B shall hand over their respective numbers to the referee.
The referee added A and B to get X=(a+b), and he thought of another number Y.
The referee told A and B about X and Y, telling them that there is a number in X and Y that is the sum of A and B given by you two, and the other is my imagination.
The referee began to ask questions (three people were present at the same time).
The referee first asked A: Do you know B's number?
At this time, A may know B's number, if A says it doesn't.
The referee then asked B: Do you know A's number?
At this time, B may know the number of A, if B says it doesn't.
The referee asked A again: Do you know B's number?
……
Ask questions in this cycle. There will always be a time when A or B can work out another number.
Q: What is the reason?
8. There are two integers, both greater than or equal to 2 and less than or equal to 200. A knows its product and b knows its sum.
A said, I can't tell what these two numbers are.
B said: I knew you couldn't judge.
A said: I know now.
B said: Then I know.
Q: What are these two numbers?
Answer: the sum of two numbers must be less than 55, otherwise they can be 53+x, and then B can't be sure that A doesn't know these two numbers.
Add this condition to the first and second conditions I mentioned earlier, and the result of adding the two numbers is:
1 1, 17,23,27,35,37,4 1,47,5 1
According to the third condition
Because: 1 1 = 2 2+7 = 4+7 or 2 3+3 = 8+3, remove 1 1 (because if two numbers are 4 and 7 or 8 and 3, A will say that he knows, but B won't know).
17 = 4+ 13 and 8+99 are not prime numbers, so 17 is reserved.
23=4+ 19, 16+7.
35 = 4+3 1, 32+3 has been deleted.
37 = 8+29, remove 32+5.
4 1 = 4+37, 8+33, 16+25, 32+9, except that 37 is a prime number, the rest are not prime numbers, so they are reserved.
47 = 4+43, 16+3 1 has been deleted.
5 1 = 4+47, 8+43 has been deleted.
So the result of this screening is that the sum of the two numbers can only be 17 or 4 1.
But for the fourth screening, my fourth screening conditions above are not complete enough, and it should be completely like this.
4. There are more than two pairs of even-odd combinations in the factorization factor of the product of two numbers, and the sum of each pair except one of these combinations can be removed by the first three steps.
Because 4 1 = 32+9. And 32× 9 = 96× 3 = 4× 72 =16×18.
Among the above decomposition factors, 96+3 >: 55,
Obviously, if the sum of two numbers is 4 1, B cannot determine whether these two numbers are 4+37 or 32+9.
The final screening result is that the sum of the two numbers is 17.
Knowing the sum of two numbers, it is easy to know that it is 4 and 13.
9. Three tigers and three people cross the river. A boat. Only two people cross the river at a time. Everyone can row a boat. Only one tiger can row a boat. Tigers eat more people than others. How to cross the river?
PS: What's on the boat also counts.
10. A mathematician took 10 liter of oil and divided it into two bottles of 5 liters of oil, but he only had a 7 liter and a 3 liter cup to divide the oil. Do you know what to do?
Tip:
5 can be composed of 1+4 and 2+3.
10, the subtraction result between 7 and 3 is 10-7=3, 7-3=4, 10-3=7, 10-3-3=4, 7-3-3 =/kloc.
What if you write down your solution steps?
10 liter oil bottle 7 liter oil bottle 3 liter oil bottle
1 10 0 0
2 7 0 3
three
Suppose,
Step 1: 10 liter oil bottle contains 10 liter oil. It can be written as: 10 0.
Step 2: You pour 10 liter of oil into a 3 liter oil bottle. You can write: 7 0 3.
1 1. A logic professor has three students, and all three students are very clever!
One day, the professor gave them a question. The professor put a note on everyone's forehead and told them that everyone had written a positive integer on the note, and the sum of some two numbers was equal to the third! Everyone can see the other two numbers, but not his own.
The professor asked the first student: Can you guess your own number? Answer: No, ask the second, the third, the first, the second, the third: I guessed right, it was 144! The professor smiled with satisfaction. Can you guess the numbers of the other two?
Answer: 36 and 108.
12. 1 yuan can buy 1 candy 1 candy paper 1 candy paper 3 pieces of candy paper can be exchanged 1 candy Q:1how many candy can 5 yuan buy?
Answer: 15 yuan can buy 15 candy, 15 candy paper, 15 candy paper, 5 candy paper, 5 candy paper, 1 candy paper, 1 candy paper.
13. Under what circumstances 0>2>5>0
A: When guessing boxing.