Tisch
Xiaohua divides the numbers 2-9 into 4 pairs, so that the sum of each pair is a prime number. How many different departments are there? Answer and analysis:
According to the topic conditions, each logarithm must consist of an odd number and an even number. In order not to omit, we choose numbers from 2, 3, …, 9 to pair.
The numbers that can be paired with 2 are 3, 5 and 9. Discuss the following in different situations:
(a)2 and 3 are paired. Then the smallest number left is 4. Of the remaining numbers, 7 and 9 can be paired with 4.
① If 4 and 7 are paired, 5 can only be paired with 6, and 8 can only be paired with 9.
② If 4 and 9 are paired, 5 can only be paired with 8, and 6 can only be paired with 7.
So there are two ways to divide this situation.
(b)2 and 5 are paired. Then the smallest number left is 3. Of the remaining numbers, 4 and 8 can be paired with 3.
① If 3 and 4 are paired, 6 can only be paired with 7, and 8 can only be paired with 9.
② If 3 and 8 are paired, 4 can only be paired with 9, and 6 can only be paired with 7.
So there are two ways to divide this situation.
(c)2 and 9 are paired. Then the smallest number left is 3. Of the remaining numbers, 4 and 8 can be paired with 3.
① If 3 and 4 are paired, 5 can only be paired with 8, and 6 can only be paired with 7.
② If 3 and 8 are paired, 4 can only be paired with 7, 5 and 6.
So there are two ways to divide this situation.
To sum up, there are six different ways to divide a * *.
There are 46 students in a class, art group 12, music group 23, and both groups are 5 students. How many students in this class have never participated in art group or music group?
Answer and analysis:
Knowing the total number of students in the class, we can find out the number of students who have participated in art or music groups from a negative perspective. Just subtract this number from the total number of students in the class to get the number of students who have never participated in art group or music group. According to the inclusion and exclusion method, the total number of students who have participated in at least one group in this class is 12+23-5=30 (people). So the students in this class who have never participated in art or music groups are 46-.
extreme
Bind exercise books with a batch of paper. If 120 volumes have been bound, the remaining paper is 40% of this batch of paper; If 185 sheets are bound, there are still 1350 sheets left. How many sheets are there in this batch? Answer and analysis:
Method 1: bind 120 copies and keep 40% paper, that is, use 60% paper.
Then to bind 185 copies, you need185× (60% ÷120) = 92.5% of paper, that is, the remaining 1-92.5%=7.5% of paper is/kloc-.
So this batch has 1350÷7.5%= 18000 sheets of paper.
Method 2: 120 books correspond to 60% of the total number (1-40%=), so the total number is 120÷60%=200 books.
When 185 books are bound, there are still 200- 185: 15 books unbound, corresponding to 1350, so each book needs 1350÷ 15=90.
That is, there are 18000 sheets of paper.
100 monks, 140 steamed buns, 1 big monk, 1 little monk. Q: How many monks are there?
Answer and analysis:
This topic comes from China's famous ancient topic "The problem of dividing steamed buns among hundreds of monks". If the big monk and the little monk are regarded as chickens and rabbits respectively, and the steamed stuffed bun is regarded as legs, then the problem of chickens and rabbits in the same cage can be solved by hypothetical methods.
Assuming that 100 people are big monks, then * * * needs 300 steamed buns, which is 300- 140= 160 (one) more than the actual situation. Now, if young monks are replaced by big monks, the total number will remain the same, but the number of steamed buns will be reduced by 3- 1 = 2 (one), because 160÷2=80, so there are 80 young monks and big monks.
100-80=20 (person).
Similarly, it can also be assumed that 100 people are young monks, so students might as well try it themselves.
I was skipping rope with Xiao. Xiao jumped for 2 minutes first, and then they each jumped for 3 minutes, a total of 780 times. It is known that Xiao Xi jumps more 12 times per minute than Xiao Le, so how many times does Xiao Xi jump more than Xiao Le?
Answer and analysis:
Using the hypothesis method, suppose that Xiao Xi's skipping speed is reduced to the same as Xiaole's, and the total number of jumps of two people is reduced 12×(2+3)=60 (below).
It can be found that Xiaole jumps (780-60)÷(2+3+3)=90 (bottom) and Xiaole * * * jumps 90×3=270 (bottom), so Xiao Xi jumps 780-270×2=240 (bottom) more than Xiaole * *.
Tisso
There are 40 identical balls in a cloth bag, of which 1, 2, 3 and 4 are numbered 10. Q: How many balls must be taken out at a time to ensure that at least three of them have the same number? Answer and analysis:
The four numbers 1, 2, 3 and 4 are regarded as four drawers, and it is necessary to ensure that there are at least three apples in one drawer. The worst case is that there are two apples in each drawer, and * * * has: 4×2=8 (one), so 1 can meet the requirements, so at least 9 apples should be taken out at a time.
Feng Chun primary school originally planned to plant poplars, willows and pagodas *** 1500. After planting trees, when 3/5 of the total number of poplars and 30 of the total number of willows were planted, 15 Sophora japonica trees were temporarily removed, which means that the remaining three trees are exactly the same. How many trees were originally planned to be planted?
Answer and analysis:
Suppose that poplar, willow and Sophora japonica are: A, B and C respectively.
a+b+c = 1500( 1-3/5)a = b-30 b-30 = c+ 15
Three kinds of trees are easy to get: 825,360,315.
Pack 15 identical yo-yos into four identical cartons, with at least one in each carton, and the number of each carton is different. Q * * * What are the packing methods?
Answer and analysis:
Because 2+3+4+5= 14, the minimum two addends can only be 1 and 2; 1 and 3; 1 and 4; 2 and 3 four cases:
(1)15 =1+2+3+9 (2)15 =1+3+4+7 (3) None (4)15 =
= 1+2+4+8= 1+3+5+6
= 1+2+5+7
So 15 yo-yo is put in different cartons * * *, with 3+2+ 1=6 different packaging methods.