answer
Decomposition of 4002 into 2×3×23×29 shows that there are two products when decomposing 4002 into two digits, that is, 4002 = 58× 69 = 46× 87; There are two ways to decompose 4002 into a product of two digits and three digits, namely, 4002 = 23×174 = 29×138; So each number can only be used once, so there is only one filling method, that is, 23× 174=58×69=4002.
One day, 945 tourists were waiting to check in at the gate of China Pavilion of Shanghai World Expo. At this time, several people come to the entrance every minute to prepare to enter the museum. In this way, if the four ticket gates are opened, all visitors can enter the exhibition hall within 15 minutes. If eight ticket gates are opened, all visitors can enter the museum within seven minutes. Now all visitors are required to enter the museum within five minutes, and _ _ _ ticket gates need to be opened.
answer
Assuming that 1 ticket gate is 1 unit, the number of new tourists in 1 minute is:
(4× 15-8×7)÷( 15-7)=0.5;
When the ticket gate is opened, the number of tourists waiting is 4×15-0.5×15 = 52.5;
The number of tourists who need to be released at the ticket gate within 5 minutes is: 52.5+0.5× 5 = 55;
So the number of ticket gates to be opened is 55÷5= 1 1.
There are three grasslands with an area of 5 15 and 25 mu respectively. The grass on the grass is as thick and grows as fast. The first grassland can feed 10 cows for 30 days, the first grassland can feed 28 cows for 45 days, and the third grassland can feed 60 days.
answer
Assuming that the daily grazing amount of each cow is 1, then:
The first piece of grassland: 5 mu of original grass +5 mu of grass for 30 days = 10×30=300 copies;
Namely: the original amount of grass per mu+the amount of grass per mu for 30 days =300÷5=60 copies.
The second grassland:15mu of original grass+15mu of grass for 45 days =28×45= 1260;
That is, the original amount of grass per mu +45 days of grass per mu =126015 = 84 copies.
Therefore, the amount of grass growing per acre per day =(84-60)÷(45-30)= 1.6.
The amount of grass per mu =60-30× 1.6= 12.
The third grassland covers an area of 25 mu, and the amount of grass in 60 days is1.6× 60× 25 = 2400;
So the third grassland can be eaten by * * (2400+ 12× 25) ÷ 60 = 45 cows for 60 days.
The fifth-grade students of Yucai Primary School are divided into three groups to visit the museum. The ratio of the first group to the second group is 5:4, and the ratio of the second group to the third group is 3: 2. It is known that the number of the first group is 55 less than the sum of the numbers of the second group and the third group. How many people are there in Grade One of Yucai Primary School?
answer
The first batch: the second batch = 5: 4 = 15: 12.
The second batch: the third batch = 3: 2 = 12: 8.
So the first batch: the second batch: the third batch = 15: 12: 8.
Assuming that the number of people in the first, second and third batches is 15, 12 and 8 respectively, then:
The first batch is less than the sum of the second and third batches 12+8- 15=5 copies.
It is concluded that the number of people in each plot is 55÷5= 1 1.
So the primary school has:1/kloc-0 /× (15+12+8) = 385 students.
A, B, C and D got the top four in the mid-term exam of our class. The sum of the scores of Party A and Party B is 108, that of Party B and Party C is 149, and that of Party C and Party D is 12 1. We know that the score of the first place is twice that of the third place.
answer
Using divisibility to solve problems
In contrast, C-A =4 1 and B-D =28.
So the first place is B or C.
(1) If b is the first, then because 149 is not divisible by 3, c is not the third, but only the second and third.
B =56 because B-D =28, but C =149-56 = 93 >; conflict
(2) If C is the first, because 149 is not divisible by 3, B can only be the second, because 12 1 is not divisible by 3, and D can only be the fourth.
So A is the third, C-A =4 1, that is, C =82, A =4 1.
Finally, the second place is b =108-41= 67.
There are more than 100 students in the sixth grade of a primary school. If there are three of them in a row, there will be one more student. If you line up five people, there will be two more; If you line up seven people, there is one more person. The number of students in this grade is _ _ _ _ _ _.
answer
The number of people who meet the first and third conditions is at least 3× 7+1= 22; The result 22 also satisfies the second condition, divided by 5, leaving 2; So it can be concluded that the minimum value satisfying these three conditions is 22; But the condition given by the title is a * * *, with more than one hundred students; Therefore, according to the nature of congruence, it can be concluded that the common multiples of 3, 5 and 7 need to be added on the basis of 22 people; The least common multiple of 3, 5 and 7 is 3×5×7= 105.
So the total number of students is 22+3×5×7= 127 (people).
There are five light bulbs, each of which is controlled by a switch. Each operation can pull two switches to change the bright and dark state of the corresponding light bulb. Can all five bulbs be dimmed after several operations?
answer
Master the use of parity to demonstrate.
When each light bulb dims, it is necessary to pull the switch odd times; Then all five bulbs are dimmed, and the switch needs to be pulled for odd times. Each operation is to pull two switches; After several operations, the number of times a * * * is pulled must be a multiple of 2, that is, even times; However, when all five bulbs are dimmed, they have to be pulled * * * times in total, so it is contradictory; So no matter how many times you operate it, it is impossible to dim the five bulbs together.