Math problems, Olympiad math problems for the fourth grade of primary school.

1. Cut a rectangular piece of paper 20 decimeters long and 15 decimeters wide into a square with a side length of 2 decimeters, and you can cut up to one.

2. Regular number filling 1 2 3 5

There are several chickens and rabbits in the cage, they have a total of 50 heads and 140 legs, and ask how many chickens and rabbits there are?

Hypothetical method. Suppose an unknown number is known, for example, suppose that 50 heads are all rabbits, then there are a total of feet (4 50 =) 200 (only), which is inconsistent with the known 140 in the question, more (200-140 =) 60 (only), the reason for more is that each chicken counts 2 more feet after the chicken is a rabbit, so the number of chickens is (60 2=) 30 (only), then the number of rabbits is (50-30=) 20 (only).

Find the difference series 3··5··7···

What are items 10 and 100?

Solution: A formula is required for the first term + (number of terms - 1) tolerance = last term.

The number of terms here is 10,100) The tolerance is the difference of (5-3).

The first term is 3 and the last term is the number that is found.

That is: 3+(10-1) (5-3)=21Q: 5, 7, 9, 11, what is item 15?

No money.

A and B two numbers of 132, the decimal point of A is moved to the left by one place is equal to B, and the number A is (120).

It's okay to buy a book casually, or search the Internet yourself, it's a lot.

Question 1: E1 is with A, D2 is with A, B, C3 is with A, B, and children.

So the child has spoken to A, B, and C 3 times.

Question 2: Six + Other = 102, Five + Other = 96, Five + Six = 114 Six + Other + Five + Other + Five + Six = 102 + 96 + 1142 (Five + Six + Other) = 312

Five + Six + Other = 156

Five = 156-102 = 54

Six = 156-96 = 60

The five times of 1 a affirm that everyone has one time. So E's time is used up, (talking to A) B is talking four times, which means that B has spoken to all but E once, so D's two times have run out (talking to A and B) and C has to be three times is the child A and B. So the child passed 3 times**(abc).

2 102 is not a 5th grade 1 2 3 4 6th grade all 102 . There are 96 people who are not in 6th grade, 1 2 3 4 5th grade, all 96 people. Add 102 to 96 and add 1 2 3 4th graders plus 5 6th graders to 198, and subtract 114 5 6th graders to double the 1 2 3 4th graders and 42 1 2 3 4th graders.

So there are 102-42 = 60 in 6th grade and 96-42=54 in 5th grade.

Question 1: 3 times.

a: b, c, d, e, the child ——— a pass five times, only in this case b: a, c, d, the child ——— e pass once, already with a, so b does not communicate with e, b passes four times, only this case.

C: A, B, children ——— the same way, C is not the same as E. D passes twice, already with A, B, so C does not communicate with D. C pass three times, and that's all it can be.

d：a,，b

E:A Question 2: 54 students in the fifth grade, 60 students in the sixth grade, 42 students in the other grades, X people in the fifth grade, Y people in the sixth grade, and Z people in other grades.

From the first sentence has: y+z=102

From the second sentence there is: x+z=96

From the third sentence there is: x+y=114

So: z=[(y+z)+(x+z)-(x+y)] 2=(102+96-114) 2=42

So x=96-z=54

y=102-z=60

Little brother, if you don't understand the process, you can ask the teacher or classmates, believe me, that's right.

Kid, your math Olympiad almost made me do it for 5 minutes.

1 Xiao Ming doesn't have to call **, because everyone has spoken to each other, so Xiao Ming called 5 times.

2 There are 60 students in 6th grade and 54 people in 5th grade, I can't express it in equations, but you don't learn equations.

Very helpless

In the first question, A and 5 other people have passed **, so E only talks to A, D only talks to A and B, and C also talks to A, B, and the child, so the child also communicates with A, B, and C, so the answer is 3 times.

In the second question, students in other grades are A, B in the fifth grade, and C in the sixth grade, so A+C=102, A+B=96, B+C=114. The solution yields b=54 and c=60.

Because A is 5 times, I talked to everyone once, and concluded that E did not talk to anyone except A. B is 4 times in addition to E and each person once, it is concluded that D has also used up the number of calls, C is 3 times only with children. So the children's call is 3 times, 2, 102-96 = 6 (people) This is the number of people in 6 classes more than 5 classes (114-6) 2 = 54 (people) This is the number of people in 5 classes 54 + 6 = 60 (people) 6 classes of people.

E1 is with A, D2 is with A, B, C3 is with A, B, and Child.

So the child has spoken to A, B, and C 3 times.

Six + Other = 102, Five + Other = 96, Five + Six = 114 Six + Other + Five + Other + Five + Six = 102 + 96 + 114 (Five + Six + Other) = 312

Five + Six + Other = 156

Five = 156-102 = 54

Six = 156-96 = 60

1，32×29-18×32-32

7. Find a pattern and fill in the blanks.

1+2，2+4，3+6，4+8，1+10，2+12，3+14，4+16+1+18，……Q: The formula for 2009 is (

Fill in the number 1Fill in the following equation with the appropriate operator symbols and parentheses to make the equation hold. 12

2.Fill in the following 9 5s with the appropriate operator notation to make the equation hold. 55

3.Add or () to the equation below to make the equation hold. 12

5.Add parentheses to the equation below to make the equation hold.

Application Question 1Jiejie asked his grandmother, "How old are you?" My grandmother said, "After subtracting 6 from my age, it shrinks by 9 times, and by adding 2 to each of them, it expands by 8 times, and it happens to be 80 years old." Do the math, how old is grandma? ”

2.Add 14 to a number, divide by 3, subtract 26, and multiply by 25 to get 100. What is this number?

3.The fourth grade students participated in the interest group, half of them more than 2 people participated in the choir group, the remaining half more than 2 people participated in the calligraphy level, the remaining half more than 2 people participated in the aviation module, and the last 5 people participated in the chess group. Q: How many people in Year 4 participate in interest groups?

4.The school was to plant 36 saplings, and Class 4 (1) first took a number of saplings, and then Class 4 (1) gave 10 trees to Class 4 (2), and Class 4 (2) did not have enough people, so they returned 6 trees to Class 4 (1), and the number of trees in Class 4 (1) was twice that of Class 4 (2). Q: How many trees did the first four (1) classes take?

5.There are some colored balls in a box, and a child takes the balls out of the box, and the one who takes them is: each time he always takes out half of the total number of colored balls in the box and then puts back one.

After the child took it 399 times according to this rule, there were still 2 colored balls in the box. Find the number of colored balls in the box at the beginning.

Hello, your answer should be wrong, right? It should be 52 pieces for 5 points, and 100 pieces for 2 points!

The equation is actually very simple, as follows: the number of 2 pieces is. (152*5-60) divided by (2+5)=100Then the number of 5 points is, 152-100=52

Hey, hope it helps!

1|The dividend is 3320, the quotient is 150, the remainder is 20, and the divisor is ( ).

3398 is the sum of 4 consecutive natural numbers, the smallest of which is ( ) 249005024 is composed of ( ) billion, ( ten thousand and ( ) one.

That's all I know, and I hope it helps.

Only 132 can write 4 formulas, because the 4 numbers 3, 5, 7, and 9 contain a relationship 3+9=7+5=12,12 11=132

In the same way, it can be 5, 6, 7, 8, because 5 + 8 = 6 + 7 = 13, 13 11 = 143, it can already be composed of 65 + 78 = 143, 56 + 87 = 143, 68 + 75 = 143, 86 + 57 = 143

For example, 2, 3, 7, 8 can be composed of 23 + 87 = 83 + 27 = 78 + 32 = 72 + 38 = 110

Because 3+7=2+8, 110=10 11

You should get the idea.