Ten Olympic math problems are urgently needed.

1. The product of three rational numbers A, B and C is negative and the sum is positive, when:

X = (absolute value of a)/A+(absolute value of b)/A+(absolute value of b+(c)/C.

What is the value of the algebraic expression (x 29-95x+ 1028)?

& lt solution: because ABC is less than 0.

So a, b and c can only be two positive, one negative or all three negative.

Because A+B+C is greater than 0.

So all three can't be negative, so there is only one negative and two positive.

Suppose a is less than 0, b is greater than 0, and c is greater than 0, then

X = (absolute value of a)/a+(absolute value of b)/a+(absolute value of b+(c) /A+(B+(c = (-1)+1.

So (x 29-95x+ 1028) = 934.

& gt

2. At what value of x, the absolute value of the formula -4X+(4-7X)- (1-3x) +4 is a constant?

& lt Solution: Remove the sign of absolute value, and after merging similar items, the formula should no longer contain the "x" item, that is, the coefficient of x is 0, (4-7X is less than or equal to 0)( 1-3X is less than or equal to 0).

So x is greater than or equal to 4/7 >

At the New Year's Eve party, the host asked a question: How to turn "2+3=8" into a real equation? No one can answer it. At this time, Xiao Li took out a mirror and solved the problem. Everyone says Xiao Li is very clever. Do you know what method Xiao Li used? (Numbers are written on the spreadsheet)

& lt Take a mirror and look at its image backwards >

4. If the product of 12 ×9 is a multiple of 6 and m+n is the minimum, then m=__, and n=__.

& ltM=3，N = 1 & gt；

5.8 Players play a game of chess, and every two players play a game. The winner gets 2 points, and the loser gets 0 points. If the draw gets 65,438+0 points, then the eight players will get different points after the game. After sorting by the score order, it is found that the score of the second player is equal to the sum of the scores of the fifth, sixth, seventh and eighth players, and the fourth player scored 9 points, so the first player won.

& lt 13。 The total number of round robin matches for 8 players is 28. The total score is 56. The last four players, regarded as four-person round robin, have to play six games, each with 2 points, and the games between these four players should accumulate 12 points, so the final total score of these four players should be at least 12 points, while the second one should at least 12 points and the fourth one should be 9 points. So the first and third place got 23 points, so the first place got 13, and the third place got 10.

& gt

6. New Year's Day in 2007 is Monday, and the next New Year's Day is Monday. This year is _ _.

& lt20 18. There is one more day in a normal year and two more days in a leap year. Leap four years, starting from 2007, * * * will have three leap years and eight average years, which will exceed 14 days and return to Monday's New Year's Day >

7. Answer the following questions and write the solution process: There is a small square with a side length of lcm in the middle of the rectangular ABCD, and the connection is shown in the figure. It is known that the upper and lower trapezoidal areas are 8cm each, and the left and right trapezoidal areas are 9cm each. So, what is the circumference of the rectangular ABCD? cm？

& lt 180.( 1) Assuming that the arrival time of 1 hour in advance is t and the speed of walking 3 kilometers less per hour is v, then

( 1) 2T= 1(V+3)

(2) 2V=3(T+ 1)

Substituting V=2T-3 from (1) into (2), it is easy to get T=9.

Similarly, V= 15.

The whole journey is 9*( 15+5)= 180 or (9+3)* 15= 180.

(2) The speed (that is, the sum of the speeds of two people) can also be reduced by 5 kilometers, and the time (that is, the meeting time) is more than 3 hours. Get: 5T=3V gives t = (3/5) v.

Let the walking speed within 3 kilometers per hour be x, and the equation is:

X+3=2*(3/5)X or

3*(3/5)X+ 1*3=2X

X= 15 can be obtained.

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8. Try to prove that if you choose two numbers from any four odd numbers, their sum or difference is zero.

& lt There are five cases of odd-numbered units, * * *: unit 1, unit 3, unit 5, unit 7 and unit 9.

According to the fact that (cell 1 and cell 9), (cell 3, cell 7) and (cell 5) are three drawers, and any four odd numbers are four apples, there must be two numbers from the same drawer. The last bit of their sum or difference must be 0 >;

9. Fold a long line in half repeatedly, and fold it m times to get a wire harness. Cut the obtained wire harness into n equal parts with scissors, and you will get two kinds of line segments with different lengths. If the number of longer line segments accounts for111.

Q: What is the maximum m? What is the corresponding n?

& lt hypothesis: after cutting according to the meaning of the question, the number of long line segments is l and the number of short line segments is s,

Then: l = 2 m- 1 (take both ends, which is exactly twice as long as the short line, but after deducting both ends of the line, it is a line rather than a ring).

S = (n- 1) * 2 m+2 (the first item is the short line in the middle of the scissors, and the second item is the first two line ends).

10L = S。

Namely:10× (2m-1) = (n-1) * 2m+2.

Simplification: (11-n) × 2m =12.

So: 2 m = 12, 6, 4, 3, 2, 1, because m is a positive integer, and the maximum m can only be equal to 2.

Correspondence: 2 m = 4, 1 1-n = 3, that is, n = 8 >

10. A class participated in the quiz, and each of the three questions ***A, B and C got full marks or zero marks, of which the full mark for A was 20, and the full mark for B and C was 25. Competition results: each student answered at least one question correctly, 1 person answered all three questions correctly, and 15 person answered two questions correctly. The total number of people who answered questions A and B correctly is 29; The total number of people who answered questions A and C correctly is 25; The total number of people who answered questions B and C correctly is 20. What is the average score of this class?

& lt Hypothesis: A: The total number of people who have done a question.

B: The total number of people who did the B question.

C: the total number of people who have done C.

X: the number of people who only did question a.

Y: the number of people who only did question B.

Z: the number of people who only did C.

A: the number of people who did questions a and B.

B: the number of people who did questions a and C.

C: the number of people who did questions b and C.

There are 1 people who have all worked out three questions.

1) According to the meaning of the question,

A+B=29

A+C=25 }= > This equation group can be solved as A= 17, B= 12 and C=8.

B+C=20

2) From the pie chart, we can get the following relationship:

x+A+b+ 1 = A = & gt； x+a+b= 16 (i)

y+a+c+ 1 = B = & gt； Y+a+c= 1 1 (2)

z+b+C+ 1 = C = & gt； z+b+c=7 (iii)

A+b+c= 15 (number of people who answered two questions correctly) (4)

(1)+(2)+(3) = & gtx+y+z+2(a+b+c)=34

D: x+y+z=4(v)

3) From the requirement that we have six unknowns but only four independent equations, we can see that it is impossible to get an exact solution, but because the ultimate goal of this problem is to get an average score, B and C can be regarded as a class, because their full marks are all 25, and A is regarded as a class. So y and z have the same properties, and a and b have the same properties.

(I)= & gt； x+(a+b)= 16

(4) = >(a+b)+c= 15

(v)= >x+(y+z)=4

(I)-(iv):x-c = 1 = & gt； x=c+ 1(vi)

(4)-(5): (a+b)-(y+z) =12 = > (y+z) = (a+b)-12 (vii)

4) Let's see what the average score should be.

First of all, the total number of students in this class = x+y+z+a+b+c+1= 4+15+1= 20.

Average score p= total score/total number of people

=(20x+25(y+z)+45(a+b)+50c+70)/20

=( 1/4)(4(c+ 1)+5(a+b- 12)+9(a+b)+ 10c+ 14)

=( 1/4)(4c+4+5(a+b)-60+9(a+b)+ 10c+ 14)

=( 1/4)( 14(a+b+c)-42)