Mathematical thoughts and methods embodied.

1.( 1998+ 1999+2000+……+2007+2008)÷2003

There is a arithmetic progression in brackets, and one * * * is eleven numbers, and the difference is 1, so the sixth number of the series is the average of the series, and the sum is equal to 2003x11÷ 2003 =11.

2 If A is one of nine numbers 1~9, how many times is A+AA+AAA+ ... +AAAAAA?

Recombine the above formula with decimal principle, a+(10a+a)+(100a+100a+a)+(100a+10a+a)+(/kloc) Other figures may also get the same result.

When the numerator of a fraction decreases by 25% and the denominator increases by 25%, how many percent is the new fraction less than the original fraction?

It is difficult for students to understand with the method of setting unknowns, and they can explore with the method of hypothetical data. Assuming that the original score is 4/ 16 (arbitrarily assumed data), according to the requirements of the topic, the numerator is reduced by 25%, the numerator becomes 3, and the denominator becomes 20. Now the score is 20% higher than before. The results of other data are the same, and the method of using hypothetical data is very close to the students' thinking level. For some difficult choices or judgments, this method can be used, assuming more data in different intervals, and actively exploring and discovering laws.

4. The least common multiple of two numbers is 180, and the greatest common divisor is 30. Suppose one number is 90, what is the other number?

There are many ways of thinking. Try it, which is commonly used by ordinary students. Because the greatest common divisor of two numbers is 30, both numbers are multiples of 30. When the other number is 30, the minimum common divisor of the two numbers is 90, which does not meet the meaning of the question. When the other number is 60, the greatest common divisor of the two numbers is 30 and the smallest common divisor is 180. So the two numbers are. In fifth grade, I knew that the result of the least common multiple and the greatest common divisor of two numbers is equal to the product of these two numbers, which means that the product of the original two numbers is equal to 180×30=5400, so the other number is 5400÷90=60.

5 If A ÷ 2009 = 2008...B, maximize the remainder b and dividend A= ().

This is a division with a remainder, and the remainder must be less than the divisor, so the maximum remainder is 2008. Then calculate dividend A=4036080 according to dividend = quotient × divisor+remainder.

6 If both P and Q are prime numbers, and 35P+ 13Q= 135, what are P and Q?

According to the fact that P and Q are prime numbers, 35P+ 13Q= 135, because 35 and 13 are both odd numbers, and 135 is also odd numbers, then one of the two products of 35P and 13Q must be even, and the other. P and q are both prime numbers, and one of them must be equal to prime number 2. Try: when P=2, Q=5 meets the problem. If Q=2, p has no integer solution. So the answer is unique, P=2, Q=5.

7.2008 can be expressed as the sum of three prime numbers, so what are these three prime numbers respectively? (Just write one)

The sum of the three prime numbers is 2008, so one of the three prime numbers is equal to 2 and the sum of the other two prime numbers is 2006. , you can choose 3 and 2003. Or 7 and 1999, etc.

(8) If A ÷ B ÷ C = 6, A ÷ B-C = 15. A-B = 17, then A+B+C=( ).A×B×C= ().

At first glance, this seems to be a complicated problem of solving equations, but it is not. It is enough to use addition, subtraction, multiplication and division to divide the relationship between the parts. According to A÷B÷C=6, we can know that A÷B=6C, combined with the formula A÷B-C= 15, we can get C = 3；; Reconsidering A÷B=6C, it is found that A is 18 times that of B. According to A-B= 17, it is found that B= 1, so A= 18. The rest can be solved by finding the sum or product of three numbers. The sum is 22 and the product is 54.

(9) Observe the following formula and find out the rules: 1×3=3, 3×5= 15, 15×7= 105, 105×9=945. Then according to the rules, the fifth formula is ().

Observe and analyze the law that the multipliers are 3, 5, 7, 9 ... In turn, then the multiplier of the fifth formula is 1 1, and the multiplicand uses the product result of the previous formula respectively, so it is inferred that the multiplicand of the fifth formula is 945, and the formula required by the topic is 945 ×1.

(10) 1 min, 2 min, 5 min and 1 min are 1 min, and there are () kinds of coins that can be used to form monetary value.

Classified thinking: only four coins 1, 2, 5, 1 are used, and two coins/3, 6, 1 and 1 are used; 7 points, 1 angle 2 points; Six kinds 1.5 points. With 3 coins, the currency value is 8 points, 1.3 points, 1.6 points; There are four kinds of 1.7 points. The monetary value of using four coins is 1.8 cents, so the number of kinds of coins is 4+6+4+ 1= 15.

Of course, there are as many as 100 questions in each pre-competition training, which embodies many mathematical ideas and methods to solve problems. As long as we are good at guiding students to observe carefully, guess boldly, verify carefully, explore actively and apply what they have learned comprehensively, our students can solve quite a few problems.