What are the contents of simple math problems in the Olympic Games?

Simple operations of multiplication and division

The core ideas of multiplication and division are "rounding" and "simplification". That is to say, in the calculation, we should flexibly use the multiplication and division algorithm, and try to make the data given in the topic into integers of ten, hundred and thousand, preferably 10, 100, 1000, and then calculate; Or by using the property of quotient invariance, the larger data in the topic are converted into smaller data and then calculated.

In order to simplify the calculation better, we should be familiar with these three groups of calculations: 5× 2 =10; 25×4= 100; 125×8= 1000。 Even multiples of 5 15 should be calculated skillfully.

Commonly used algorithms in multiplication and division are:

Multiplicative commutative law: a× b = b× a;

The law of multiplicative association: (a× b )× c = a× (b× c);

Multiplication and distribution law: (a b) × c = a× c b× c;

The operational nature of division: a ÷ b ÷ c = a ÷ (b× c);

The properties of quotient invariance: ① a ÷ b = (a× c) ÷ (b× c);

②a÷b =(a÷c)÷(b÷c).

B and c are not equal to 0. )

In the classroom, children have mastered the operational nature of multiplication and exchange law, associative law and division. Through this lecture, they should learn to use it flexibly and popularize it. Multiplicative commutative law and associative law are often used together. The multiplicative commutative law can be summarized as "signed shift" at the same level (we define multiplication and division as the second level operation).

Title:

( 1)47600÷25; (4)6 17×958+6 17× 1043-6 17。

Analysis:

Before solving the problem, carefully observe the data in the topic, and then choose the appropriate method to calculate.

Question (1): Divider 25 times 4 can make up 100. There are two algorithms:

①47600÷25 ②47600÷25

=(47600×4)÷(25×4) =47600÷25÷4×4

= 190400÷ 100 =47600÷(25×4)×4

= 1904 =476×4

= 1904

If you are not familiar with the properties of quotient invariance, you can use the second solution.

Question (4), first think clearly, and finally subtract 6 17, that is, subtract 1 6 17, and then extend the distribution law of multiplication to the distribution of multiplication to subtraction.

6 17×958+6 17× 1043-6 17

=6 17×(958+ 1043- 1)

=6 17×200

= 1234000

Title:

( 1)9999×2222+3333×3334; (2) 1234× 1000 1000 1。

Analysis:

Question (1), when you see this formula, your first thought should be to use the law of multiplication and distribution, but the same multiplier was omitted in the first two multiplication calculations. Therefore, we must find a way to increase the same multiplier:

9999×2222+3333×3334

=3333×3×2222+3333×3334

=3333×6666+3333×3334

=3333×(6666+3334)

=3333× 10000

=33330000

Question (2), the two data in this question are not close to an integer number, but each bit of 1000 1000 1 is 1. Numbers like this can be decomposed into integer numbers with the highest digit of 1, which is convenient for calculation:

1234× 1000 1000 1

= 1234×( 100000000+ 10000+ 1)

= 1234× 100000000+ 1234× 10000+ 1234× 1

= 1234 1234 1234

Title:

Calculate 456×567567-567×456456.

Analysis:

Observing the data in the topic is very regular, and it is easy to associate it with the law of multiplication distribution, but there is no same multiplier in the left and right multiplication calculation, so it is necessary to decompose the data in the topic and calculate the same multiplier:

456×567567-567×456456

=456×567× 100 1-567×456× 100 1

=0

Title:

How many zeros are there at the end of the result of finding 99…99 × 99…99+ 199…99?

1999 9 1999 9 1999 9

Analysis:

To observe the data in the topic, we must first decompose the last addend into the sum of the previous multiplier and another number, and then observe while calculating. After completing one or more of the previous steps, you can find many simple calculation methods:

99…99 × 99…99 + 199…99

1999 9 1999 9 1999 9

=99…99 × 99…99 + 99…99 + 100…00

1999 9 1999 9 1999 9 1999 0

=99…99 ×( 99…99 + 1)+ 100…00

1999 9 1999 9 1999 0

=99…99 × 100…00 + 100…00

1999 9 1999 0 1999 0

=(99…99 + 1)× 100…00

1999 9 1999 0

= 100…00 × 100…00

1999 0 1999 0

= 100…00

3998 zeros

So the result of this question has 3998 zeros at the end.

According to the structure of formulas and the characteristics of numbers, flexible use of operation rules, laws, properties and some formulas can make some complex elementary arithmetic simple and difficult.

In the process of calculation, we first analyze the characteristics of the formula as a whole, and then make some transformations to create conditions to simplify the calculation by multiplication and division. This way of thinking is very useful in the four operations.

In the fractional operation, we should not only keep in mind the operation rules and nature, but also carefully examine the questions, carefully observe the characteristics of operation symbols and numbers, and rationally disassemble or merge the numbers involved in the operation to make them conform to the operation rules, so as to facilitate oral calculation and simplify the operation.

In front of us, we introduced some ingenious and simple calculation methods by using the characteristics of laws, properties and numbers. Next, we briefly introduce how to calculate the score by split method (also called split term method and split term method).

Solving problems by splitting is mainly to make some scores after splitting cancel each other out, thus simplifying the operation. Generally speaking, it is shaped like

The score of 1/(a×(a+ 1)) can be

Disassemble it into1/a-1/(a+1); 1/(a×(a+n)) can be decomposed into1/n× (1/a-1(a+n)). You can use examples to think about the law.