Solution: Reverse thinking: Because 225=25×9, and 25 and 9 are coprime, as long as the modified number can be divisible by 25 and 9 respectively, this number can be divisible by 225. Let's examine the divisibility of 25 and 9 respectively.
According to the feature that the number can be divisible by 25 (the last two digits can be divisible by 25), the last two digits of the modified six-digit number may be 25 or 75.
According to the feature that numbers can be divisible by 9 (the sum of numbers in each bit can be divisible by 9), 9+7+0+4+5 = 25, 25+2 = 27, 25+7 = 32.
So the revised six-digit number is 970425.
7. In a three-digit number, the unit, the tenth place and the hundredth place are all the squares of a number.
Answer 48
There are three options to solve the hundred digits: 1, 4, 9, and four options to solve the ten digits and the single digits: 0, 1, 4, 9. The three-digit * * * that satisfies the meaning of the question is
3× 4× 4 = 48 (pieces).
12. The product of each digit of a given three-digit number is equal to 10, so the number of three digits is _ _ _ _.
Answer 6
Because 10 = 2× 5, these three digits can only be composed of 1, 2,5, so * * * has = 6.
12. There are five triangles in the figure below, and the sum of the three numbers in each triangle is equal to 50, where A7 = 25, A 1+A2+A3+A4 = 74, A9+A3+A5+A 10 = 76. What is the sum of A2 and A5?
Answer 25
The solution is A 1+A2+A8 = 50,
A9+A2+A3=50，
A4+A3+A5=50，
A 10+A5+A6=50，
A7+A8+A6=50，
So a1+a2+A8+A9+a2+a3+a4+a3+a5+a10+a5+a6+a7+A8+a6 = 250,
That is, (a1+a2+a3+a4)+(a9+a3+a5+a10)+a2+a5+2a6+2a8+a7 = 250.
There are 74+76+A2+A5+2 (A6+A8)+A7 = 250, and there are A6+A7+A8 = 50 in triangle A6A7A8, where A7 = 25, so A6+A8 = 50-25 = 25.
Then A2+A5 = 250-74-76-50-25 = 25.
It is suggested that the above deduction is completely correct, but we lack a sense of direction and overall grasp.
In fact, when we see such a digital array, our first feeling is that the five 50s here do not refer to the sum of 10 numbers, but the sum of this 10 number plus the number of the inner circle 5. This is the most obvious feeling and an important equal relationship.
Then "look at the problem and set the direction" and find the sum of the second number and the fifth number.
It is related to the other three numbers in the inner circle. The sum of the sixth number and the eighth number is 50-25 = 25.
Look at the third number. When the numbers 1, 2, 3 and 4 of two straight lines are added with the numbers 9, 3, 5 and 10, the third number will be counted repeatedly.
The play begins:
74+76+50+25+ second number+fifth number = 50× 5
So the second number+the fifth number = 25.
First, fill in the blanks:
1 How many filling methods satisfy the following formula?
Word of mouth-word of mouth = word of mouth
Answer 4905.
You can know the answer from the correct formula. This problem is equivalent to how many formulas are there to find out that the sum of two digits A and B is not less than 100.
When a= 10, b is between 90 and 99, and there are 10 species;
When a= 1 1, b is between 89 and 99, with 1 1 species;
……
When a=99, there are 99 kinds of B between 1 99. * * * Yes
10+11+12+... 99 = 4905 (species)
It is suggested to combine the mystery of formula with the counting problem, which is an example. The analogy association of mathematical models is the key to solving problems.
There are pentagons and hexagons on the football surface (see the top right). Each pentagon connects five hexagons, and each hexagon connects three pentagons. Then the simplest integer ratio of pentagon and hexagon is _ _ _ _ _.
The answer is 3: 5.
This solution has x pentagons. Each pentagon is connected with five hexagons, so there should be 5X hexagons, but each hexagon is connected with three pentagons, that is, each hexagon counts three times, so there are six hexagons.
Second, answer questions:
1. Xiaohong went to the store and bought a box of flowers and a box of white balls. The number of two boxes of balls was equal. The original price of flower balls is 3 in 2 yuan and 5 in 2 yuan. When the new year's goods were on sale, the price of both balls was 4 yuan and 8, and as a result, Xiaohong spent less 5 yuan. So, how many balls did she buy?
150 answers
solve
Use a rectangular diagram to analyze, as shown in the figure.
Easy to get,
Solution:
So 2x= 150.
2.22 Parents (father or mother, they are not teachers) and teachers accompany some pupils to take part in a math contest. It is known that there are more parents than teachers, more mothers than fathers, more female teachers than mothers, and at least one male teacher. So how many of these 22 people have fathers?
Answer 5 people
Know parents and teachers ***22 people, parents are more than teachers, parents are not less than 12 people, teachers are not less than 12 people, mothers and fathers are not less than 12 people, mothers are more than fathers, and mothers are not less than 7 people. There are two more female teachers than mothers, and the number of female teachers is not less than 7+2 = 9 (people). But the title points out that male teachers have at least 1, so male teachers have 1 and female teachers do not exceed 9. It has been concluded that there are no fewer than 9 female teachers, so there are 9 female teachers and 7 mothers, so the number of fathers is: 22-9- 1-7 = 5 (.
Wonderful tip, this question uses the thinking method of maximum problem many times, and skillfully borrows the half-difference relationship to get the range of inequality.
The method of combining positive and negative discussions is also reflected.
3. The sum of the ages of Party A, Party B and Party C is 1 13 years old. When Party A is half the age of Party B, Party C is 38 years old. When Party B is half the age of Party C, Party A is 17 years old. How old is Party B now?
The answer is 32 years old
The solution is shown in the figure.
X years later, A is 17 years old, so:
The solution is x= 10,
At a certain moment, A is 17- 10=7 years old, B is 7×2= 14 years old, C is 38 years old and 59 years old.
So now everyone has to add (113-59) ÷ 3 =18 (year).
So b is now 14+ 18=32 years old.
7. The number of students in Class A and Class B is equal, and some students take math courses. The number of people in class A who take mathematics is exactly 1/3 of class B, and the number of people in class B who take mathematics is exactly 1/4 ... So how many more people are not in class A than in class B?
answer
Solution: 4x people in Class A didn't attend, and 3y people in Class B didn't attend.
Then the number of participants in class a is y, and the number of participants in class b is X.
According to the conditions, the number of students in the two classes is equal, so 4x+y = 3y+x.
3x=2y x:y=2:3
So 4x:3y=8:9, so the number of people who didn't attend class A is the number of people who didn't attend class B.
In addition, to solve the linear equation: it can be assumed that the number of people in both classes is "1". If Class A participates in X, then Class A does not participate (1-X); Then what Class B didn't attend was 3x, and what Class B attended was (1-3x). The equation can be listed as follows: (1-x)/4= 1-3x Find x = 3/1.
It is a good question to suggest equation calculus, setting up without seeking, and quantifying ideas.
Target class
Famous school true volume seven
First, fill in the blanks:
3 1 How many filling methods * * * satisfy the following formula?
Word of mouth-word of mouth = word of mouth
Answer 4905.
You can know the answer from the correct formula. This problem is equivalent to how many formulas are there to find out that the sum of two digits A and B is not less than 100.
When a= 10, b is between 90 and 99, and there are 10 species;
When a= 1 1, b is between 89 and 99, with 1 1 species;
……
When a=99, there are 99 kinds of B between 1 99. * * * Yes
10+11+12+... 99 = 4905 (species)
It is suggested to combine the mystery of formula with the counting problem, which is an example. The analogy association of mathematical models is the key to solving problems.
There are pentagons and hexagons on the football surface (see upper right). Each pentagon is connected with five hexagons, and each hexagon is connected with three pentagons. Then the simplest integer ratio of pentagon and hexagon is _ _ _ _ _.
The answer is 3: 5.
This solution has x pentagons. Each pentagon is connected with five hexagons, so there should be 5X hexagons, but each hexagon is connected with three pentagons, that is, each hexagon counts three times, so there are six hexagons.
36 Cut out a figure with an area of 4 with square paper, which can only have the following seven shapes:
If four of them form a square with an area of 16, then the sum of the four numbers is the largest _ _ _ _.
Answer 19.
Solution In order to get the maximum sum of numbers, we must first use numbers with large numbers, so that we can spell out: (7), (6), (5), (1); (7),(6),(4),( 1); (7), (6), (3) and (1) form a square with an area of 16:
Obviously, the largest sum of numbers is the number 1, and the sum of numbers is 7+6+5+ 1 = 19. Check again, there are no other spellings.
Pay attention to the thinking method based on results. Let's draw a square with an area of 16, and color it first (6), (7) and then (5). After proper transformation, we can see that only (1) can be used.
In other cases, if (6), (7) and (4) are used, only (3) and (5) are considered.
40 Suppose the answer number is A, the unit number of A is B, and seven consecutive natural numbers are filled in seven circles, so that the sum of the numbers in every two adjacent circles is equal to the known number on the line, then the circle written A should be filled in _ _ _ _ _ _ _ _ _.
Answer a = 6
The solution is as shown in the figure:
B=A-4，
C = B+3, so c = a-1;
D = c+3, so d = a+2;
And a+d =14;
So a = (14-2) ÷ 2 = 6.
It is suggested that the main point of this problem is to deduce the difference between two circles separated by a circle.
So as to get the final sum-difference relationship to solve the problem.
43 A natural number is divided by 187 and 52, so the remainder of this natural number divided by 22 is _ _ _ _ _.
Answer 8
After subtracting 52 from this natural number, it can be divisible by 187 and 188. For convenience of explanation, the number obtained by subtracting 52 from this natural number is expressed as m, because 187 =17×11,and m can be18. Because m can be divisible by 188 and m can also be divisible by 2, m can also be divisible by 1 1× 2 = 22. The original natural number is M+52, because m can be divisible by 22. When considering the remainder of M+52 divided by 22, we only need to consider 52 divided by 22.
56 There is a pile of balls. If it is a multiple of 10, divide it into 10 piles on average and take away 9 piles. If it is not a multiple of 10, add a few more balls (no more than 9) to make this pile of balls a multiple of 10, then divide these balls into 10 piles evenly and take away 9 piles. This process is called surgery. If the initial number of balls in this pile is
1 2 3 4 5 6 7 8 9 1 0 1 1 1 2…9 8 9 9.
Continue to operate until there are 1 balls left, and then * * * operate for times; * * * added a ball.
Answer 189 times; 802.
Solving this number * * has 189 bits, and each operation is reduced by one bit. After operating 188 times, there are 2 times left. After the operation again, 1 remains. * * * Operation 189 times. The sum of this 189 number is
( 1+2+3+…+9)20=900。
From the operation process, we can know that the number of balls added is equivalent to adding each digit of the original number of balls to 9, plus 1 ball. So * * * add the ball.
1899-900+ 1=802 (pieces).
60 has the simplest true fraction, and the product of their numerator and denominator is 693. If all these scores are in descending order, the second score is _ _ _ _ _.
answer
The solution decomposes 693 into prime factors: 693 = 3× 3× 7× 1 1. In order to ensure that numerator and denominator cannot be reduced (otherwise the product of numerator and denominator after reduction is not 693), the same prime factor is either in numerator or denominator, and numerator is smaller than denominator. Molecules from large to small are 18.
68 Select some numbers from 1, 2, …, 1997, so that the sum of every two numbers of these numbers can be divisible by 22, then you can choose _ _ _ _ _ at most.
Answer 9 1
There are two ways to choose the solution: (1) choose all integer multiples of 22, namely: 22, 22×2, 22×3, …, 22× 90 = 1980, * * 90; (2) Select all odd multiples of 1 1, namely: 1 11+22×1,1. ...
Second, answer questions:
1. Xiaohong went to the store and bought a box of flowers and a box of white balls. The number of two boxes of balls was equal. The original price of flower balls is 3 in 2 yuan and 5 in 2 yuan. When the new year's goods were on sale, the price of both balls was 4 yuan and 8, and as a result, Xiaohong spent less 5 yuan. So, how many balls did she buy?
150 answers
solve
Use a rectangular diagram to analyze, as shown in the figure.
Easy to get,
Solution:
So 2x= 150.
2.22 Parents (father or mother, they are not teachers) and teachers accompany some pupils to take part in a math contest. It is known that there are more parents than teachers, more mothers than fathers, more female teachers than mothers, and at least one male teacher. So how many of these 22 people have fathers?
Answer 5 people
Know parents and teachers ***22 people, parents are more than teachers, parents are not less than 12 people, teachers are not less than 12 people, mothers and fathers are not less than 12 people, mothers are more than fathers, and mothers are not less than 7 people. There are two more female teachers than mothers, and the number of female teachers is not less than 7+2 = 9 (people). But the title points out that male teachers have at least 1, so male teachers have 1 and female teachers do not exceed 9. It has been concluded that there are no fewer than 9 female teachers, so there are 9 female teachers and 7 mothers, so the number of fathers is: 22-9- 1-7 = 5 (.
Wonderful tip, this question uses the thinking method of maximum problem many times, and skillfully borrows the half-difference relationship to get the range of inequality.
The method of combining positive and negative discussions is also reflected.
3. The sum of the ages of Party A, Party B and Party C is 1 13 years old. When Party A is half the age of Party B, Party C is 38 years old. When Party B is half the age of Party C, Party A is 17 years old. How old is Party B now?
The answer is 32 years old
The solution is shown in the figure.
X years later, A is 17 years old, so:
The solution is x= 10,
At a certain moment, A is 17- 10=7 years old, B is 7×2= 14 years old, C is 38 years old and 59 years old.
So now everyone has to add (113-59) ÷ 3 =18 (year).
So b is now 14+ 18=32 years old.
11.The number of students in Class A and Class B is equal, and some students take math courses separately. The number of math elective courses in class A is just 1/3 of the number of students who don't attend math elective courses in class B, and the number of math elective courses in class B is just 1/4 of the number of students who don't attend math elective courses in class A, so how much more are there in class A than in class B?
answer
Solution: 4x people in Class A didn't attend, and 3y people in Class B didn't attend.
Then the number of participants in class a is y, and the number of participants in class b is X.
According to the conditions, the number of students in the two classes is equal, so 4x+y = 3y+x.
3x=2y x:y=2:3
So 4x:3y=8:9, so the number of people who didn't attend class A is the number of people who didn't attend class B.
In addition, to solve the linear equation: it can be assumed that the number of people in both classes is "1". If Class A participates in X, then Class A does not participate (1-X); Then what Class B didn't attend was 3x, and what Class B attended was (1-3x). The equation can be listed as follows: (1-x)/4= 1-3x Find x = 3/1.
It is a good question to suggest equation calculus, setting up without seeking, and quantifying ideas.
Analysis Series 7 of Entrance Examination Papers for Key Middle Schools in 2007
24. The famous mathematician Stephen Barnach died on August 3, 20051,1945. The age of the year in which he lived is exactly the arithmetic square root of that year (the year of that year is the square of his age). Then the year of his birth is _ _ _ _ _ _ _ _ _ _.
Answer1892; 53 years old.
The solution first finds the complete square number less than 1945 and greater than 1845, where 1936 = 442, 1849 = 432, obviously only 1936, so1.
Then the year of his birth is 1936-44 = 1892.
He died at the age of 1945- 1892 = 53.
The key points are: determine the scope and pay attention to the "subtext": when the year corresponds to the corresponding age, there is a year-age = year of birth.
36. A primary school will hold a sports meeting soon. One * * *, there are ten competitions, and each student can enter two competitions. Therefore, it is necessary for _ _ people to sign up for the sports meeting to ensure that two or more students sign up for the same event.
Answer 46
Each student can sign up for two of the ten competitions, so there are 45 different ways to sign up.
Then, according to the pigeon hole principle, 45+ 1 = 46 people encountered problems when registering.
37.
43. As shown in the figure, ABCD is a rectangle, BC=6cm, AB= 10cm, AC and BD are diagonal lines. In the figure, the shadow part rotates around CD, so what is the volume of the solid swept by the shadow part? (π=3. 14)
Answer 565.2 cubic centimeters
Let's assume that the solid volume obtained by rotating the triangle BOC around CD is S, and S is equal to the volume of a cone with a height of 10 cm and a bottom radius of 6 cm minus the volume of two cones with a height of 5 cm and a bottom radius of 3 cm minus the volume of two cones with a height of 5 cm and a bottom radius of 3 cm. Namely:
s =×62× 10×π-2×32×5×π= 90π，
2S= 180π=565.2 (cubic centimeter)
It is suggested that S can also be regarded as the volume of a truncated cone with a height of 5 cm and a radius of 3 or 6 cm at the top and bottom minus the volume of a cone with a height of 5 cm and a radius of 3 cm at the bottom.
4. As shown in the figure, point B is the midpoint of line segment AD, and the lengths of all line segments composed of four points A, B, C and D are integers. If the product of the lengths of these line segments is 10500, the length of line segment AB is.
Answer 5
The line segments composed of four points, A, B, C and D, are AB, AC, AD, BC, BD and CD. Since point B is the midpoint of line segment AD, it can be assumed that the lengths of line segments AB and BD are both X and AD=2x, so there must be x3 in the product.
Factorizing 10500 into prime numbers;
10500=22×3×53×7,
Therefore, x = 5, ab× BD× ad = 53× 2, AC× BC× CD = 2× 3× 7,
Therefore, AC=7, BC=2, CD=3, AD= 10.
5. The distance between Party A and Party B is 60 kilometers, and bicycles and motorcycles run from Party A to Party B at the same time. Motorcycles arrive four hours earlier than bicycles. It is known that the speed of motorcycles is three times that of bicycles, so the speed of motorcycles is _ _ _ _ _.
The answer is 30 km/h.
If it takes "1" for a motorcycle to reach B and "3" for a bicycle, and 4 hours corresponds to "3"-"1"= "2", then it takes 4 ÷ 2 = 2 hours for a motorcycle to reach B, and its speed is 60 ÷ 2 = 30 km/h.
It is suggested that this is the most essential application of proportional relationship in the trip, and attention should be paid to the idea of corresponding number of copies.
6. It takes a car to transport goods from the urban area to the mountainous area, and it takes 20 hours to go back and forth. The time to go is 1.5 times that of coming back, which is slower than coming back12km. The car traveled * * * kilometers.
Answer 576
The time you left was "1.5", so the time you came back was "1".
So the return time is 20 ÷ (1.5+ 1) = 8 hours, and the departure time is 1.5× 8 = 12 hours.
According to the inverse relationship, when the round-trip time ratio is 1.5: 1 = 3: 2, the round-trip speed is 2: 3.
According to the proportional distribution, we know that the walking speed is 12 ÷ (3-2) × 2 = 24 (km).
Therefore, the round trip distance is 24× 12× 2 = 576 (km).
7. There are 70 numbers in a row. Except for the two numbers at both ends, three times of each number is exactly equal to the sum of the two numbers on both sides. Suppose that the first two numbers are 0 and 1, and the remainder of the last number divided by 6 is _ _ _ _ _.
Answer 4
Solution Obviously, we only involve the remainder divided by 6, including 0, 1, 3, 2, 3, 1, 0, 5, 3, 3, 5, 0, 1, 3, …
Every number 12 has a cycle from the number 1 to the remainder of 6.
Because 7012 = 5 ...10,
So the 70th number divided by 6 is the 10 number in the cycle, which is 4.
The generation of original data is also the key, and the details determine success or failure.
8. The teacher wrote a natural number on the blackboard. The first classmate said, "This number is a multiple of 2." The second classmate said, "This number is a multiple of 3." The third classmate said, "This number is a multiple of 4." ..... The fourteenth classmate said, "This number is a multiple of 15." Finally, the teacher said, "Of all the statements of 14, only two consecutive statements are wrong." The smallest natural number written by the teacher is.
Answer 60060
Solutions 2, 3, 4, 5, 6 and 7 are 4, 6, 8, 10, 12, 14 twice. If this number is not a multiple of 2, 3, 4, 5, 6, 7, then this number is not 4, 6, 8, 654. So this number is a multiple of 2, 3, 4, 5, 6, 7. It can be inferred that this number is also a multiple of (2×5=) 10, (3×4=) 12, (2×7) 14 and (3×5=) 15. Of the remaining 8, 9, 1 1 3, only 8 and 9 are continuous, so this number is not a multiple of 8 and 9. The minimum common multiple of 2, 3, 4, 5, 6, 7, 10,12, 13, 14, 15 is 22× 3× 5×.
16. Xiao Wang and Xiao Li usually like playing cards and have strong reasoning ability. One day, they were playing cards with Professor Hua around the table, and Professor Hua gave them a reasoning problem. Professor Hua drew the following 18 playing cards from the table:
Hearts a, q, 4 and spades j, 8, 4, 2, 7, 3, 5.
Cao Hua K, Q, 9, 4, 6, lO Plaza A, 9
Professor Hua draws a card from 18, tells Xiao Wang the number of points in this card and tells Xiao Li the color of this card. Then Professor Hua asked Xiao Wang and Xiao Li, "Can you infer what this card is from the known points or colors?
Xiao Wang: "I don't know this card."
Xiao Li: "I know you don't know this card."
Xiao Wang: "Now I know this card."
Xiao Li: "I know, too."
Excuse me: What kind of card is this?
Answer box 9.
Wang knows the number of points in this card. Xiao Wang said "I don't know this card", which means that the number of points in this card can only be one of A, Q, 4 and 9, because all other points are just one card.
If the number of points in this card is not A, Q, 4 and 9, then Xiao Wang knows this card, because all points except A, Q, 4 and 9 are spades and flowers. If this card is spades or grass flowers, Xiao Wang may know this card, so Xiao Li said, "I know you don't know this card", indicating that the color of this card is hearts or diamonds.
The problem now focuses on the five cards of hearts and diamonds.
Because Xiao Wang knows the number of points in this card, Xiao Wang said, "Now I know this card", which means that the number of points in this card is not A, otherwise Xiao Wang still can't tell whether it is a heart or a diamond.
Because Xiao Li knows the color of this card, Xiao Li said "I know it too", which means that this card is diamond 9. Otherwise, if the suit is a red heart, you can't tell whether it is a red heart Q or a red heart four.
It is suggested that in logical reasoning, it should be noted that a proposition points to a conclusion, and its inverse proposition is also a clear conclusion.
10. Two natural numbers from 1 to 100 are taken out at one time. If the sum of the two is greater than 100, there are _ _ _ _ ways to take * *.
Answer 2500
There are two numbers a and b in the solution, a < b,
When a is 1, b can only be 100 and 1.
When a is 2, b can be 99, 100, which can be taken in two ways;
When a is 3, b can be 98,99, 100, and there are three ways to take it;
When A is 4, B can be 97, 98, 99, 100. There are four methods.
When A is 5, B can be 96, 97, 98, 99, 100. There are five methods.
…… …… ……
When A is 50, B can be 5 1, 52, 53, …, 99, 100, 50;
When A is 5 1, B can be 52, 53, …, 99, 100, 49;
When A is 52, B can be 53, …, 99, 100, 48;
…… …… ……
When a is 99, b can be 100 and 1.
Therefore, * * has1+2+3+4+5+…+49+50+49+…+2+1= 502 = 2500.
How many different ways are there to take two different numbers from 1- 100 and make their sum a multiple of 9?
Considering that the remainder is divided by 9, we know that the sum of two different numbers divided by 9 is 9. Through calculation, it is easy to know that there are 12 kinds of remainder divided by 9 1,1/species has 2-8 remainder, and1/species has 0 remainder, but there are1/kloc-0. And the remainder is 1, which means 12, which means 1 1 more. In this way, we can see that there are 1- 100, and each number corresponds to 1 1 situation.
1/kloc-0 /×100 ÷ 2 = 550 species. Divide by 2 because 1+8 and 8+ 1 are the same.
14. The product of each digit of a given three-digit number is equal to 10, so the number of three digits is _ _ _ _.
Answer 6
Because 10 = 2× 5, these three digits can only be composed of 1, 2,5, so * * * has = 6.
12. There are five triangles in the figure below, and the sum of the three numbers in each triangle is equal to 50, where A7 = 25, A 1+A2+A3+A4 = 74, A9+A3+A5+A 10 = 76. What is the sum of A2 and A5?
Answer 25
The solution is A 1+A2+A8 = 50,
A9+A2+A3=50，
A4+A3+A5=50，
A 10+A5+A6=50，
A7+A8+A6=50，
So a1+a2+A8+A9+a2+a3+a4+a3+a5+a10+a5+a6+a7+A8+a6 = 250,
That is, (a1+a2+a3+a4)+(a9+a3+a5+a10)+a2+a5+2a6+2a8+a7 = 250.
There are 74+76+A2+A5+2 (A6+A8)+A7 = 250, and there are A6+A7+A8 = 50 in triangle A6A7A8, where A7 = 25, so A6+A8 = 50-25 = 25.
Then A2+A5 = 250-74-76-50-25 = 25.
It is suggested that the above deduction is completely correct, but we lack a sense of direction and overall grasp.
In fact, when we see such a digital array, our first feeling is that the five 50s here do not refer to the sum of 10 numbers, but the sum of this 10 number plus the number of the inner circle 5. This is the most obvious feeling and an important equal relationship.
Then "look at the problem and set the direction" and find the sum of the second number and the fifth number.
It is related to the other three numbers in the inner circle. The sum of the sixth number and the eighth number is 50-25 = 25.
Look at the third number. When the numbers 1, 2, 3 and 4 of two straight lines are added with the numbers 9, 3, 5 and 10, the third number will be counted repeatedly.
The play begins:
74+76+50+25+ second number+fifth number = 50× 5
So the second number+the fifth number = 25.
13. There are three sets of numbers below.
( 1) , 1.5, (2)0.7, 1.55 (3) , , 1.6,
Take a number from each group and multiply the three numbers. What is the sum of the products of three numbers in different ways?
Answer 720
Pave it into a 6×5 grid, and fill in 0, 1, 3, 5, 7 and 9 in the top row in turn; Fill in 0, 2, 4, 6 and 8 in the leftmost column in turn, and the numbers in each other's cells are equal to the sum of the leftmost number in the same row and the top number in the same column. Q: What is the sum of these 30 numbers in turn?
The solution is the same as the original problem. (2+4+6+8)×6+( 1+3+5+7+9)×5=245
Because the original problem is more complicated, we can also talk about this problem first and then talk about the original problem.
Solution = 16× 2.25× 20 = 720.
It is suggested to deduce this part, but don't forget to help students review a formula for finding the sum of all divisors. The opportunity to master comprehensive learning has come.
Family work
1.
answer
The numerator denominator decomposition factor: 9633 = 3× 32 1 1, 35321=1× 321.
It is suggested that it is better to divide the points by turns.
14. Party A and Party B start from A and B at the same time, face to face, and the speed ratio is 3: 2 when starting. After their first meeting, Party A's speed increased by 20% and Party B's speed increased by 30%. In this way, when A arrives at B, B is still 14 km away from A, so A and B,
The answer is 45 kilometers.
Let's assume that the distance between a and b is 5 segments. According to the speed ratio of the two people, when we meet for the first time, A takes 3 paragraphs and B takes 2 paragraphs. After that, A has to walk 2 paragraphs and B has to walk 3 paragraphs. When a and b increase the speed respectively, the ratio is:
The topic of tip is old-fashioned, but considering the flexibility of the method, we can practice it in different ways.
This problem can also be solved by general ratio (or even ratio).
14 ÷ (27-13) × (27+18) = 45 (km)
20. At the New Year's party, 2 1 students from Class 1, Grade 6 took part in the guessing activity, and they guessed 44 riddles correctly. Then at least _ _ _ _ _ of 2 1 students answered as many riddles correctly.
Answer 5
We should make the number of riddles guessed as evenly as possible, including:
0+0+0+ 1+ 1+ 1+ 1+2+2+2+3+3+3+4+4+4 = (0+65438+
So at this time, five people guessed as many riddles correctly, all four.
It is not difficult to verify that at least five people guessed the same number of riddles correctly.
The difficulty of this problem lies in the starting point, that is, the way of thinking. Students can speak, and their speeches can lead to problems, so that students can fully express their opinions, and then under the inspiration of teachers, correct and solve problems. This statement is better than the teacher directly cutting into the problem.
Note that if there is no limit on the number of people, "at least" here should be 1 person. Combine 2 1 person, and you should find the direction.
26. Project A can be completed by one person in 50 days, and Project B can be completed by one person in 75 days. Now they are cooperating, but B left for a few days because of something. The project will be completed 40 days after the start of construction, and B will leave the company for _ _ _ _ _ _ _.
Answer 25
Xie B left halfway, but A worked for 40 days from beginning to end, and the amount of work completed was 40× = pieces of the whole project.
Then the remaining 1-= was completed by B. It took B ÷ = 15 days to complete, so B left for 40- 15 = 25 days.
Seeking adoption is a satisfactory answer.