1, here is a series of numbers arranged according to the law. What number is 1995?

Answer: 2,5,8, 1 1, 14, ... According to the law, this is a arithmetic progression, the first term is 2, and the tolerance is 3, so the term 1995 = 2+3× (/kloc-0.

2. Of the natural numbers starting from 1, what is the100th number that cannot be divisible by 3?

Answer: We found that every three numbers in 1, 2, 3, 4, 5, 6, 7, ... are grouped from 1, and the first two of each group cannot be divisible by 3. If there are two groups, 100 will have 100 ÷ 2.

3. If 1988 is expressed as the sum of 28 consecutive even numbers, what is the largest even number?

Answer: 28 even numbers are grouped into 14 groups, and 2 symmetrical numbers are grouped into groups, that is, the minimum number and the maximum number. The sum of each group is:1988 ÷14 =142, and the difference between the minimum number and the maximum number is 28- 1=27 tolerance.

4. In the integer greater than 1000, find all the numbers with equal quotient and remainder after division by 34, so what is the sum of these numbers?

Answer: Because 34× 28+28 = 35× 28 = 980 < 1000, there are only the following figures:

34×29+29=35×29

34×30+30=35×30

34×3 1+3 1=35×3 1

34×32+32=35×32

34×33+33=35×33

The sum of the above numbers is 35× (29+30+31+32+33) = 5425.

5. There is a 1, 2, 3, ... which says 134 and 135 respectively. Take some cards out of the box at random, calculate the sum of the numbers on these cards and divide by 17, then write the remainder on another yellowcard and put it back in the box.

Answer: It is difficult to grasp several times at a time. It is better to consider the whole situation and take a step back to a simple situation analysis: suppose there are two numbers, 20 and 30, and divide their sum by 17 to get the number of yellow cards. If calculated separately, it is 3 and 13, and then the sum of 3 and 13 is divided by 65436. That is to say, no matter how many numbers are added, the remainder of the sum divided by 17 remains unchanged, and we return to the topic1+2+3+...+134+135 =136×136. The remainder is 0, and19+97 =1116 ÷17 = 6 ... 65438+.

6. The following formulas are arranged regularly:

1+ 1, 2+3, 3+5, 4+7, 1+9, 2+ 1 1, 3+ 13, 4+ 15, 65438

Solution: First find out the law: each formula is added by two numbers, the first number is the cycle of 1, 2, 3, 4, and the second number is the continuous odd number starting from 1 Because 1992 is even, the second of the two addends must be odd, so the first one must be odd, so it is 1 or 3. If it is 1: then the second number is1992-1=191. 199 1 is (1 991+1) ÷ 2 = 996, and1is always odd and inconsistent, so this formula is 3+/kloc-0.

7. As shown in the figure, the upper and lower rows in the table are arithmetic progression, so what is the minimum difference (reduction of large numbers) between two numbers in the same column?

Answer: From left to right, their differences are: 999, 992, 985, ... 12, 5. From right to left, their differences are: 1332, 1325,13/kloc.

8. There are 19 formulas:

So what are the results on the left and right sides of equation 19?

Answer: Because the left and right sides are equal, we might as well just consider the situation on the left and solve two problems: How many are used in the first 18 formula? All numbers are 5, 7, 9, ..., 18, and 5+2× 17 = 39, 5+7+9+...+39 = 396, so the equation of 19 starts from 397; How many numbers are added to the formula 19? The numbers on the left are 3, 4, 5, ... and 19 should be 3+ 1 × 18 = 2 1, so the result of 19 formula is 397+398+399+...+.

9. The number of two known columns: 2, 5, 8, 1 1, …, 2+(200-1) × 3; 5、9、 13、 17、……、5+(200- 1)×4。 They are all 200 projects. How many pairs of the same number of items are there in these two columns?

A: It is easy to know that the first such number is 5. Note that in the first series, the tolerance is 3, and in the second series, the tolerance is 4. That is to say, the second logarithm minus 5 is a multiple of 3, which is a multiple of 4, so the number converted into arithmetic numbers is calculated with the tolerance of12,5,17,29, ... The maximum number of the second sequence is 5+(200-1) × 4 = 80. The maximum number of new series cannot exceed 599, because 5+ 12× 49 = 593, 5+ 12× 50 = 605, so there are 50 pairs of * *.

10, as shown in the figure, has a lower triangle with a side length of 1 m. Starting from the vertex of each side, take a point every 2 cm, and then take these points as the endpoint to divide the large regular triangle into many small regular triangles with a side length of 2 cm as parallel lines. Find (1) the number of small regular triangles with a side length of 2 cm, and (2) the total length of parallel lines.

Answer: (1) From top to bottom, * * has 100÷2=50 lines, the first line has 1, the second line has 3, the third line has 5, ... and the last line has 99, so * * * has (/kloc. There are three parallel lines, which are the same. There are 49 lines * * * in the horizontal direction, the first one is 2cm, the second one is 4cm, the third one is 6cm, ..., and the last one is 98 cm, so the length of * * * is (2+98)×49÷2×3=7350 cm.

1 1, a factory 1 1 is busy in October and does not rest on Sundays. And from the first day, the same number of workers are sent from the general factory to the branch factory every day. By the end of the month, there were 240 workers left in the main factory. If at the end of the month, the workload of workers in the general factory is 8070 working days (one person works 1 working day) and no one is absent, how many workers will the general factory send to work in the branch factory this month?

Answer: 165438+ 10 has 30 days. According to the meaning of the question, the number of people in general factories is decreasing every day, and finally it is 240. The number of people every day constitutes arithmetic progression. According to the nature of arithmetic progression, the sum of the number of people on the first day and the last day is equivalent to 8070÷ 15=538, that is to say, there are 538-240=298 workers on the first day, and (298-240) II is scheduled every day.

12, Xiao Ming read an English book. When he first read it, he read 35 pages on the first day, and then read 5 more pages every day than the day before. As a result, he only read 35 pages on the last day. The second time, I read 45 pages on the first day, and then I read 5 more pages every day than the day before. As a result, I only need to read 40 pages on the last day. How many pages are there in this book?

Answer: The first scheme: 35, 40, 45, 50, 55, ... 35 The second scheme: 45, 50, 55, 60, 65, ... 40 The secondary scheme is adjusted as follows: the primary scheme: 40, 45, 50, 55, ... 35+35. The second plan: 40, 45, 50, 55, ... (The last day is on the first day) So the second plan must be 40, 45, 50, 55, 60, 65, 70, ***385 pages.

13 and 7 teams * * * planted trees 100, and the number of trees in each team is different. Among them, the team with the most trees planted 18 trees, and the team with the least trees planted at least how many trees?

Answer: We know that the other six teams planted 100- 18=82 trees in order to make Li Ru-Nan's "What's the value of neon" stamp? As many as possible, including:17+16+15+14+13 = 75 trees, so the minimum team should plant at least 82-75=7 trees.

14. Line up 14 different natural numbers from small to large. It is known that their total number is 170. If the maximum and minimum numbers are removed, the remaining total is 150. What is the second number in the original arrangement order?

A: The sum of the maximum and minimum numbers is 170- 150 = 20, so the maximum number is 20- 1 = 19. When the maximum number is 19, there are19+18+17+16+15+14+13+. Yes18+17+16+15+13+12+/kloc-0.

Periodic problem

Basic exercises

1, (1) ○□□□□□□□□……… The 20th number is (□).

(2) The thirty-ninth chess piece is (sunspot).

2. Xiaoyu practices calligraphy. She wrote the sentence "I love the great motherland" over and over again, and the 60th word should be written (big).

Class 3.2 (1) participated in the tug-of-war competition in the school. Their participating teams are arranged in a row according to "three men and two women", and the 26th student is (male).

4. There is a column number: 1, 3,5, 1, 3,5, 1, 3,5 ... The 20th number is (3), and the sum of these 20 numbers is (58).

5. There are three kinds of beads with the same size *** 100, which are discharged continuously according to the requirements of 3 red and 2 white 1 black.

……

(1) The 52nd one is a (white) bead.

(2) There are (17) white beads in the first 52 beads.

6.a Q B: Today is Friday, and it will be Sunday in 30 days.

B Q A: If 16 is Monday, then the 3 1 day of this month is Tuesday.

May 1 2006 is Monday, so the 28th of this month is Sunday.

A, B, C and D play poker. Party A puts "Wang" in the middle of 54 cards, and the 37th card is counted from top to bottom. Party C thought about it, confidently grabbed the card first, and finally caught the "king". Do you know how C is worked out? ※? (37 ÷ 4 = 9 ... 1 The first person to get the card must catch the "king". )

answer

1、( 1)□。

(2) sunspots.

2. It's very big.

3. Male students.

The 20th number is (3), and the sum of these 20 numbers is (58).

5、

(1) The 52nd one is a (white) bead.

(2) There are (17) white beads in the first 52 beads.

6. (days) (2). (days).

(37 ÷ 4 = 9 ... 1) The first person to get the card will definitely catch the "king". ※

Improve practice

1, (1) ○□□□□□□□□……… The 20th number is (□).

(2) ○○○○○○○○○○○○○○○○○○○○○○○○○○○○967

2. There is a row of colorful flags on the sports field, with 34 faces, arranged according to "three reds, one green and two yellows", and the last face is (green flag).

3. "Love mathematics since childhood, love mathematics since childhood ..." The 33rd word is (love).

4. The class (1) participated in the school tug-of-war. The teams in their competition are arranged in the order of "three men and two women", and the 26th student is (male).

5. There is a column number: 1, 3,5, 1, 3,5, 1, 3,5 ... The 20th number is (3), and the sum of these 20 numbers is (58).

6.a Q B: Today is Friday, and it will be Sunday in 30 days.

B Q A: If 16 is Monday, then the 3 1 day of this month is Tuesday.

May 1 2006 is Monday, so the 28th of this month is Sunday.

A, B, C and D play poker. Party A puts "Wang" in the middle of 54 cards, and the 37th card is counted from top to bottom. Party C thought about it, confidently grabbed the card first, and finally caught the "king". Do you know how C is worked out? ※?

37 ÷ 4 = 9 ... 1 (The first person to get the card will definitely catch the "king"). ※

answer

1、( 1)□。

(2)○。

2. Green flag.

3. Love.

4.( 1) male students.

5. The 20th number is (3), and the sum of these 20 numbers is (58).

6. (days) (2). (days).

37 ÷ 4 = 9 ... 1 (The first person to get the card will definitely catch the "king"). ※

Fast Skill Calculation of Decimals (2)

First, the vacuum problem.

1. Calculate 4.75-9.64+8.25-1.36 = _ _ _.

2. Calculate 3.17-2.74+4.7+5.29-0.26+6.3 = _ _ _.

3. Calculate (5.25+0. 125+5.75) 8 = _ _ _ _.

4. Calculate 34.5 8.23-34.5+2.77 34.5 = _ _ _.

5. Calculate 6.25 0.16+264 0.0625+5.26.25+0.625 20 = _ _ _.

6. Calculate 0.035935+0.035+30.035+0.0761.5 = _ _ _.

7. Calculate19.9837-199.81.9+19980.82 = _ _ _.

8. Calculate13.59.9+6.510.1= _ _ _.

9. Calculate 0.125 0.25 0.564 = _ _ _.

10. Calculation11.8 43-860 0.09 = _ _ _.

Second, answer the question.

1 1. Calculation 32.14+64.28 0.5378 0.25+0.5378 64.28 0.75-8 64.28 0.125 0.5378.

12. Calculate 0.888 125 73+999 3.

13. Calculation1998+199.8+19.98+1.998.

14. There are two decimal places below:

a=0.00…0 125 b=0.00…08

1996 0 2000 0

Try a+b, a-b, a b, a b.

———————————— Answer the case ————————

1.2

Original formula =(4.75+8.25)-(9.64+ 1.36)

= 13- 1 1

=2

2. 17

Original formula = (3.71+5.29)+(4.7+6.3)-(2.74+0.26)

=9+ 1 1-3

= 17

3.89

Original formula =(5.25+5.75+0. 125) 8.

=( 1 1+0. 125) 8

= 1 1 8+0. 125 8

=88+ 1

=89

4.345

Original formula =34.5 (8.23+2.77- 1)

=34.5 10

=345

5.62.5

Original formula = 6.25 0.16+2.64 6.25+5.26.25+6.25 2

=6.25 (0. 16+2.64+5.2+2)

=6.25 10

=62.5

6.35

7. 1998

8. 199.3

The original formula =13.5 (10-0.1)+6.5 (10+0.1).

= 13.5 10- 13.5 0. 1+6.5 10+6.5 0. 1

= 135- 1.35+65+0.65

=( 135+65)-( 1.35-0.65)

=200-0.7

= 199.3

9. 1

Original formula =0. 125 0.25 0.5 (8 4 2)

=(0. 125 8) (0.25 4) (0.5 2)

= 1 1 1

= 1

10.430

Original formula = 1 1.8 43-43 20 0.09

= 1 1.8 43-43 1.8

=43 ( 1 1.8- 1.8)

=43 10

=430

1 1.

Original formula = 32.14+64.28 0.5378 (0.25+0.75-8 0.125)

=32. 14+64.28 0.5378 0

=32. 14

12.

The original formula = 0.11(8125) 73+11(93).

= 1 1 1 73+ 1 1 1 27

= 1 1 1 (73+27)

= 1 1 1 100

= 1 1 100

13.

Original formula = (2000-2)+(200-0.2)+(20-0.02)+(2-0.002)

=2222-2.222

=2222-( 10-7.778)

=2222- 10+7.778

=22 19.778

14.a+b, a+B, A has 1998 digits after the decimal point, and B has 2000 digits after the decimal point. The decimal addition needs to be aligned and then calculated according to the integer addition rule, so

a+b = 0.00…0 12508 = 0.00…0 12508

2000 bits 1996 zero

, method with a+b, digit alignment, pay attention to abdication to fill zero, because

A=0.00…0 125, b=0.00…08, from 12500-8= 12492, so

1998 bit 2000 bit

a-b = 0.00… 12492 = 0.00…0 12492

2000 bits 1996 zero

A b, A b should have 1998+2000 digits after the decimal point, but 125 8= 1000, so

a b = 0.00…0 1000 = 0.00…0 1

1998+2000 bits, 3995 zeros

A b, expand A and B at the same time 100…0 times, and get

2000 zero

a b= 12500 8= 1562.5

Calculation of geometric knowledge area

1. The playground of Renmin Road Primary School is 90 meters long and 45 meters wide. After transformation, the length is increased by10m, and the width is increased by 5m. How many square meters has the playground area increased now?

Subtract the original area of the playground for thinking navigation to get the increased area. The current area of the playground is (90+ 10)×(45+5)=5000 (square meters), and the original area of the playground is: 90×45=4050 (square meters). So now it is 5000-4050=950 square meters more than before.

(90+10) × (45+5)-(90× 45) = 950 (square meter)

Exercise (1) has a rectangular board with a length of 22 decimeters and a width of 8 decimeters. If the length and width are reduced by 10 decimeter and 3 decimeter respectively, how many square decimeters will the area be reduced?

Exercise (2) A rectangular plot is 80 meters long and 45 meters wide. If the width is increased by 5 meters, how many meters should the length be reduced to keep the area unchanged?

2. If the width of a rectangle remains the same and its length increases by 6 meters, its area will increase by 54 square meters. If the length remains the same and the width is reduced by 3 meters, its area will be reduced by 36 square meters. What is the original area of this rectangle?

The train of thought navigation shows that its width is 54÷6=9 (meters), which means: "The width is unchanged, the length is increased by 6 meters, and its area is increased by 54 square meters"; From "the length is constant, the width is reduced by 3m, so its area is reduced by 36m", we can know that its length is 36÷3= 12 (m), so the area of this rectangle is 12x9 = 108 (m). (36÷3)×(54÷9)= 108 (square meter)

Exercise (1) a rectangle. If its width remains the same and its length is reduced by 3 meters, its area will be reduced by 24 square meters. If its length remains the same and its width increases by 4 meters, its area will increase by 60 square meters. What is the original area of this rectangle?

Exercise (2) If the width of a rectangle remains the same, its length will be increased by 5 meters, and its area will be increased by 30 square meters; If the length remains the same and the width increases by 3 meters, its area will increase by 48 square meters. What is the original area of this rectangle?

Exercise (3) A rectangle, if its length is reduced by 3 meters, or its width is reduced by 2 meters, then its area is reduced by 36 square meters. Find the original area of this rectangle.

3. The picture below shows a rectangular chicken farm surrounded by a fence16m long. How big is the floor space?

According to the meaning of the question, because a wall is used on one side, two lengths plus one width is equal to 16m, and the width is 4m, so the length is (16-4)÷2=6 (m). Therefore, it covers an area of 6×4=24 (square meters).

(16-4)÷2×4=24 (square meter)

Exercise (1) The picture below shows a professional poultry farmer enclosing a rectangular chicken farm with a fence13m long. How big is the chicken farm?

Exercise (2) Use a 56-meter-long wooden fence to form a rectangle with a length or width of 20 meters. How to maximize the enclosed area by using the fence on one side?

4, a square steel plate, first cut off the rectangle of 5 decimeters wide, and then cut off the rectangle of 8 decimeters wide (as shown in the figure below). The area is reduced by 18 1 square decimeter compared with the original square. What is the side length of the original square?

Cut out the shadow, put two small squares (as shown below) together, and add a small rectangle with a length of 8 decimeters and a width of 5 decimeters respectively. The area of this splicing rectangle is:181+8× 5 = 221(square decimeter), and the length is the original square. Therefore, the side length of the original square is 22 1÷ 13= 17 (decimeter).

(181+8× 5) ÷ (8+5) =17 (decimeter)

Exercise (1) One side of the square is reduced by 6 decimeters, and the other side is reduced by 10 decimeter to become a rectangle. The area of this rectangle is 260 square decimeters smaller than that of the square. Find the side length of the original square.

Exercise (2) If the length of a rectangular plate is reduced by 5 decimeters and the width is reduced by 2 decimeters, its area is reduced by 66 square decimeters, and the rest is just a square. Find the area of the original rectangle.

Exercise (3) After a square piece of glass is cut off by 8 cm, the remaining square is 448 cm less than the original one. What is the original area of this square glass?

Reply: Han 52 1- Beginner in Jianghu Level 2 6-23 13:07

1 A number is 36, the least common multiple of A and B is 288, and the greatest common divisor is 4. How to find number B?

The greatest common divisor of numbers A and B is 6, and the least common multiple is 282. How to find a and b?

3 Cut a piece of iron with a length of 252 and a width of 120 cm into a square iron piece with the same area and a whole side length of 1cm. There is nothing left. Should I at least cut a few pieces?

The four gears have 96 teeth and the pinion has 36 teeth. After biting at point A, how many times did you bite at point A?

5 Two natural numbers A and B greater than 300, their greatest common divisor is 132, their least common multiple is 1890, and A+B =? Ask that. . We haven't learned any of the differences mentioned here. . .