The TV station will broadcast a 30-episode TV series. If the number of episodes scheduled to be broadcast every day is not equal to each other, how many days can the TV series be broadcast at most?
A square paper box can just hold a cylinder with a volume of 628 cubic centimeters. What is the volume of this carton? (Pi =3. 14)。
There is a basket of apples. After dividing them into three equal parts, there are still two apples left. Take out two, divide them into three equal parts, and there are two left. Then take out two, divide them into three parts, and there are two left. Q: How many apples are there in this basket?
5. Calculation:
6. The circumference of rectangular ABCD is 16m. Draw a square with this side as the length of each side. It is known that the sum of the areas of these four squares is 68m2. Find the area of rectangular ABCD.
7. The "Hua" Golden Cup Junior Mathematics Invitational Tournament was held in 1986, the second in 1988, the third in 199 1 year, and every two years thereafter. The sum of the figures of the first "Huabei" year is: A 66.
The sum of digits in the first two sessions is: A2 =1+9+8+6+1+9+8+8 = 50.
Q: What was the number of the first 50 "China Cup"? A50= =?
8. Arrange natural numbers in the following order:
[blockquote]
1 2 6 7 15 16 …
3 5 8 14 17 …
4 9 13 …
10 12 …
1 1 …
[/blockquote]
In this arrangement, the number 3 is in the first column of the second row and 13 is in the third column of the third row. Q: In which ranks is 1993?
9. Try to fill in the eight numbers 1, 2, 3, 4, 5, 6, 7 and 8 in the small circle shown in the figure below, so that the difference between the numbers filled in the two small circles connected by line segments in the figure (big numbers minus numbers) is exactly 1, 2, 3, 4, 5, 6 and 7.
10.
What is the remainder of dividing by 3? Why?
Six players (1 1. A, B, C, D, E, F) Play a single-cycle table tennis match (each player plays one game with other players), and each player plays one game on three tables at the same time every day. It is known that on the first day B plays D, on the second day C plays E, on the third day D plays F, on the fourth day B plays C, Q: fifth. Who is playing with whom on the other two tables?
12. There are several thin strips with the lengths of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 and1cm respectively, and the number is sufficient, so three strips can be properly selected as three sides to form one.
13. Color the circle in the picture below with red or blue at will. Q: Is it possible to make the number of red circles on the same line all odd? Please explain the reason.
14.a and B have special training on the same oval track: they start from the same place and run in the opposite direction at the same time. After everyone reached the starting point after running the first lap, they immediately turned back and accelerated the second lap. On the first lap, the speed of B is the speed of A..
On the second lap, A ran faster than the first lap.
B's speed increased on the second lap
It is known that the second intersection of A and B is 90 meters away from the first intersection/kloc-0. How long is this oval runway?
15. In the figure below, the area of the square ABCD is 1, and m is the midpoint of the AD side. Find the area of the shaded part in the picture.
16. For a party of four people, each person brought two gifts for two of the other three people. It proved that there were at least two couples, and each pair gave gifts to each other.
answer
[blockquote] 1。 1 2.7 3.8 4.23 5.
6. 15 7.629 8. Line 24, column 40
9. At places A, B, C, D, E, F and H, fill in 1, 3, 8, 2, 7, 4, 5, 6 10. 1 in turn in a small circle.
1 1. On the fifth day, A played against B, and the other two tables, C played against D and E played against F 12. 36 13. Impossible.
14. Runway length is 400 meters 15. The shaded area in the figure is
.
16. After giving gifts, there are eight gifts for four people, with an average of two for each person. If a person has more than two, it must be three, and each of the other three people has one except himself. So this man and two people who got gifts themselves formed two pairs. If four people each get two gifts from others, then his own two gifts can't be given to only one person. Therefore, he and the person who received his gift were divided into two groups.
The fourth rematch of China Cup
1. Molecules with original molecular formula =
=
=
Denominator of the original formula =
[blockquote]
=
=
=
=
=
[/blockquote]
So the original formula is equal to 1.
2. If it is broadcast for more than 8 days, at least 1+2+3+4+5+6+7+8 = 36 episodes due to the different number of episodes broadcast every day.
Therefore, it is impossible to broadcast more than 8 days and 30 episodes as required. Conversely, 1+2+3+4+5+6+9 = 30.
So it can be broadcast for 7 days at most, and the number of episodes broadcast per day is 1, 2, 3, 4, 5, 6 and 9 respectively.
3. The height and the diameter of the cylinder of the solution are equal to the side length of the cube, that is, 6.28 = 3. 14 × side length×
So (side length)
=
× 4 = 8, that is, the volume of the carton is 8 cubic centimeters.
4. Solution If four apples are added, the first time it is exactly divided into three parts, and each part has two more apples than the original. The same is true for the second time and the third time. There are two more apples in each part of the third time, and because the total number of two apples in the second time is even, the number of apples in each part of the third time is even, so there are at least four apples in each part of the third time.
It is also explained that if you add four apples, the first time you only need to divide them into three parts (two more than the original), the second time you take two parts (four more than the original two), then you only need to divide them into three parts (two more than the original two), and finally you take two parts (four more than the original two), which is exactly three parts. Because the last time you divided them, the total was even (because you took two parts), so each part was even.
= 6 (pieces). Last time each piece was at least 6×
= 9 (pieces), originally 9× 3 = 27 (pieces), originally this basket of apples was at least 27-4 = 23 (pieces).
5. Solution formula = (1+3+5+7+9+1+13+17)+(
)
[blockquote][blockquote]=
=8 1+
=
[/blockquote][/blockquote]6。 Take it as an illustration.
Extend to the right,
Extend upward, pass through point E, and then the square.
The area is equal to the square of the half circumference of the rectangular ABCD, that is, 64 square centimeters.
Congruent and square.
and
The sum of the areas is equal to half of the sum of the areas of the four squares given in the question, that is
× 68 = 34 square centimeters. 64-34 = 30 square centimeters should be equal to twice the area of rectangular ABCD. So the area of ABCD is
× 30 = 15 cm2.
7. According to the given law, there were 7 events in the 20th century and 43 events in the 20th century.
It is known in the 20th century that A2 = 50, and the sum of digits in other five years is: 5× (1+9+9)+(113+5+7+9) = 95+25 =120.
Therefore a [sub] 7 [/sub] = a [sub] 2 [/sub]+120 =170.
2 1 The sum of the figures in the first 45 years of the century is: 2× 45+(1+2+3+…+8 )× 5+(1+3+5+7+9 )× 9 = 495.
The sum of the figures in the first 43 years is: 495-2-8-7-2-8-9 = 459.
So a [sub] 50 [/sub] =170+459 = 629.
8. The number of odd diagonal lines increases from bottom to top, and the number of even diagonal lines increases from top to bottom.
What is the largest number on the nth diagonal?
n(n+ 1)
The largest number on the diagonal of article 62 is
× 62× 63 = 1953. The largest number on the 63rd diagonal is 1953+63 = 20 16. So 1993 is located on the 63rd diagonal. The median in the 63rd diagonal line increases from bottom to top, and the first digit on the left is 193. 1993 is located on the 63rd diagonal line, and the number is (1993-1954+1) = 40, that is, (63-40+ 1)= 24 rows and 40 columns of the original array.
Answer: 1993 Row 24 Column 40.
9. There are many solutions, as shown below:
10. When the solutions 3[sup]3[/sup], 6[sup]6[/sup] and 9[sup]9[/sup] are divided by 3, the remainder is 0, so just look at the expression1[sup]1[/sup].
Note: If A is divided by 3 and the remainder is A [sub]1[sub], and B is divided by 3 and the remainder is B [sub]1[sub], then the remainder obtained by dividing a×b by 3 is A [sub]1[/sub ]× B [sub]/kloc-.
Because the remainder when 4 and 7 are divided by 3 is 1, when 4[sup]4[/sup] and 7[sup]7[/sup] are divided by 3, the remainder is also 1.
Because 5 and 8 are divided by 3, the remainder is 2, so 5[sup]5[/sup] and 8[sup]8[/sup] are divided by 3, and the remainder is the same as 2 [sup] 5 [/sup] and 2 [sup] 8 [/sup] divided by 3. And 2 [sup] 4 [/sup] = Therefore, 2 [sup] 5 [/sup] = 2 [sup] 4 [/sup] x 2 divided by 3 is 2, and 2 [sup] 8 [/sup] = 2 [sup] x 2 [sup] 4 [/sup] divided by 3 is/sup.
So1[sup]1[/sup]+2 [/sup]+4 [sup]15 [sup]+5 [/sup]+7 [sup]+8 [sup] 8.
1 1. Solve the problem that B can't be right for A on the second day, otherwise B can be right for A, D can be right for F, D can be right for F on the third day, so B should be right for F, and A should be right for D. 。
On the third day, B can't be against A, otherwise C is against E, and the next day, C is against E, but B is against E (not against C on the fourth day), A is against C, and D is against E on the fourth day, so on the fifth day, B is against A, D is against C, and E is against F.
12. To solve a triangle, the sum of the lengths of any two sides is longer than the other side. In this problem, if the other two sides of a triangle with a base of 1 1 cm are A and B, then there must be11< a+B. In addition, for the sake of accuracy, a can be set.
( 1 1, 1 1); ( 10， 1O)，( 10， 1 1)； (9,9),(9, 10),(9, 1 1); (8,8),(8,9),(8, 10),(8, 1 1); (7,7), (7,8),(7,9),(7, 10),(7, 1 1); (6,6),(6,7),(6,8),(6,9),(6, 10),(6, 1 1); (5,7), (5,8),(5,9),(5, 10)(5, 1 1); (4,8),(4,9),(4, 10),(4, 1 1); (3,9),(3, 10),(3, 1 1); (2, 10),(2, 1 1); (1, 1 1)***36 species.
Answer: It can form 36 different triangles.
13. The solution assumes that each row has an odd number of red circles, and the sum of five red circles is still odd.
On the other hand, when the number of red circles on five lines is added, because each circle is on two lines, it is calculated twice, so the added sum should be even, and the result is contradictory, so it is impossible to make the number of red circles on the same line odd.
14. Solutions
Let's draw two pictures (above). Let the speed of A be A at first, then the speed of B is
A. If the runway length is L, then the point where A and B meet for the first time is the distance from the starting point according to the direction of A..
A finished the first lap and B ran away.
, b Run the rest again.
, A has turned back, and with a (1+
)=
A ran at a speed, so by the time B finished the first lap, A had turned around and ran.
At this time, B will take it back.
One (1 10
)=
The speed of A is running. Since then, the speed ratio of A and B has been
one
a=
So when they met for the second time, A ran away from the others.
about
And b escaped.
That is, the second meeting started from the starting point.
×
=
As you can see, the distance between two intersections is (
-
L = 190 (m), i.e.
= 190 (m),
L = 400 meters
A: The runway is 400 meters long.
15. When AM‖BC is needed for solving, △GAM is directly proportional to the edge correspondence of △GCB.
that is
,
therefore
=2,
=2.
Because the side length of the square ABCD is 1.
=
× 1×
=
,
=
× 1×
=
,
therefore
=
=
×
=
,
=
=
×
=
.
+
=
+
=
That is, the area of the shaded part is
.
16. Explain that these four people are represented by four points. If two people give gifts, they will connect a line. Because everyone gives two gifts, the number * * * has eight lines. Because everyone gives gifts to two people, there is at most 2 between two points (=1+65438).
Note that there are 6 kinds of socks, each with no more than 2 socks. If you take out eight pairs of socks, there must be two kinds of socks. This is essentially the same as this question.
[/blockquote]，3。 Help me find some math problems in the sixth grade of elementary school (try to be from China Cup).
Find some application problems of fractions.