The 8th National Mathematics Knowledge Application Contest 7th Grade Preliminary Round Paper B Answe

As long as the answer is different.

That's a hard question!! ）

1 corresponds to f, 2 corresponds to t, and 3 corresponds to b

4 corresponds to c, 5 corresponds to i, and 6 corresponds to e

7 corresponds to H, 8 corresponds to D, and 9 corresponds to G

10 corresponds to a

7.China's total of gold medals is 51, the United States' total of silver medals is 38, and Russia's total of bronze medals is 28

Hint, count the American first. ）

9.Yes and no.

Yes South Korea , b is France , C is Japan , D is United States 11Correct answers: 2 true, 5 true, 1 false, 3 false, 4 true, 6 false, so D scores 30

Number of pieces a b c

There are 8 kinds in total, and there are 0 that are not counted.

2) The more A, the more B is the most economical.

So choose the last 812.

Probably done wrong.

A to B 5 pieces, B to C 2 pieces, C to D 3 pieces, E to D 4 pieces, a total of 14 pieces 15It's so simple, just make it up.

There are a total of 1,000 people in our school, 10% of them participate in the competition, ask how many people participate in 1,000*10%=100 people.

Solution: When n is an odd number, n-1 is an even number, the sum of the original n lamps is n (each lamp is recorded as 1), and the sum of each operation is n-1, then, no matter how many times it is pulled, it is impossible to appear: odd number = even number * any number. Therefore, when n is an odd number, this cannot be realized.

When n is an even number, n-1 is an odd number, then as long as n times is pulled, it will appear: even number = odd number * even number.

Here's an example.

For example, n=2, you can turn it all off twice, and turn the switch on.

Close n=4, four times is fine.

Open open open open. Close, close, close, open.

Turn off the switch. The switch is on.

Customs, customs, customs. n=6, six times is fine.

Open open open.

Close, close, close, open.

Turn off the on-off switch.

The on/off switch is off/off.

Close-off-off.

Turn on the switch on.

Guanguan

Wait a minute. As long as n is an even number, you can achieve all off.

Yes. There is also 1 light on for the first time. I turned it off for the second time, and there were n-2 lights in total. In this case, when n-1 is pulled, there are exactly n-1 lights on. So pull n times to turn off the lights completely.

n can't be closed, it has to be more than the three, and the answer he provided is correct.

This kind of analogy can be demonstrated with examples, for example, it is impossible when n=1! n=2 twice is fine! n=3 is also not true, and moving two at the same time will only result in one on, two offs and all open!

So it's impossible to close them all! And so on, and so on, and the latter cannot be closed entirely! It's only possible twice a time!

A total of 12 + 14 + 13 - 5 - 4 - 7 + 3 = 26 people participated in the competition.

So there are 30-26 = 4 people who didn't participate in 3 people.

12 Mathematics 5 Mathematics + 4 Mathematics.

14 Essays 5 Essays + 7 Essays.

13 English, 4 English + 7 English.

All 3 conditions are met.

So, the number of people left = 30-3 all participants -14

Draw three rings with ** All overlap with each other to represent English, math, and composition Then fill in the numbers, all three overlap and fill in 3 Math Composition Add 2 , English Math Add 1, Composition English Add 4, English Add 13 -4-3-1 =5 Math Add 12-3-2-1 = 6 , Essay Add 14-2-3-4=5 Statistics All the numbers in the ring are combined into 23 Last 30-23 =7

Your question is not very clear, so there are 4+7+3=14>13 for those who participate in English, is there an original question?

The minimum value of 5 is demonstrated as follows:

It should be remembered that divisible by 10 means that the sum of any number taken from n numbers can only be 10 20 30 40

1) First of all, use a special method to exclude: n takes 1, 2, 3, 4 does not meet the conditions: if n takes 4, then take these four numbers as 9, 8, 7, 6 Among these 4 numbers, 10 "the sum of any two numbers< 20, 30 "the sum of any three numbers< 40, so no matter how you take it, it is impossible to take out the number divisible by 10, and for n to take 1, 2, 3, because 4 is no longer satisfied, it can be excluded (for example, if n takes 3, you take 9, 8, 7).

2) Then use the grouping method to determine that 6, 7, 8, and 9 are all desirable n values: group these 9 numbers 19 28 37 46 5, and among the n numbers taken arbitrarily, as long as the two numbers of the same group are taken at the same time, this n meets the requirements of the problem, and n takes 6 means brushing off 3 numbers, and taking 7 means brushing off 2 numbers. In order to avoid getting the number of a group at the same time, at least 4 numbers must be brushed off, so 6, 7, 8, and 9 all meet the requirements.

3) to study the assumption that n is 5: if you group with (3), you must brush off 4 numbers in 4 groups, that is, 5 must be the number taken. Since 5 is taken, then the mantissa of the sum of the two numbers can be found to be 5 (in fact, it can only be 5 or 15), then n=5 is feasible.

Now drill the horns as much as possible so that n=5 doesn't work.

For this game goes into the following rules: 5 has been decided; If 1 is taken, neither 9 nor 4 can be taken; If 2 is taken, neither 8 nor 3 can be taken; If 9 is taken, neither 1 nor 6 can be taken; If 8 is taken, then neither 2 nor 7 can be taken The only way to analyze and meet this requirement is the following 2x2=4 group of methods:

Corresponding to each group: 2+1+7 5+9+4+2 5+1+6+8 9+8+3

Therefore, n=5 cannot drill the tip of the horns

So, the minimum value of n is 5 Proof Complete!

2.First, 6 positive integers a1, a2....a6, take 3 numbers with a total of c(3,6)=20 ways to take them, so for a given a1,a2....a6,f(i,j,k) can have 20 possible values.

f(i,j,k)<=(1 3)+(2 2)+(3 1)=13 3.

We segment 0 to (13 3):

0 to (1 2) 1 1, (1 2) to 1, 1 to (3 2).

.4 to (13 3) paragraphs.

It is divided into 9 sections.

f(i,j,k) takes the value in these 9 paragraphs, and for the given a1, a2....a6,f(i,j,k) can have 20 possible values.

Since 20 = 9 * 2 + 2, there must be 3 f(i, j, k) falling in the same segment.

That is, the absolute value of the difference between the three f(i,j,k) pairs is less than.

Set a seven-ton car x, 8 tons y, the total cost w, 7x+5y=73

y=73/5-(7/5x)

w=50y+65x=730-5x

So, when x is the largest, w is the least.

But 73 tons of cargo must be transported, and the car is a whole number.

When x=9 and y=2, the maximum saving is 685 yuan.

Another way of thinking:

Only the mantissa of an integer multiple of 7 may be 3, so 7 tons should be 9 cars, 7 * 9 = 63 and the remaining 10 tons are 2 5 tons of trucks.

Freight: 2 * 50 + 9 * 65 = 685 (yuan).

With 2 5-ton trucks and 9 7-ton trucks, a total of 2 * 50 + 9 * 65 = 685 (yuan) is required

2010 National Junior High School Mathematics Competition.

As shown in the figure, in the square ABCD, E is the moving point on the edge of the CD, the angle EAF=45, AF crosses BC at the point F, when DE=3,EF=8, what is the length of BF??? One question.

What is the drawer theorem.

For example, a deck of playing cards with the same back, after removing the queen and the queen of the remaining 52 cards, four suits, 13 cards of each suit, wash the back side up, draw cards at will, at least how many cards must be drawn, in order to ensure that at least 4 cards of the same suit.

Each question needs to be explained in detail.

I have the answer. The process is the most important.

Thank you!! Solution: Cross the point A as AG AE, and cross the extension line of CB at the point G (note: draw the following figure by yourself).

The quadrilateral ABCD is a square.

ab=ad,ab⊥bc，ad⊥cd，ba⊥da

abg=∠ade

Again ag ae

eag=∠bad

eag-∠eab=∠bad-∠eab

ABG is equal to ADE (ASA).

ae=ag，de=bg=3

and eag=90°, eaf=45°

fag=∠eaf=45°

and af=af(common edge).

fag is equal to fae

gf=ef=8

bf=gf-bg=8-3=5

bf=52.At least 13 cards should be drawn.

Basic drawer principle.

There are two things: (1) If x+k(k 1) elements are placed in x drawers, then at least one drawer contains 2 or more elements.

Like the one above, think of suits as drawers and cards as elements.

500 students is the exact value, 8 30 is the approximate value, so choose: c