Mathematics Olympiad for the first grade of junior high school, and ask for Olympiad questions for t

If the plane takes x hours to sail, it takes (, there are two cases here:

When the departure is downwind, then there is 600x=550(, and the solution is x=so the plane should return when it flies out of the kilometer.

When there is a headwind when sailing, there is 550x=600(, and the solution is x=so the plane should return when it flies out of the kilometer.

Therefore, regardless of whether the flight is downwind or headwind, the aircraft should return after flying 1,320 kilometers.

Set the flight x hours to return home, and the return time is hours.

575+25)x=(575-25)*(

600x=2530-550x

1150x=2530x=

It takes x hours to go and y hours to return.

x+y=575＋25）x＝（575-25）y

Solve x,y

Then calculate (575+25) x 1320km

So the plane should return after flying 1320km.

Solution: If you fly x kilometers, you have to return.

Then the meaning of the question can be listed: x (575 + 25) + x (575-25) = solution x = 1320

A: The plane should return when it has flown 1,320 km.

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Let the two numbers be 3m and 7n, respectively

3m+7n=89

m=25 when n=2

The least common multiple of 3 and 7 is 21

That is, 3 7's and 7 3's can be converted to each other.

Correspondingly, m=18 when n=5

When n=8, m=11

n=11, m=4

The two numbers that meet the conditions are:

The value range of 3x+7y=89 x is 1--27[(89-7) 3=27......1], the value of y ranges from 1 to [89-3) 7 = 12......2], respectively, substitute for the request;

In addition, 89 divided by 3 remainder is 2, 1 7 divided by 3 remainder is 1, so to achieve a remainder of 2, then the number divisible by 7 should be a multiple of 7, so there are 25 * 3 + 2 * 7 = 89 18 * 3 + 5 * 7 = 89 11 * 3 + 8 * 7 = 89 4 * 3 + 11 * 7 = 89

Let a1=mn, a2=mn+1....an=mn+n-1;

b1=pn,b2=pn+1...bn=pn+n-1;

Since n is a positive and even number, the remainders of a(n 2+1) and b(n 2+1)de are n 2, and the remainders of a(n 2+2), b(n 2+2)de are n 2+1....

then a1+b1,a2+b2,..a(n 2) + b (n 2) remainders are.

a(n/2+1)+b(n/2+1),a(n/2+2)+b(n/2+2),a(n/2+3)+b(n/2+3)..The remainders of an+bn are also 0, 2, 4, 6, 8 respectively....n-2;

Therefore the remainder of the harvest shall be the same.

Something wrong with your writing.

Let's start with a simple example, that is.

If n=4, the remainders of these four numbers are 0,1,2,3,0+1+2+3=6,6 4=1....2

If n=6, the remainders of these six numbers are 0, 1, 2, 3, 4, 5 respectively

It can be seen that if a1, a2 ,..., divide an by n, and the remainders obtained are different from each other, then a1 plus a2 ......The remainder of the sum added to an and divided by n is half of n, i.e., n 2.

The same goes for b1 plus b2 ......The remainder of the sum added to bn and divided by n is half of n, i.e. n 2.

So, a1 + b1 plus a2 + b2 ......The sum added to an+bn divided by n should be divisible (the previous two remainders are added together, n 2 + n 2 = n, so it becomes divisible).

As can be seen earlier, if the resulting remainders are different from each other, the remainder should be n 2. Now it is divisible, so it is inevitable that there will be a situation where the resulting remainders are not the same.

Modify what you wrote earlier

Prove that n is positive and even, and n 2 is an integer.

a1,a2,…, an divided by n, the resulting remainders are different from each other, so these n remainders are exactly 0,1,...，n-1.Thus a1+a2+....+an≡0+1+…+n-1)=[0+（n-1）]*n/2

（n-2）/2]*n+n/2≡n/2(mod n)

Note: The above uses the sum of the difference series, if you don't understand it, you can understand it according to the previous simple example).

Same b1 + b2 + ....+bn≡ n/2(mod n)

But (a1+b1)+(a2+b2)+....an+bn)= a1+a2+…+an)+(b1+b2+…+bn)

0(mod n)

So a1+b1, a2+b2 ,..., an+bn divided by n, the resulting remainder must be the same.

We only need to observe each "small triangle" in the figure An (the side in it has been removed), and when the figure An becomes the figure A(N+1), the two sides of each small triangle of an are divided into three equal parts, and then a triangle with smaller rows of clumps is "grown". The total length of each small triangle after the change is 1 3 longer than the original. i.e. a(n+1)=an*(1+1 3)=an*(4 3).

So it is a proportional series with a1 as the first term and the common ratio is 4 3. It is easy to find: an=a1*(4 3) (n-1), and a1=3, so an=3*(4 3) (n-1), (n 1).

The answer you gave is wrong, obviously it can't be an=3+(4 3) (n-2), it should be a multiplier, not a plus! Mathematics is verifiable. You can just put n=3 generations into the number of pictures or laughs.

a1=3a2=a1+(1/3)*3=a1+1

a3=a2+(1/3)^2*3*4=a2+(4/3)^1

a4=a3+(1/3)^3*3*16=a3+(4/3)^2

a5=a4+(1/3)^4*3*64=a4+(4/3)^3

an=a(n-1)+(4/3)^(n-2)

Add the two sides of the equation to Li Zhengchen to get:

a1+a2+……an=a1+a2+……a(n-1)+3+1+(4/3)^1+(4/3)^2+……4/3)^(n-2)

i.e. an=3+1+(4 3) 1+(4 3) 2+......4/3)^(n-2)

1, (4 3) 1, (4 3) 2, which Zen ......, 4 3) (n-2) is the first proportional series with a term of 1 and a common ratio of 4 3.

There are (n-1) terms, and the sum formula sn=a1(1-q n) (1-q) of the ratio is used to obtain the following

sn=1*[1-(4/3)^(n-1)]/1-4/3)=-3+3*(4/3)^(n-1)

So: an=3+sn=3-3+3*(4 3) (n-1)=3*(4 3) (n-1).

1.Let the uphill from A to B be A, the flat land is B, and the downhill is C

a/72+b/63+c/56=4 7a+8b+9c=4*7*8*9

a/56+b/63+c/72=14/3 9a+8b+7c=7*8*9*14/3

Add 16 (a + b + c) = 7 * 8 * 9 * (4 + 14 3).

a+b+c=273

2.Let the ** number be abcdefgh, then.

abcd abc

efgh + defgh

From the latter one can be sure of d = 1, substitute the previous rank h=4, substitute the last one rank c = 6, substitute the top ten rank g = 4, substitute the last ten rank (note that there is a carry) b = 2, substitute the top 100 rank (with carry) f = 1, and then substitute the last hundred rank a = 8 e = 6

**Numbers are 82616144

3.The number of people is xIf there is one person in the car who does not get on the bus, it is A

x a = 30 more than 1

One less car, still 30 people per car x (A-1) = 30 more than 30 + 1, and the remaining people are just a multiple of A-1 car.

So 31 (a-1) = n n is a natural number and n is less than or equal to 10 (each car can only have 40 people, there are already 30, at most ten can be added), 31 is a prime number. So n can only be 1.

So a=32 x=961 is 961 people.