A primary school math Olympiad problem with causes and steps .

Because 67% of the pigs raised by the Wang family are boars, the number of pigs raised by the Wang family should be an integer multiple of 100, (because only in this way can his pigs be an integer number, and the pigs cannot be heads, right?)

In the same way, the pigs raised by the Li family should also be pigs in multiples of 13 such as 13, 26, 39 (because only in this way can his pigs be whole heads).

Set up 100 pigs of the Wang family, and the Li family is 521-100 = 421 heads, so the boars of the Li family are 421 * 1 13 = heads, impossible! (Or that 421 is not a multiple of 13).

There are 200 pigs in the Wang family, and the Li family is 521-200 = 321 heads, so the Li family's boar is 321 * 1 13 =, impossible! (Or 321 is not a multiple of 13).

Set up 300 pigs in the Wang family, and the Li family is 521-200 = 221 heads, then the Li family's boar is 221 * 1 13 = 17, probably! (Or 221 is a multiple of 13).

Set up 400 pigs of the Wang family, and the Li family is 521-400 = 121 heads, then the boar of the Li family is 121 * 1 13 =, impossible! (Or 121 is not a multiple of 13).

And so on, you can see:

The Wang family raised 300 pigs, and the Li family raised 221 pigs!

Because the pig must be an integer, the number of pigs in the Wang family must be a multiple of 100, and the number of pigs in the Li family must be a multiple of 13.

The Wang family raised 100 heads, and the Li family raised 421 heads, and 421 is not a multiple of 13.

The Wang family raised 200 heads, and the Li family raised 321 heads, and 321 was not a multiple of 13.

The Wang family raised 300 heads, and the Li family raised 221 heads, and 221 is a multiple of 13.

The Wang family raised 400 heads, and the Li family raised 121 heads, and 121 is not a multiple of 13.

The Wang family raised 500 heads, and the Li family raised 21 heads, and 21 is not a multiple of 13.

Because it is impossible to have half a pig or a fraction of a pig, the Wang family has 100x and the Li family has 13y.

100x+13y=521

1<=x<=5

x=1 y=score.

x=2 y=fractions.

x=3 y=17

x=4 y=fraction.

x=5 y=score.

[The question conditions should be more precise, Xiao Ming Xiaoqiang can buy up to 17] "Xiaohong brought 1 yuan, you can buy up to 2" Description, the ** of popsicles is between and. , and the money brought by Xiao Ming is an integer, so the money brought by Xiao Ming can only be 4 yuan or 5 yuan.

1) If Xiao Ming's money is 4 yuan, then the ** of each popsicle is 4 11 = the approximate value method of removing the tail], because it is also more than 6 yuan, so the yuan does not meet.

2) If the money brought by Xiao Ming is 5 yuan, then the ** of each popsicle is 5 11 = the approximate value method of removing the tail].

So Xiao Ming Xiaoqiang's money is 8 yuan, and Xiaoqiang has 8-5=3 yuan.

So the ** of the popsicle is yuan, the money brought by Xiao Ming is 5 yuan, and the money brought by Xiao Qiang is 3 yuan.

Let each popsicle be x points, x is obviously an integer, Xiaohong has one yuan, buy a maximum of 2, then 2x<=100,x< 50,3x>100=33*3+1,x>33,x> 34

Xiao Ming bought a maximum of 11, Xiao Ming brought less than 12 50 = 600 points, at least 34 * 11 374 points, three people are integer yuan, Xiao Ming brought 4 yuan or 5 yuan, assum, Xiao Ming 4 yuan, at this time.

12x>400 12*33+4,x>=34,11x<=400 11*36+4,x<=36, Xiao Ming, Xiaoqiang's money is still not enough to buy 18, 18x<=36*18 648 17x>=34*17 578, Xiao Ming Xiaoqiang's total money is 6 yuan, Xiaoqiang 2 yuan.

At this time, 18x>600=33*18+6, x>=34, 17x<=600=17*35+5, x<=35, popsicle ** is yuan or yuan.

If Xiao Ming is 5 yuan, there is one at this time.

12x>500 12*41+8,x>=42,11x<=500 11*45+5,x<=45, Xiao Ming, Xiaoqiang's money is still not enough to buy 18, 18x<=45*18 810, 17x>=42*17 714, Xiao Ming Xiaoqiang's total money is 8 yuan, Xiaoqiang 3 yuan.

At this time, 18x>800=44*18+8, x>=45, 17x<=800=17*45+15, x<=45, and the popsicle ** is the yuan.

Because Xiaohong can buy up to two, that is to say, the most expensive is the yuangen, assuming that it is the yuangen. Xiao Ming buys up to 11 roots, it costs yuan, because it is an integer yuan, so Xiao Ming brings a maximum of 5 yuan, Xiao Ming and Xiao Qiang's money together can't buy 18 roots, assuming that you can buy 18 roots, on the basis of Xiao Qiang yuan, then Xiao Ming has to bring at least yuan, so Xiao Ming brought a maximum of 3 yuan. Now, let's assume that the x-root is rooted, so we can get:

18x>8 11x<5 So the value range of x is 4 9 < x < 5 11 So a reasonable ** should be the root.

Xiaohong has one yuan, buy 2 more, that is to say, a popsicle is less than or equal to yuan, Xiao Ming buys a maximum of eleven, 11 yuan, and because the popsicle is less than or equal to yuan, three people are integer yuan, so Xiao Ming is 5 yuan, in the same way, Xiaoqiang is 3 yuan. If the popsicle is $x each, then 2x 1, 11x 5, 17x 8. There is x, x, x.

And because Xiao Ming Xiaoqiang's money doesn't add up to 18, it's 18x 8, so x, so the ** of the popsicle is yuan.

Yuan, the three of them each brought Yuan.

Each popsicle costs 45 cents.

Set up A, B, C, ABC people.

8a:4b:3c=5:3:1

Require abc to be an integer, expand 5:3:1 to an integer multiple of 8a:4b:3c, and the least common multiple of 8,4,3 is 24

then 8a=120, a=15

4b=72,b=23

3c=24,c=8

First of all, from Xiaohong analysis, 1 yuan can buy up to 2, and the ** of the popsicle is an integer point, so the ** of the popsicle must be 4 cents or 5 cents, if it is 4 cents Xiao Ming with yuan, Xiao Qiang with 3 yuan, if 5 cents a piece, Xiao Ming with yuan, Xiao Qiang with how to analyze is not an integer yuan? Is the analysis wrong, ask the great god for an answer?

Suppose the length and width of the B rectangle are a and b, respectively

a+b=4a/9+5b/2

10a=27b

Because the side lengths are all whole numbers.

then a=27b=10

Area of B = 27 10 = 270

The length of the nail = 27 4 9 = 12

Width = 10 5 2 = 25

Area of A = 25 12 = 300

The sum of the area of A and B is at least 270 + 300 = 570

Solution: Let the difference between the age of ginkgo trees and cypress trees be (10a+b), then the difference between the ages of cypress trees and pine trees is (10b+a), then:

2/3(10a+b)-4=10b+a

2(10a+b)-12=30b+3a

17a=12+28b

a=(12+28b)/17

a is an integer, and when b=2, a=4

That is, the difference between the age of ginkgo biloba and cypress is (10a+b)=40+2=42.

Let the age of ginkgo biloba be x, cypress tree y, and pine tree z. Two digits are ab.

x-y=10a+b

y-z=10b+a

y-z = (10a+b)*2 3-4=10b+a gets: 28b 3=17a 3-4

17a-28b=12

It can be known that a must be greater than b, otherwise you can't get a positive difference.

Determine the unit digits of the division.

b=1, a cannot be a natural number.

b=2，a=4

b=3, a is not equal to an integer.

b=4, a must be greater than 10, and it is not true.

Therefore, only a=4 and b=2.

The number is 42.

It's simple! The single digit of this number is x, and the ten digits are y

x+10y)x2\3=10x+y+4

The solution yields 28x+17y=12

Let the difference between the age of ginkgo biloba and cypress be 10*a+b, and the difference between the age of cypress and pine is 10*b+a, then 10*b+a=(10*a+b)*2 3-4, since a, b are positive integers 0 9, the last a=5, b=3

The difference between the age of ginkgo biloba and cypress is 10*5+3=53