Interesting math problems, interesting math problems

There is a person who leaves an inheritance: the eldest son takes 100 yuan, and takes the remaining tenth (the remainder, which refers to the total property minus 100 yuan) The old 2 takes 200, and takes the remaining tenth, and the remaining refers to the total property minus the money taken by the eldest child, and the rest below 200 is the meaning) The old 3 takes the remaining tenth of 300. And so on.

Q: How many children does this man have, how much inheritance?

Answer. You have the famous question of Euler's inheritance, we may as well assume that this father has n sons, the last son is the nth son, and the penultimate is the (n-l) son. Through the analysis, it can be seen that:

The first son's share of property 100 1 tithes of the remainder;

The second son's share of property 100 2 tithes of the remaining property ;

The third son's share of property 100 3 tithes of the remaining property;

(n 1) of the share of the son 100 (n 1) of the remainder of the property;

The nth son received 100n of property.

Since each son received an equal share of the property, i.e. 100 (n 1) 100 tenths of the remaining property, the tenth of the remaining property was 100 n 100 (n 1) 100 dollars.

Then the remaining property is 100 tenths of 1,000 yuan, and the last son gets 1,000 100,900 yuan. Thus, the father had (900 100) 9 sons, leaving a total of 900 9 8100 yuan.

Ten trees, plant five rows, 4 in each row, and ask how they should be planted.

Recommend a book "Mathematics Beauty Picks Up" by Yi Nanxuan.

It's funny. The fun of math is not limited to math games. There is something very practical about mathematics, which contains profound mysteries, thought-provoking and surprising things.

There are different levels and realms of fun in mathematics, and the fun that math masters see will be different from the fun that elementary school students see. In the case of this series, different readers will find different pleasures and benefits from it. It can be flipped through casually as a leisure and entertainment sketch, which helps to relieve work fatigue and mundane troubles; It can be used as a reference material for teachers, which helps to enliven the classroom atmosphere and enlighten students' minds. It can be used as extracurricular reading materials for students, which helps to broaden their horizons, increase their knowledge, and exercise their logical thinking skills.

Even for college students, graduate students, and even mathematics research authors with relatively high mathematical attainment, it will be beneficial to open the book.

There are a lot of interesting questions in it.

(1)c9 2/c10 2=(9*8/2)/(10*9/2)=4/5=80%

2) There are x defective products.

c(10-x) 2 / c10 2=1/3[(10-x)(9-x)/2]/(10*9/2)=1/3(10-x)(9-x)=30

x^2-19x+60=0

x-4)(x-15)=0

x1 = 4x2 = 15>10 (round).

So there are 4 defective products.

Hello! The first problem is solved by the probability method, because there is a defective product in ten products, then there are nine qualified products, and now in order to be qualified, we must extract two pieces from these nine qualified products as the sampled, and because these ten products are not numbered, so according to this idea, there will be 9 * 8 2 = 36 kinds of situations, and then so many situations are punished in the total possible number of ten products to extract two pieces of 10 * 9 2 = 45 kinds, which is the pass rate of the sample 36 45 = 4 5 The second question, according to the same idea as the first question, we can quickly conclude that the numerator of this fraction should be 15, so, according to the same method, we can get 6*5 2=15, so, among these ten products, there are six qualified products and five unqualified products. That's it!

I don't know if you know?

1-9/（10*9/2）=

1 3=(10-x)*(9-x) (10*9) solution x=15 or x=4 Therefore, there are 4 defective products.

The original formula is ax-x+ay+2y+5-2a=0

i.e. a(x+y-2)=x-2y-5

Since there is a common solution for any a, x+y-2=0, x-2y-5=0 solves x=3, y=-1

It's so much fun!

Isn't the landlord playing tricks, he doesn't know how to do it.

Little A, little missing, B and little C went to the shoe store to buy shoes together.

Three pairs are worth 300 yuan.

The boss felt that he had three pairs of spring envy, so he asked the guy to return 50 yuan to the three.

Halfway through, the guy felt that fifty yuan was not easy for three people, so he drew twenty yuan.

One person will pay back ten dollars.

So for three people, a pair of leather shoes is 90 yuan.

90 * 3 + 20 yuan for 290 yuan.

And ten more dollars of Volson money ran there.

Maybe it's simple, it's even a little naïve, but it's still quite interesting, and the typical topic steals the concept, and in the end: the three of them together come up with 270 yuan, the boss gets 250 yuan, and the guy gets 20 yuan. 270=250+20

If you have to count 300 as a number.

Then 100x3 (all paid) = 300 = 250 (boss gets) + 10x3 (customer gets) + 20 (guy gets).

In other words, the 20 yuan is included in the 270 miles.

Hello: A football sewn with black and white leather, the black leather is a regular pentagon, the white leather is a regular hexagon, and each black leather is sewn with 5 pieces of white leather around it. It is known that there are 12 black skins in the whole football, and the number of white skins is (20).

There are 12 pieces of black, each piece has 5 sides, and there are 5 * 12 = 60 sides of the solution: let the white skin have x blocks.

6x÷2=12*5

3x=60x=20

Answer: There are 20 pieces of white skin.

Good luck with your studies!

Solution: Let A have sheep x and B have sheep y, and the equation is obtained

x+9=2（y-9）

x-9=y+9

Solution: {x=63.}

y=45 means that A has 63 sheep and B has 45 sheep.

And man, I can't understand it all the time, you typed it wrong, it's "said" not "said". I thought about it for a long time...

Analysis and solution.

1 B says that A has 9 animals, and the two of them have the same number, and A has 9 fewer animals, and B has increased 9 at the same time, that is, A must have 9 9 18 more animals.

2 When A says that B has 9, it is twice as many as B, and A increases by 9 and B decreases by 9.

That is, A had 9,9,18 more sheep than B.

From the analysis of 1, we know that A originally had 18 more sheep than B, so A had 18 18 36 more sheep than B at this time, and A was twice as many as B at this time, so 36 sheep were equivalent to twice that of B, so at this time, A had 36 2 72 sheep and B had 36, and A had 72 9 63 sheep and B had 36 9 45 sheep.

Question 1: This is the long-lost Professor Kuibel. Professor Kuibel:

I've come up with another question for you. Of the animals I raise, all but two are dogs, all but two are cats, and all but two are parrots, how many animals do I have in total? Did you figure it out?

Answer: Professor Kuibel has only three animals: a dog, a cat, and a parrot. All but two are dogs, all but two are cats, and all but two are parrots.