Problems with the math drawer principle!! Application of the drawer principle Please answer in detai

1. Draw at least 4 balls. Reason: There are 3 colors of the ball (red, blue, yellow), if you touch three balls of different colors first, and then touch the fourth ball, it must be the same color as one of the previous balls.

2. At least 367 people. Reason: Since there are 366 days in 2012, if all 366 students are not born on the same day, then the 367th student must be born on the same day as one of the previous 366 students.

1.At least 4 of them.

In the worst-case scenario, if the first time is red, the second blue, and the third time yellow, then the fourth time will have two colors of the same color anyway.

2.Again, in the worst-case scenario, since 2012 is a leap year with 366 days, then assuming that there are 366 students with different birthdays, then the 367th place will definitely be heavy, so the answer to this question is at least 367 people.

one, because of the worst case, you have to draw 4.

2.A: 367, the worst is one day per person, 2012 is a leap year, there are 366 days, so there must be 367 people.

Created by yourself!

The first question is four balls.

After drawing one of the red, blue and yellow, as long as you touch one casually, you can ensure that there are 2 balls of the same color.

The second question is 366 people.

The new students in 2012 are basically 6 years old, and they were born in 2006 or 2005, both of which are 365 days old.

If 365 new births happen to be born every day, another one will satisfy that two new births are born on the same day.

This question can be constructed in this way, if there are 8 chopsticks of 4 colors in each of the 4 drawers, then you go to find the chopsticks, 1. Ensure that at least 2 chopsticks are of the same color: the worst situation is that 3 of the drawers are drawn to 1, and only 1 drawer is drawn to 2 to slow the year.

So at least 5 of them should be found.

2. Ensure that at least 4 chopsticks are of the same color: the worst case is that 3 of them are drawn to 3 drawers, and only 1 drawer is drawn to 4.

So at least 13 of them should be found.

Touch at least 11 balls to ensure that you can touch two colored balls, because it is possible to touch 10 balls of the same color at the same time.

Answer: Because there are 4 kinds of balls in red, yellow, green and white, of course, you can only make sure that there must be two balls of the same color by touching out (4 + 1 = 5) balls.

10 may be of the same color, so at least 11 can be found.

5 Because there are four colors, the first 4 may be different colors, and the 5th must be the same color as one of the previous 4.

At least 15 balls of the same color should be drawn at one time.

When there is 1 ball missing, at least 14 balls are of the same color.

There are up to 14 red balls, 14 green balls, 12 yellow balls, 14 blue balls, 10 white balls, and 10 black balls.

With 1 more ball, there must be 15 balls of the same color.

Touch at least 75 balls from the bag.

Considering the worst-case scenario, 14 red balls, 14 green balls, 12 yellow balls, 14 blue balls, 10 white balls, and 10 black balls are taken, and any one more ball is taken to satisfy at least 15 balls of the same color.

Therefore, at least 14 + 14 + 12 + 14 + 10 + 10 = 74 should be taken.

There are 14 red balls, 14 green balls, and 12 yellow balls. There are 14 blue balls, 10 white balls, 10 black balls, and if you take any more ball from the bag, there will be 15 balls of the same color.

So 14 + 14 + 12 + 14 + 10 + 10 + 1 = 75 (pcs).

The minimum is 1 and the maximum is 11.

Because at least one cage has more than 4 animals, let's say 4 cages, and the other cages are all one.

The number is (14-4) 1+1=11.

At least one, put it all into that one.

The maximum is 10 because the first cage has the least.

You have to have 5 14 5 9 at most, 9 at most, 9, 9.

1. There are only 52 weeks in a year, and the maximum span is 53 weeks. Therefore, there are at least 2 students on the same week's birthday2, because the difference between any 1 of the 13 numbers and the other 12 numbers is different, and if the difference between the 12 is not divisible by 12, there are only 11 remainder cases, and the remaining 1 and 2 ...... remainder. If there must be two differences divided by 12 and the remainder is the same, then the difference between the numbers corresponding to the two differences must be an integer multiple of 12.

3. Assuming that the distance between all points is greater than 1 9, the total length must be greater than 1 meter.

4. There are 4 kinds of numbers, and the most important thing to take out is 3 * 2 + 3 = 9 pieces of 5, because any 1 of the 7 numbers is different from the difference between the other 6 numbers, if the 6 differences are not divisible by 3, then the remainder is only 5 kinds, the remaining 1, the remaining 2 ....... If there must be two differences divided by 6 and the remainder is the same, then the difference between the numbers corresponding to these two differences must be an integer multiple of 6.

6. Because the difference between any one of the four numbers and the other three numbers is different, if the difference between the three numbers is not divisible by 3, there are only 2 kinds of remainders, 1 and 2. There must be two differences divided by 3 and the remainder is the same, then the difference between the numbers corresponding to these two differences must be an integer multiple of 3 (a1 a2 a3 a4, the corresponding difference b2 b3 b4 must have 2 remainders divided by 3 are the same, assuming b2, b4, then the difference between a2 a4 must be a multiple of 3, because a2-a4=a1+b2-(a1+b4)=b2-b4).

7. Assuming that the distance between all points is greater than 1 8, the total length must be greater than 1 meter.

The first one is right (52 weeks a year).

The second 13 consecutive natural numbers include 12, and the last subtract should be a multiple of 12.

The third one takes 10 points, which is equivalent to 11 parts, two more than 9, and the fourth should be 7 pieces, (4 + 3) points, 4 numbers + the same 3, the fifth is the same, and the second is the same.

I don't understand the rest of the world, I'm sorry.

It can be seen from the title.

The goal is to find the method with the most steps required to complete the "Draw three cards with consecutive numbers" task.

When drawing cards, draw the same number discontinuous (the first draw to 1, the second time to draw 1), draw to the interval number discontinuous (the first draw to 1, the second draw to 3, the third draw to 5, in short, do not let him consecutive).

Therefore, the method with the most steps should be in the case of repeated and spaced draws, and any one more drawn will make up three consecutive numbers.

Therefore: 7 * 4 = 28 sheets.

When drawing the 29th card, it is inevitable that "draw three cards with consecutive numbers" will be a visual example: after drawing four cards each, any one more will be drawn, so choose option B.

Apply the drawer principle. 2 pairs are a total of 4, red, black, white socks are 3 drawers, each drawer is put 3, and then take out one arbitrarily and put it in one of the drawers, that is, 4, so at least 3 3 + 1 = 10 socks to ensure that there must be 2 pairs of the same color.

I don't know if it's right.