Make a high score bounty kneel, process and result High score bounty

1 out of 13, and the remaining 4 in a group into three groups, will be weighed first, there are two possibilities.

1) Equal, you can know that the unqualified is in or the one that is taken out, that is, all are qualified, and you can put it.

If it is equal to the scale, it is not acceptable to take out the one; If it is not equal, it means that the unqualified is in, the two are divided into two groups, and two qualified ones are weighed with one of the groups, if they are equal, the remaining two will have one unqualified; If there is no equal, there is a disqualified group; Therefore, finally obtain a qualified and a unqualified 2 parts, take any qualified part and one of them can be called unqualified, so far a total of 4 times.

2) If it is not equal, it can be known that the unqualified are in the group, then they are all qualified, and if they are equal, the unqualified ones are in ; If you don't wait, you won't be qualified; Divide the unqualified group into two groups, take two qualified ones and weigh one of them, if they are equal, the remaining two will have one unqualified; If there is no equal, there is a disqualified group; Therefore, finally obtain a qualified and a unqualified 2 parts, take any qualified part and one of them can be called unqualified, so far a total of 4 times.

Hungry... I meditated for half an hour, how to weigh it 4 times, but maybe 3 times can be weighed, I can't think of it, you can refer to me.

If you only say that the weight is unqualified, there is no way to weigh it, and if you give how it is unqualified, then it can be solved. For example, the part is too heavy, or too light. Otherwise, there is no solution.

For example, this part is too heavy.

The thirteen parts, take out one, the remaining twelve are divided into two parts, six each, one on the left and right sides of the scale, if the weight is equal, then the one taken out is a substandard product; If one side is heavy, the defective product is in that part.

Divide this one into two equal parts, three on the left and three on the right, and the heavy side has defective products; Take out two more scales, one on the left and one on the right, and the heavy one is defective; If it is the same weight, then the one taken out is defective.

I didn't expect the first floor to find the same answer as me, hehe, I thought he was his!

I just read other people's explanations on the know, and I thought about it for a long time, and I don't know if it's too light or too heavy. So don't drill the horns anymore, everyone thinks so, there is a 90 percent chance that it is right, but I support you to continue**, and I will not stop**, if you find (preferably prove), remember to find me.

The balls are numbered a1, a2, a3, a4, b1, b2, b3, b4, c1, c2, c3, c4, c5

The first scale: a1a2a3a4 is compared with b1b2b3b4, and there are three results:

If 1:a1a2a3a4=b1b2b3b4, the ball is in c1c2c3c4c5.

Second scales: C1C2 and C3A1

If c1c2=c3a1, then that ball is c4 or c5. The third scale can be called C4 and A1, which is equal, indicating that the difference is C5, and the difference is C4.

If C1C2 is greater than C3A1, it means that one of the balls in C1C2 is heavier or C3 is lighter. The third scale is called C1 and C2, and if the two are the same, it means that C3 is lighter, and if the two are different, the heavier one is different.

Following the first scale, the other assumptions are analyzed.

If 2: a1a2a3a4 is greater than b1b2b3b4 (the ball of class A is heavy, or the ball of class B is light).

Second scale: a1a2b1 and a3a4c1

If a1a2b1=a3a4c1, then the ball is in b2b3b4, the third scale can be called b2 and b3, equal then b4 is lighter, and unequal is lighter.

If A1A2B1 is greater than A3A4C1, it can be concluded that a ball in A1A2 is heavier (it will not be that a ball in A3A4 is lighter, nor will it be a B1 ball), and the third scale compares A1 and A2, and the heavier one is different.

If a1a2b1 is less than a3a4c1, it can be concluded that b1 is lighter or a3a4 is heavier, and the difference is among the three. The third scale compares A3 and A4, the two are the same, indicating that B1 is lighter and different, and the two are different, which proves that the heavier one is different and heavier than other balls.

If 3:a1a2a3a4 is less than b1b2b3b4, change the number of the a1a2a3a4 quadruple balls to b1b2b3b4 and the number of the b1b2b3b4 quadruple balls to a1a2a3a4, and the reasoning is the same as in the second assumption.

Lack of conditions, to say whether it is lighter or heavier than ** (assuming it is heavy).

First divided into 5 pieces, 5 pieces, and 3 pieces.

Weigh two groups of 5 pieces together, and the heavy one will have a fake one.

Divide the 5 pieces into 2 pieces, 2 pieces, and 1 block.

Weigh two groups of 2 pieces together, and the heavy one will have a fake one.

Finally, divide the two pieces into 1 block and 1 block.

Weigh it again to distinguish the authenticity.

In the case of ignorance, 12 are the limit, 13 theories can be reached, but not in practice.

But if the parts, which are similar and can be split, it's different.

If you have this question in the exam, you just write that it is impossible to achieve.

Divide the thirteen parts into three groups, 4, 4, 5

The rest is simple, you can first consider the following 5 parts called twice (the rest of the conditions are unchanged) This problem is an extension of it, you think about 5 first, and then do this problem to know, if you really can't think of it, send me a message and I'll tell you,

Do you know the weight of the qualified product.

Solution: Let A take (4-k) and take (x) times, then take 4 (15-x) times, where 0 x 15

0＜k＜4 4-k＜4

That is, the more times (4-k) is taken, the fewer times A takes them, and x(4-k)+4(15-x) 16(6-k) +6 4x-kx +60 -4x 96-16k +6-kx+60 102-16k

kx≤42-16k

0＜k＜4 x≥ （16k-42）/k

i.e.: x 16 - 42 k

x is a natural number, 0 k 4

From 16k-42 0 get: k 3

That is: 3 k 4, and k is a positive integer.

k = 3 substituting k = 3 in: x 16 - 42 k in: x 2 2 x 15

Then: k = 3 B: 102-16k = 102-16x3 = 54 (sheets) A also took 54 sheets.

54 + 54 = 108 (sheets).

A: There are at least 108 cards.

I'm glad to solve the above problems for you, I hope it will help you in your studies! 】≤

The least number of cards means that A and B have just finished drawing, and when the value of the cards drawn is the least, the cards reach the minimum A draw value range: 15 60 (15 * 1 15 * 4) B draw value range: 54 102 (6 + 16 * 3 17 * 6) and because the last 2 people draw the same number of cards, and meet the 2 draw value range only:

54 60 makes the least number of cards that take the value of 54, 2 people draw cards, then the total number of cards is: 54 * 2 = 108

8。Solution: Let a(a,0); b(b，0)；c(c，0)；o(m，n)；So:

oa=(a-m，-n)；ob=(b-m，-n)；bc=(c-b，0)；

Obtained by x oa + xob = 2bc

x²(a-m)+xb=2(c-b)..1)

x n-xn=-xn(x+1)=0, so x=0 or x=-1

Substituting x=0 into (1) gives c-b=0, that is, c=b, which does not match the title, because a, b, and c are three different points, i.e., c≠b, so x=0

should be discarded; In substituting x=-1 into (1) formula, oa-ob=2bc is established, so x=, d. should be chosen

9.Solution: f(x)=e x + x is an even function with the image opening facing upwards, symmetrical with respect to the y-axis, and the lowest point is (0,1). And.

A straight line parallel to the x-axis y=k has two intersections11

It can be seen that the original question is wrong! Unable to solve].

Calculation (1).

The first term is divided by the third item 2

Calculation (2) is preceded by the formula infinitely squared.

This is followed by the squared difference formula.

After calculating (3) squared, move the term again.