Quick !! Classic Question 5, Quick Quest !! Classic question

This is indeed a classic question, there are many answers to this question, and it is not limited to one expression, and this is my answer:

The first time is called left and right

The second time is called left and right

The third time is called left and right

Judgment principle: If the 1st left weight, the 2nd left weight, and the 3rd right weight, then 1 weight.

Conversely, if the 1st right weight, the 2nd right weight, and the 3rd left weight, then 1 light.

For the cleaning of the layout, the following is abbreviated, and the results of the record in turn can not be messed up.

Note: Balance is the opposite or balance).

If it is left and right, it is 2 heavy, and vice versa is light.

If the left and right sides are flat, it is 3 heavy, and vice versa.

If the left is flat and left, it is 4 heavy, and vice versa.

If it is left and right on the right, it is 5 heavy, and vice versa.

If the right side is flat, it is 6 heavy, and vice versa.

If the right side is flat, it is 7 heavy, and vice versa is light.

If the right is flat and the left is flat, it is 8 heavy, and vice versa.

If the left is flat, it is 9 heavy, and vice versa.

If it is flat on the right and left, it will be 10 heavy, and vice versa will be light.

If it is flat and left, it will be 11 heavy, and vice versa.

If it is flat on the right, it is 12 heavy, and vice versa.

This is my discussion on QQ on this issue, and I look forward to your criticism and advice!

1: Take 4 of the 8 balances on each side, it will be easy if they are equal (one of the remaining 4 is problematic), otherwise go to the second step.

2: The balance is unbalanced, there is no problem with releasing three from the heavy side and putting the three on the upper step at the same time, and at the same time, the other one that is not replaced on the heavy side is exchanged with the light side of the scale, called the second time, there will be three situations:

1) If the balance remains the same as before, the heavy side is still heavy, and the problem is not changed on the previous light side, and there is a light one in it, and the rest is easy to weigh out...

2) If the balance is reversed, the light side is heavier, it can only be because the heavy side is replaced by a heavy ball or the one that is replaced by a light ball, the problem lies in these two balls, and the remaining time can be weighed.

3) If the balance is balanced, it means that there is a problem with the three balls that are exchanged from the heavy side, and there is a heavy ball in it, and the remaining one can be done...

It's a simple question.

Draw the curve of f(t)=sint to find out.

In monotonically increasing in the range of [- 2+2k , 2+2k ], then -2+2k 6-2x 2+2k

f(t) decreases monotonically in the interval [ 2+2k ,3 2+2k ], then 2+2k 6-2x 3 2+2k

The axis of symmetry is t= 2+k, then 6-2x= 2+k

The center of symmetry of f(t) is t=k, then 6-2x=k

3.When t= 2+2k and f(t)max=1, then 6-2x= 2+2k and f(x)max=4

When t=- 2+2k, f(t)min=-1, then 6-2x=- 2+2k, f(x)min=-4

The above k are integers.

This is a common recipe.

For reference, please smile.

Solution: (1) multiples of 9;

2) Let the original two-digit number be a and the single-digit number be b, then the new two-digit number is (10b+a) and the original two-digit number is (10a+b).

(10b+a) (10a+b).

10b+a﹣10a﹣b

9b﹣9a9（b﹣a）

a, b is an integer, so b a is an integer, so the difference between the new two-digit number and the original two-digit number is a multiple of 9 I hope it will help you.

a b, 10a+b)-(10b+a)=9 (a-b) power when a(10a+b)-(10b+a)=-9 (b-a) power.

I'll answer right away, but you're going to give me the money.

Let the radius be r squared x (2r) = 628 r cubic = 100