Test question
A, fill in the blanks (each small 10 points, * * *)
, is such a four-digit number, and the sum of its digits is; Like this, there are always four digits whose sum is * * *.
Let the series satisfy:, and any three consecutive terms have:. Then there is the generic term.
Take a point on the parabola as a right-angled vertex and two right-angled triangles inscribed on the parabola as sum, then the intersection coordinates of the sum of line segments are.
, set, the maximum value of the function is.
、 .
The base length of a regular triangular pyramid is, the side length is, and the intersection point is a section that intersects with the side. So, the minimum circumference is.
A set of positive integers satisfying.
, the sum of numbers represents a positive integer, and then.
Second, answer questions (* * * questions, total score)
, (20 points), set and satisfied value:.
(points) As shown in the figure, the hearts are respectively
The midpoint and inscribed circle of are tangent to the edge respectively; Proof: three lines * * * points.
(Point) Give a regular polygon on the computer screen with its vertices painted black and white respectively; A program performs such an operation: one continuous vertex of a polygon can be selected at a time (here, it is a fixed positive integer less than), and this vertex will be "black and white upside down" as long as the mouse button is pressed, that is, the black point turns white and the white point turns black;
It is proved that if it is odd, all vertices can turn white after a limited number of operations, or all vertices can turn black after a limited number of operations.
When it is an even number, can it be done a finite number of times so that all vertices become the same color? Prove your conclusion.
answer
Tip: this four-digit number is the number of global solutions of indefinite equations that meet the conditions; That is, the number of non-negative integer solutions, in which it is easy to know that there are four such solutions, that is, there are always four such numbers. (Note: It can also be listed directly. )
Hint: obtained from conditions,
therefore
Therefore, and;
therefore
From this
.
Tip: If it is set, then
The linear equation is
That's because, then
that is
Generating equation
So the point is in a straight line;
Similarly, if set, the equation is
That is, the points are also on a straight line, so the coordinates of the intersection point are.
Hint: by
So,
that is
Get an equal sign when, which is when.
,. Hint:
.
Tip: Make a side expansion diagram of a triangular pyramid, it is easy to know the ∑, and you can get the * * * line from the smallest circumference, so you are isosceles.
That is to say,
So, by then,
.
Tip: Because there are several shapes, it must be odd, but it must be even. Let and get it for others.
that is
. ①
Even, odd, set, and then
From (1),
, ②
It is an odd number, and exactly one of them is a multiple. If, it is an odd number, and only, ② becomes.
That is, then;
If, for an odd number, and only, ② becomes, that is, it has no global solution;
So this is the only solution:
(In addition, you can also start from an even number and make
Is a multiple, then it is a multiple, so it is an even number of shapes. Take them in turn and check the corresponding six numbers. )
Tip: adding natural numbers will not change the nature of the problem; First of all, consider the numbers from to, all of which are represented by three digits, and get a set. It is easy to know that for each one, the first number has exactly one "three digits":
Therefore, the sum of the first number of all three numbers is
.
Then the first two digits of each number in in are exchanged into, and the set of 1000 numbers obtained is still,
The first two digits and the last two digits of each number in in are interchanged into a set of 1000 numbers, so we can know.
.
Now consider four digits: in the middle, the first digit (thousand), * * * has a thousand, and in
, the first (thousand), * * * has a thousand, so.
Secondly, it is easy to calculate, so,
.
, by
that is
Square gain
therefore
that is
therefore
.
As shown in the figure, we meet and connect at a point, because the midline is equal to the bisection, then, so, because, you get a * * * circle. Pay attention to the heart, and then
.
Lian, yes, because of the tangent, so
So the three-point line is the three-line point.
It is proved that because it is a prime number, and then, according to Peishu theorem, there is a positive integer, which makes
, ①
So when it is odd, it is odd and even in ①.
If it is even, it is odd, then ① is rewritten as:
Make, the above formula becomes, here are odd numbers and even numbers.
In short, there are odd numbers and even numbers, which makes the formula 1 hold; According to 1,
, ②
Now, perform the following operations: select a point, start from the beginning, operate a vertex clockwise, and then operate the next vertex clockwise ... When this operation is performed many times, it can be seen from ② that the color of this point has changed odd times, thus changing the color, while the state of all other vertices has changed even times (times), and the color remains unchanged; This secondary operation is called "one-round operation", because each round of operation only changes the color of a point, so this operation can turn all black points into white points after a limited number of rounds, so that all vertices of the polygon become white; You can also turn all white points into black points after a limited number of rounds, so that all vertices of the polygon turn black.
When it is an even number, you can also do this a limited number of times so that all vertices of the polygon become the same color. Specifically, we will draw the following conclusions:
If a given regular polygon has an odd number of black spots and an even number of white spots at the beginning, after a limited number of operations, all vertices of the polygon can become all black, but not all white; On the other hand, if a given regular polygon has an odd number of white points and an even number of black points at the beginning, after a limited number of operations, all vertices of the polygon can become all white, but not all black;
Therefore, the assignment method is adopted: the white point is changed to ""and the black point is changed to "". Changing the color once is equivalent to multiplying the assignment and changing the color of each point, which is equivalent to multiplying (even number), because;
Therefore, when the product of all vertices of a polygon is, that is, there are odd black spots or even white spots in the total * * *, the product of assignment is still, so no matter how many operations, all vertices can't turn white.
But at this time, it can be completely black. This is because, for even numbers, ① ② is odd, and let it be two adjacent vertices of a polygon. From this point, operate one vertex clockwise, and then operate the next vertex clockwise ... When repeating this operation, ② know that the color of this point has changed even times (times), so the color remains unchanged, while all other vertices have changed odd times (times). From this point on, the color of this point remains unchanged, and all other vertices change their colors after the same operation. Therefore, after the above operation, only two adjacent vertices of the polygon change their colors, while the colors of all other points remain unchanged.
Now, such sub-operations are merged and called "one-round operation"; Every round of operation can make the colors of adjacent two points in black and white exchange, so after a limited round of operation, points of the same color can always become continuous vertices of polygons;
Therefore, when a polygon always has an even number of white points at the beginning, every round of operation can turn two adjacent white points into black points, so that all vertices of the polygon become black after a finite round of operation.
Similarly, if a given regular polygon has an odd number of white points and an even number of black points at the beginning, after a limited number of operations, the vertices of the polygon can be all white, but not all black; Just assign black dots to ""and white dots to "",and the proof will be exactly the same.