It can be said to you that the combination maximum problem is an unconventional problem, and there is no unified method to solve this kind of problem at present; Otherwise, this kind of question will not be the finale of the annual national senior high school mathematics entrance examination.

Different topics need different analysis strategies and methods, and can only be analyzed in detail.

To find the maximum (minimum) value of a combination, the steps are generally consistent. ① structure; ② Proof

Take your question as an example. Question (1)

First, construct n numbers so that the sum of any three numbers is a prime number.

② It is proved that the sum of three of any number (n+ 1) is not a prime number.

And then it's over

The analysis process of this problem is as follows

If you take n positive integers, then it is best to be odd; Otherwise, if there is an even number in n, then I take two odd 1 pairs, and their sum is even, obviously not prime. And the sum of the three odd numbers is all odd numbers, which can be prime numbers.

Think again, if the sum of any three numbers is a prime number, then consider the remainder of each number divided by three (remainder 0, 1, 2).

There can only be two numbers with the same remainder at most (if there are three, the sum of the three numbers can be divisible by three).

If I take a number 5, you will find that there are numbers 0, 1 and 2, and the sum of these three numbers can also be divisible by 3.

So the number of five is definitely not enough.

Then, you can only take four at most.

This completes the "② proof" step, and then "① construction" is needed.

If the sum is a prime number, then four numbers can be set as A, B, C, D, A < b<c<d

satisfy

a + b + c = x

a + b + d = y

a + c + d = z

b + c + d = w

Then, from 100, x, y, z, w ∈ {3 3,5,7,1,13, 17, 19.

You can try a little if you like. As far as this problem is concerned, adding up the four equations, there are

3 (a + b + c + d) = x + y + z + w

That is to say, if the sum of the four prime numbers (x, y, z, w) you choose is a multiple of 3, then the solved a, b, c and d must also be integers, and then verify whether they are positive integers.

For example, if you choose (5,7, 1 1 3), then (a, b, c, d) = (- 1, 1, 5,7) is irrelevant.

Try again, choose (1 1, 13, 17, 19), and the solution is (a, b, c, d) = (1, 3, 7.

Of course, there is more than one solution to this problem.

Another example is (13, 17,19,23), and the solution is (a, b, c, d) = (1, 5,7, 1 1).

So this problem is solved.

The independent variables in combinatorial extremum problems are usually "integer", "set", "graph" and other structures, and they are often some discrete quantities. Therefore, the maximum problem is rarely solved by a functional relationship or an analytical formula, which determines the different characteristics of the combination problem.

Moreover, "construction" and "demonstration" need different ideas and need to jump from one idea to another. Therefore, this kind of questions examines observation, reasoning, construction and "mathematical literacy." Therefore, it has also become a favorite test center for the alliance and NATO (NATO just likes to test mathematics literacy, but Warsaw Pact is different).

This kind of problem is to try more, do more, see more and feel more. You will gradually become bolder.

If you want to see it, I suggest watching congruence, divisibility, tolerance principle and pigeon coop principle. Don't look too hard, especially "congruence", which is very helpful to solve this kind of problem.

The second problem is that it cannot be classified by parity (think about why), so it is classified by the remainder of module 3.

{ 1, 4, 7, 10, ..., 2008, 20 1 1}

{2, 5, 8, 1 1, ..., 2009, 20 12}

{3, 6 ,9, 12, ..., 20 10}

If you take any two of the 1 th set, the difference must be a multiple of 3, and the sum must not be a multiple of 3. Therefore, sum cannot be divisible by difference.

That's 67 1

① Construction has been completed, and then ② need to prove that 672 is not allowed!

Divide every three of these numbers into

{ 1, 2, 3}, {4, 5, 6}, {7, 8, 9}, ……{2008, 2009, 20 10}, {20 1 1, 20 12}

So there is 67 1 set.

If you take 672 numbers, then there must be two numbers in the same set, and the sum of these two numbers can definitely be divisible by the difference. In other words, the number 672 is unacceptable. (The pigeon hole principle is used above)

So the maximum number can only be 67 1.

Therefore, there is no unified method for the best combination of values for this problem, doing more, feeling feelings, integer problem, observing more.

Pay attention to the problem of divisibility and consider it from the perspective of congruence.