Given n positive integers x1, X2, X3 , xn satisfies X1 + x2 + X3 + +Xn = 2008, find the maximum value of the product of these n numbers Why x1, X2, X3 In xn, there is no?

Given n positive integers x1, X2, X3 , xn satisfies X1 + x2 + X3 + +Xn = 2008, find the maximum value of the product of these n numbers Why x1, X2, X3 In xn, there is no?

There are two ways to solve this kind of problem, one is to give n, the other is not to limit n. your n here should not be limited. At this time, if an integer of 4 or larger appears in the partition, it can be further divided into two numbers to make the product larger (at least not smaller)

N positive integers x1, X2 Xn satisfies X1 + x2 + +Xn = 2008, find the maximum value of the product of these n positive integers

First of all, the answer should be 668 times 4
prove
Because any number greater than 3 is less than (only 4 is equal to) the product of half of its decomposition, such as 2n

Given that n different numbers X1 x2 X3.. xn are positive integers 1.2.. try to find the maximum value of | x1-1 | + | x2-1 | +... + | xn-n | for any permutation

+|X2-1 | + should be + | x2-2 | +, right?
So the conclusion should be: when n is even, the maximum value of sum is n ^ 2 / 2; when n is odd, the maximum value of sum is (n ^ 2-1) / 2

Given that X1 = 2 is the root of the equation x square + mx-6, find the value of M and the other root of the equation x2

Substituting it into M = 1, that is, X2 + X-6 = 0, that is, (X-2) (x + 3) = 0. I don't need to say more about the remaining root