Note: If you don't know anything about number theory, then don't look at the following. Mathematics emphasizes rigorous deduction, which is sometimes even unacceptable.

Solution: The prices of general commodities have two decimal places, so we think that the form is limited to the form of X.XX and the four prices cannot be greater than 7. 1 1 (otherwise,

The sum must be greater than 7. 1 1), so there cannot be three less than 1 (otherwise, the other one must be greater than 7. 1 1 to make the product 7.1), that is, the sum of any three numbers cannot be.

1, so these four numbers cannot be greater than 6. 1 1. We can multiply the prices of four items by 100 to form an integer.

Let the prices of four commodities be multiplied by 100, which are A, B, C and D respectively, then A+B+C+D = 71,and A * B * C * D = 71100.

This is a typical number theory problem: integer solution of higher order equation. The more restrictive conditions there are for this kind of problem, the more difficult it is to have infinite solutions, and the less the number of finite solutions is.

For this kind of problem, the trial and error method is generally adopted. The operation process is: convert all the constraints into mathematical conditions, and then guess the answers one by one and check them. Step one:

The transformation into mathematical conditions is directly related to the complexity of the second step of checking calculation.

Firstly, 7 1 1000000 is decomposed into prime factors, 711000000 = 26 * 32 * 56 * 79.

The most striking number here is 79. List multiples of 79 as follows:

0*79=0, 1*79=79,2*79= 158,3*79=237,4*79=3 16,

5*79=395,6*79=474,7*79=553,8*79=632,9*79=7 1 1

Because the price of one of the four items must be a multiple of 79, that is, a number in the above table, and the sum of 7 1 1 is also a multiple of 79, so the other three items.

The sum of commodity prices is also a multiple of 79. This relationship is expressed as follows:

7 1 1(9 * 79)= M * 79+N * 79

Where M*79 is the price of a commodity, let A=M*79, N*79 be the sum of the prices of the other three commodities, and let B+C+D=N*79.

Obviously M+N=9

Let's filter the values of m and n in the multiple table of 79 above, because no three numbers can be less than 100, N≠0, N≠ 1, so M≠8, m ≠ 9; And 7 didn't appear.

In the prime factor set of 7 1 1000000, so M≠7, N≠2. Of course, M≠0 is obvious, so N≠9. At this point, the values that M can take are 1, 2, 3, 4, 5, 6, of course.

This is not enough, we need to further filter the value of m.

For further screening and later work, we introduce a famous theorem: if the sum of several positive integers is certain, then if and only if they are equal, take their product.

Maximum value.

Using this theorem, the following table gives the required values of B+C+D and B*C*D and the theoretical maximum value of B*C*D ((b+c+d)/3) 3 when m takes 1~6:

M B+C+D B*C*D ((B+C+D)/3)^3

1 632 9000000=2*2*2*2*2*2*3*3*5*5*5*5*5*5 9349480.296

2 553 4500000=2*2*2*2*2*3*3*5*5*5*5*5*5 626342 1.37

3 474 3000000=2*2*2*2*2*2*3*5*5*5*5*5*5 39443 12

4 395 2250000=2*2*2*2*3*3*5*5*5*5*5*5 2282587.963

5 3 16 1800000=2*2*2*2*2*2*3*3*5*5*5*5*5 1 168685.037

6 237 1500000=2*2*2*2*2*3*5*5*5*5*5*5 493039

It can be seen that when M=5 and M=6, the theoretical maximum value of B*C*D can't meet the required value, so M=5 and M=6 can't meet the requirements of the topic. At this time, the value of m can only be 1, 2, 3, 4.

We notice that among the prime factors of 7 1 1000000, only the four prime numbers of 2, 3, 5 and 79 are assigned to A.

B, C and D can only use prime factors in 2, 3 and 5. Because of the multiple of 5, the last digit can only be 0,5, so we can take 5 as the starting point of the problem. Because M≠5, so

The six fives all belong to B, C and D, and are discussed in three situations: