7. 1 1 is the name of an American chain store, which deals in food and some daily necessities. One day, the general manager of the store asked a question. He asked, "A customer bought four small commodities. The total price of these four commodities is exactly $ 7. 1 1, and the product of these four commodities is exactly $ 7. 1 1. What are the prices of these four commodities? "

[Answer: According to the original question, we can write such an indefinite equation:

A+B+C+D=7. 1 1

A×B×C×D=7. 1 1

The indefinite equation consists of two equations and four unknowns, and it is impossible to get the solution of the unknowns by the general solution equation method (which is why this equation is called indefinite equation). To solve the indefinite equation, we need to use the explicit or implicit conditions given in the title to help solve it

People are not used to the operation of decimals, so equations can be converted into integers:

A+B+C+D=7 1 1

A×B×C×D=7 1 1000000

First of all, we should start with 7 1 100000. 71/000000 is equal to 79× 5× 5× 5× 5× 3× 2× 2× 2× 2, and ABCD must be. The known conditions implied here are: ABCD, all positive integers ranging from 1 to 7 1 1 (exactly, each number of ABCD is not less than 1 and not more than 708).

Note that among the above-mentioned decomposed multipliers, the prominent number is 79, which only appears once and is also the largest, and it is the most obvious goal in solving the case. In ABCD, one of them must contain 79 (a multiple of 79). As we said above, any number of ABCD, including the number containing 79, cannot be greater than 7 1 1, so the number containing 79 is less than 7 1 1. There are six possible values, namely 79×3×2=474 and 79×5=395. You see, we narrowed the detection range to six numbers at once, and the number 79 in the answer to this question is among these six numbers.

Let's see if these six possible numbers can meet the requirements as solutions of the equation.

First look at 474. 7 1 1000000 After removing 474(79×3×2), the remaining number is 5× 5× 5× 5× 5× 3× 2× 2. These numbers should be combined into three numbers, and the sum of these three numbers should be equal to 765438+. We know that when several numbers are closest, their sum is the smallest. For example, when 2×2×2×2×2×2× 2 is combined into two numbers, their sum is the smallest, which is 16, and any other combination of the two numbers is greater than 16 (for example, 2×2×2×2). We can see that 5× 5× 5× 5× 2. No matter how they form three numbers, they can only be greater than 2 17, but can't meet the crime/problem-solving condition equal to 2 17. So what's the problem? The problem is that 79×3×2=474 can't be the solution of this problem, that is, 474 is not any one of ABCD, because if one of ABCD is 474, then the other numbers can't satisfy those two equations in any combination. So we can rule out 474.

Second, look at 395(79×5). In the same analysis, we can see that after dividing 7 1 100000 by 395, the remaining three and the smallest number are 120, 125, and the sum is 365, which is greater than the required 765438+. Similarly, 395 can also rule out suspects.

Third, look at 3 16(79×2×2), and of course use the same analysis method. Ha, guess what will happen this time? Hehe, we are really lucky this time. The flowerpot on the balcony accidentally fell and hit the head of the thief who tried to break into the house downstairs. 7 1 100000 After removing 3 16, the remaining array and the smallest three numbers are 120, 125, 150,120+/kloc. As a result, when we ruled out the suspect, we accidentally caught the guy who was committing the crime. 3 16, 120, 125, 150 are just a set of solutions that satisfy the original conditions. Moreover, when a number is 3 16, all three numbers except 120, 125 and 150 are greater than 395. So when a number is 3 16, there is only one set of solutions.

A group of criminals have been arrested, but are there any other criminals? In mathematical language, is this set of solutions the only solution?

Of the six possible values of 79, we analyzed three and there are still three left. We still need to check the remaining three figures.

Fourth, look at 237(79×3). This time, the above method will not work, because the following three numbers, divided by 7 1 1000000, make up the minimum sum of three numbers, which can be less than 7 1 1 minus the difference of this number. This time, we use a new method. If one of the four numbers is 237, the sum of the remaining three numbers should be 7 1 1-237 = 474. Let's look at 7 1 1000000 divided by 237 and get 5× 5× 5× 5× 3× 2× 2× 2× 2× 2× 2× 2. Pay attention to six of them. If all three values contain 5, their sum must also be divisible by 5. But 474 is not divisible by 5, which means that at least one value does not contain 5. Is it possible that only one value contains 5? We see that the product of six fives is equal to 15625, which is far greater than the required sum of three values of 474, so these six fives cannot be completely in one value. Similarly, a numerical value cannot be multiplied by five (3 125) and four (625). Therefore, it is only possible that in two numerical values containing 5, one numerical value has three 5' s and the other numerical value has three 5' s. In this way, these two numbers can only be 125 or 125×2 (it can't be 125×3 because 125×3+ 125 is greater than 474). So we only have two possible values, one is 125, 125, 192, and the other is125,250,96. The sum of these two sets of values is not 474, and neither of them is our solution. Rule out!

Fifth, look at 158(79×2). 158 is not divisible by 5, so you can still use the above method. The process is not verbose, and the four possible values are 125 and125,288 respectively; 125,250, 144; 250,250,72; 125,375,96。 Similarly, the sum of no groups is equal to 7 1 1- 158, so 158 is also innocent.

Sixth and finally, look at 79. 79 is not divisible by 5. We can draw a gourd according to the same pattern and omit the process, and get six sets of values, namely: 125,125,576; 125,250,288; 250,250, 144; 125,500, 144; 125,375, 192; 250,375,96。 We are glad to see that none of them meet the requirements (the sum of the three is equal to 7 1 1-79), so 79 is also innocent.

Looking back, among the six possible values including 79, only 3 16 satisfies the condition, and a set of solutions, 3 16, 120, 125, 150, are found, and they are unique.

Don't forget, for the convenience of calculation, we have removed the decimal point and we have to add it back.

The final answer: the prices of these four commodities are: 3. 16 USD, 1.20 USD, 1.25 USD and 1.50 USD respectively. ]