The original question is this:
Fill in the blanks with numbers to make the equation hold.
For this problem, I think it stumped many people, including doctors, because people mistook it for a math problem. In fact, this is a difficult problem. The difference between mathematical problems and intellectual problems is that the former should be solved by mathematical knowledge and methods, while the latter should be solved by logical thinking without special mathematical knowledge.
So this problem should be solved like this:
The original formula can be regarded as x+13× y÷z+a+12× b-c-1+d× e÷f-10 = 66.
According to the four operation rules, it can be seen that the left side of the equal sign is composed of these parts divided by black squares, as follows
x+■ 13×y \u Z ■+ A+■ 12×B■―C■― 1 1+■d×e \u F ■- 10 = 66
The left side of the equal sign contains-1 1 and-10. Without these two items, the number is equal to 87, that is, the original formula can be regarded as the following formula.
x+■ 13×y \u Z■+A+■ 12×B■―C+■d×e \u F■= 87
At present, there are six parts left on the left side of the equal sign, among which three simple parts are X, A and C, and three more complicated parts are 13×Y÷Z, 12×B and D × e ÷ f, and these six parts are all connected by addition and subtraction signs, so their order can be changed at will. So the original formula can be regarded as the following formula:
x+A―C+■ 13×y \u Z ■+■ 12×B ■+■d×e \u F■= 87
It can be further seen simply that the left side of the equal sign is composed of two parts, that is, the sum of the two terms. The following formula:
x+A―C+■ 13×y÷Z+ 12×b+ d×e÷F = 87
X, a and c in the formula are all unknowns and will be determined by us. X+A-C in the answer needs to be equal to several, and we can make it equal to several by choosing a number. That is to say, no matter what number13× y ÷ z+12× b+d× e ÷ f is, we can make the equation of this problem hold by adjusting the three numbers of x, a and c, in other words, if we don't limit the number of digits filled in, this problem will have infinite solutions. However, considering that the original intention of this question should be to fill in only one digit, that is, nine numbers from/kloc-0 to 9, and the students in the third grade of primary school have not learned negative numbers yet, the maximum value of X+A-C can only be 1 7, and the minimum value cannot be less than 0, so13× y÷z+/kloc-. For the same reason, the value of 12×b+d×e \f cannot be less than 13, so the value of 13×y \z y \z cannot be greater than 74.
Now, we can start to assume the answer according to the above conditions. Set the values of y and z first, and according to the condition that the value of13× y÷zshould not be greater than 74, we can know that as long as y÷zisis less than 6, the value of13× y÷zisis a positive integer. Then y and z are 9,3 or 8,4 or 8,2 or 6,3 or 6,2 or 5, 1 or 5,5 or 4,2 or 4, 1 or 3,3 or 3, 1 or 2,2 or 2, 1 or 650, respectively.
For example, if Y is 9, Z is 3, B is 2, D is 9, E is 9, F is 9, X is 8, A is 8, C is 1, or B is 3, D is 9, E is 9, F is 9, X is 2, A is 2 and C is 1, ...
Simplicity and sincerity
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