AMC8 2022

AMC8 2022 · Q20

AMC8 2022 · Q20. It mainly tests Word problems (algebra), Optimization (basic).

The grid below is to be filled with integers in such a way that the sum of the numbers in each row and the sum of the numbers in each column are the same. Four numbers are missing. The number $x$ in the lower left corner is larger than the other three missing numbers. What is the smallest possible value of $x$?

下面的网格要填入整数，使得每行每列的数字和相同。有四个数字缺失。左下角的数字$x$比其他三个缺失数字都大。$x$的最小可能值是多少？

(A) -1 -1

(B) 5 5

(D) 8 8

(E) 9 9

Answer

Correct choice: (D)

正确答案：（D）

Solution

The sum of the numbers in each row is $12$. Consider the second row. In order for the sum of the numbers in this row to equal $12$, the two shaded numbers must add up to $13$: If two numbers add up to $13$, one of them must be at least $7$: If both shaded numbers are no more than $6$, their sum can be at most $12$. Therefore, for $x$ to be larger than the three missing numbers, $x$ must be at least $8$. We can construct a working scenario where $x=8$: So, our answer is $\boxed{\textbf{(D) } 8}$.

每行的数字和是$12$。考虑第二行。为了使这行的和等于$12$，两个阴影数字的和必须是$13$：如果两个数字的和是$13$，其中一个至少是$7$：如果两个阴影数字都不超过$6$，它们的和最多是$12$。因此，为了使$x$比其他三个缺失数字大，$x$至少是$8$。我们可以构造一个$x=8$的工作方案：所以，答案是$\boxed{\textbf{(D) } 8}$。

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