Question

六位数 $\underline{2}\,\underline{0}\,\underline{2}\,\underline{1}\,\underline{0}\,\underline{A}$ 只有在唯一一个数字 $A$ 下是质数。求 $A$ 。

The six-digit number $\underline{2}\,\underline{0}\,\underline{2}\,\underline{1}\,\underline{0}\,\underline{A}$ is prime for only one digit $A.$ What is $A?$

$1$

$3$

$5$

$7$

$9$

答案：E

知识点：整除性数字质数

难度评级：1140

解答：

注意 $A$ 不能是偶数，否则这个数能被 $2.$ 整除。

$A$ 也不能是 $5,$ 否则这个数能被 $5.$ 整除。

若 $A$ 等于 $1$ 或 $7,$ 则这个数的数位和分别为 $6$ 和 $12$ 。

这样这个数能被 $3,$ 整除，所以排除了这两个 $A$ 值。

最后，若 $A$ 等于 $3,$ 整个数为 $202103.$ 交错数位和之差为 $2 + 2 - 1 - 3 = 0,$ ，所以这个数能被 $11.$ 整除。

除 $9$ 外，每个选择都会使这个数成为合数。题目说明恰有一个数字符合条件，所以这个数字必须是 $A=9.$ （事实上，试除所有不超过 $\sqrt{202109}<450$ 的质数，可确认 $202109$ 是质数。）

所以正确答案是 E。

Note that $A$ cannot be even, as then the number would be divisible by $2.$

$A$ also cannot be $5,$ as that would make the number divisible by $5.$

If $A$ equaled $1$ or $7,$ then the sum of the digits of the number would be $6$ and $12$ respectively.

This would make the number divisible by $3,$ so that rules out $A$ equaling either of these numbers.

Finally, if $A$ equals $3,$ then the whole number becomes $202103.$ If we look at the difference of the sums of alternating digits, we get $2 + 2 - 1 - 3 = 0,$ which means the number is divisible by $11.$

Every choice except $9$ makes the number composite. Because the problem states that exactly one digit works, that digit must be $A=9.$ (Indeed, trial division by the primes at most $\sqrt{202109}<450$ confirms that $202109$ is prime.)

Thus, E is the correct answer.

2021 AMC 10A Fall 第 5 题

答案：E

难度评级：1140

解答：

其他年份的第 5 题