Question

在 $100$ 到 $999$ （含端点）之间，有多少个整数具有如下性质：它的数字经过某种排列后，是一个从 $100$ 到 $999$ 的 $11$ 的倍数？例如， $121$ 和 $211$ 都具有这个性质。

How many integers between $100$ and $999,$ inclusive, have the property that some permutation of its digits is a multiple of $11$ between $100$ and $999?$ For example, both $121$ and $211$ have this property.

$226$

$243$

$270$

$469$

$486$

答案：A

知识点：整除性数字排列分类讨论

难度评级：2380

解答：

可以分析所有 $11$ 的倍数，并看它们各自贡献多少排列。我们可以按数字中不同数字的个数分类。

情形 $1$ 三个数字全相同

这不可能。由 $11$ 的整除规则可知，首位与末位之和减去中间位必须能被 $11$ 整除。

若三个数字全相同，上述表达式就等于那个数字，不可能被 $11$ 整除。

情形 $2$ 两个数字相同

可以把它分为含有数字 $0$ 的数和不含数字零的数。

不含数字 $0$ 的 $11$ 的倍数有 $8$ 个：以及 $979$ 。 $121, 242, 363, 484, 616, 737, 858,$

这些数各贡献 $3$ 个排列，所以这一情形有 $8 \cdot 3 = 24$ 个数。

含数字 $0$ 的 $11$ 的倍数有 $9$ 个： $880$ 和 $990$ 。 $110, 220, 330, 440, 550, 660, 770,$

对这些数， $0$ 不能作为百位，所以每个只贡献 $2$ 个排列，总共 $9 \cdot 2 = 18$ 。

情形 $3$ 三个数字都不同

$100$ 到 $999$ 之间共有 $81$ 个 $11$ 的倍数。三位都不同的个数为与情形 $2$ 一样，还要特别处理含数字 $0$ 的数。这样的数有 $8$ 个：以及 $902$ 。 $81 - 8 - 9 = 64.$ $209, 308, 407, 506, 605, 704, 803,$

每个这样的数给出 $2 \cdot 2 = 4$ 个排列，但首末位交换会得到集合中已有的另一个数，所以要除以 $2$ 。

因此这些数总共提供个不同排列。 $8 \cdot 4 \div 2 = 16$

现在还剩 $64 - 8 = 56$ 个需要计入的 $11$ 的倍数。

每个这样的数有 $3! = 6$ 个排列。不过和上面一样，任意一个数首末位交换后仍会得到这个集合中的另一个数。

这可以用 $11$ 的整除规则看出。若 $ABC$ 能被 $11$ 整除，则 $A + C - B$ 能被 $11$ 整除。

这意味着 $C + A - B$ 能被 $11$ 整除，也就意味着 $CBA$ 也能被 $11$ 整除。

因此这些数还贡献个排列。 $56 \cdot 6 \div 2 = 168$

所有情形合计共有个数。 $24 + 18 + 16 + 168 = 226$

所以正确答案是 A。

We can analyze all the multiples of $11$ and see how many permutations each of them contribute. We can do this by casing on the number of unique digits in the number.

Case $1:$ all the digits are the same

This cannot happen. We can see this by the divisibility rule for $11,$ which says that the sum of the first and last digit minus the middle digit must be divisible by $11.$

If all the digits are the same, then the above expression evaluates to that digit, which cannot be divisible by $11.$

Case $2:$ two of the digits are the same

We can split this up into the numbers that have the digit $0$ and those that don't.

There are $8$ multiples of $11$ that do not have the digit $0:$ $121, 242, 363, 484, 616, 737, 858,$ and $979.$

Each of these numbers contributes $3$ permutations, so this scenario has $8 \cdot 3 = 24$ numbers.

There are $9$ multiples of $11$ that have the digit $0:$ $110, 220, 330, 440, 550, 660, 770,$ $880,$ and $990.$

For these numbers, $0$ cannot be the hundreds digit, so each of them only contributes $2$ permutations, for a total of $9 \cdot 2 = 18.$

Case $3:$ all the digits are different

There are a total of $81$ multiples of $11$ between $100$ and $999.$ The number of these with all different digits is $81 - 8 - 9 = 64.$ As in case $2,$ we have to specially account for the numbers with $0$ as a digit. There are $8:$ $209, 308, 407, 506, 605, 704, 803,$ and $902.$

Each of these gives us $2 \cdot 2 = 4$ permutations, but we overcount by a factor of $2$ since flipping the first and last digits creates another number already in the set.

Therefore, these numbers provide a total of $8 \cdot 4 \div 2 = 16$ unique permutations.

There are now $64 - 8 = 56$ multiples of $11$ that we need to account for.

We know that each of these provides $3! = 6$ permutations. As above, however, note that flipping the first and last digit of any number in this set produces another number in this set.

We can see this by using the divisibility rule for $11.$ If $ABC$ is divisible by $11,$ then we have that $A + C - B$ is divisible by $11.$

This means that $C + A - B$ is divisible by $11,$ which means that $CBA$ is also divisible by $11.$

Therefore, these numbers contribute $56 \cdot 6 \div 2 = 168$ more permutations.

Over all the cases, we have a total of $24 + 18 + 16 + 168 = 226$ numbers.

Thus, A is the correct answer.

2017 AMC 10A 第 25 题

答案：A

难度评级：2380

解答：

其他年份的第 25 题