Question

数 $2013$ 有这样的性质：它的个位数字等于其他数字之和，即 $2+0+1=3$ 。大于 $1000$ 且小于 $2013$ 的整数中，有多少个具有这个性质？

The number $2013$ has the property that its units digit is the sum of its other digits, that is $2+0+1=3.$ How many integers less than $2013$ but greater than $1000$ have this property?

$33$

$34$

$45$

$46$

$58$

答案：D

知识点：数字分类讨论三角形数

难度评级：1420

解答：

给定前三位数字时，如果它们的和不超过 $9$ ，就能确定一个满足条件的数。

按千位数字为 $1$ 的情况讨论。

若十位和百位，即第 $2$ 位和第 $3$ 位数字之和不超过 $8$ ，设其和为 $s$ ，则有 $s+1$ 种取法，因为其中一个数字可以从 $0$ 到 $s$ 。

因此这一段共有第 $9^{th}$ 个三角数： $\dfrac{9\cdot 10}2 =45.$

从二千到 $2013$ 之前的数中，只有 $2002$ 满足条件，所以总数为 $46$ 。

所以正确答案是 D。

Given the first three numbers, if their sum is less than or equal to $9,$ it creates one number with the property.

Now, we can case on the $1$ st digit.

If it is 1, then the sum of the $2$ nd and $3$ rd digit must be less than or equal to $8.$ For each possible sum $s,$ there are $s+1$ ways to choose the other numbers as the 2nd number can be anywhere from $0$ to $s.$

Thus, the total is the $9^{th}$ triangular number: $\dfrac{9\cdot 10}2 =45.$

If it is 2, then the only way we can get a number that works less than $2013$ is $2002,$ making a total of $46$ cases.

Thus, the correct answer is D .

2013 AMC 10B 第 18 题

答案：D

难度评级：1420

解答：

其他年份的第 18 题