Question

Cozy the Cat 和 Dash the Dog 正在爬一段有某个步数的楼梯。不过，它们不是一步一步走上去，而是跳上去。Cozy 每次跳上两级台阶（但如果有必要，它最后只会跳上最后一级）。Dash 每次跳上五级台阶（但如果剩下少于 $5$ 级，必要时它最后会只跳完剩下的台阶）。假设 Dash 到达楼梯顶部所用的跳数比 Cozy 少 $19$ 次。令 $s$ 表示这段楼梯所有可能步数的和。 $s$ 的各位数字之和是多少？

Cozy the Cat and Dash the Dog are going up a staircase with a certain number of steps. However, instead of walking up the steps one at a time, both Cozy and Dash jump. Cozy goes two steps up with each jump (though if necessary, he will just jump the last step). Dash goes five steps up with each jump (though if necessary, he will just jump the last steps if there are fewer than $5$ steps left). Suppose that Dash takes $19$ fewer jumps than Cozy to reach the top of the staircase. Let $s$ denote the sum of all possible numbers of steps this staircase can have. What is the sum of the digits of $s?$

$9$

$11$

$12$

$13$

$15$

答案：D

知识点：取整函数分类讨论

难度评级：2170

解答：

一段 $t$ 级的楼梯需要 Cozy 跳 $\left\lceil \tfrac{t}{2} \right\rceil$ 次，Dash 跳 $\left\lceil \tfrac{t}{5} \right\rceil$ 次，我们需要二者差为 $19$ 。

检查可能情形，有效值为 $d+1$ ， $64$ ，和 $66$ ，所以 $s = 63 + 64 + 66 = 193$ 。它的各位数字和是 $1 + 9 + 3 = 13$ 。 $t$ $5d+1$ ， $5d+2$ ， $5d+3$ ， $5d+4$ ， $5d+5$ 。 $d+20$ $t$ $2d+39$ $2d+40$ 。 $d$ $t=63,66,64$ 。 $t=63$ ， $\begin{gathered} 5d+3=2d+39,\\ 5d+1=2d+40,\\ 5d+4=2d+40. \end{gathered}$

因此，正确选项是 D。

A staircase of $t$ steps takes Cozy $\left\lceil \tfrac{t}{2} \right\rceil$ jumps and Dash $\left\lceil \tfrac{t}{5} \right\rceil$ jumps, and we need the difference to equal $19.$

Suppose Dash makes $d+1$ jumps. Then $t$ is one of $5d+1,$ $5d+2,$ $5d+3,$ $5d+4,$ $5d+5.$ Cozy makes $d+20$ jumps, so $t$ is either $2d+39$ or $2d+40.$ Equating these two lists gives an integer $d$ only in the three cases $\begin{gathered} 5d+3=2d+39,\\ 5d+1=2d+40,\\ 5d+4=2d+40. \end{gathered}$ These yield respectively $t=63,66,64.$ Thus the valid values are $t=63,$ $64,$ and $66,$ so $s = 63 + 64 + 66 = 193.$ Its digit sum is $1 + 9 + 3 = 13.$

Thus, the correct answer is D.

2015 AMC 12B 第 21 题

答案：D

难度评级：2170

解答：

其他年份的第 21 题