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考题21 2015 AMC 10B

2018-08-06 重点归纳

AMC10数学竞赛是美国高中数学竞赛中的一项，是针对高中一年级及初中三年级学生的数学测试，该竞赛开始于2000年，分A赛和B赛，于每年的2月初和2月中举行，学生可任选参加一项即可。不管是对高校申请还是今后在数学领域的发展都极其有利！那么接下来跟随小编来看一下AMC10的官方真题以及官方解答吧：

Problem 21

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 the 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 $s$ ?

$考题21 2015 AMC 10B$

Solutions

Solution 1

We can translate this wordy problem into this simple equation:

$考题21 2015 AMC 10B$

We will proceed to solve this equation via casework.

Case 1: $\left\lceil \frac{s}{2} \right\rceil = \frac{s}{2}$

Our equation becomes $\frac{s}{2} - 19 = \frac{s}{5} + \frac{j}{5}$ , where $j \in \{0,1,2,3,4\}$ Using the fact that $s$ is an integer, we quickly find that $j=1$ and $j=4$ yield $s=64$ and $s=66$ , respectively.

Case 2: $考题21 2015 AMC 10B$

Our equation becomes $\frac{s}{2} +\frac{1}{2} - 19 = \frac{s}{5} + \frac{j}{5}$ , where $j \in \{0,1,2,3,4\}$ Using the fact that $s$ is an integer, we quickly find that $j=2$ yields $s=63$ . Summing up we get $63+64+66=193$ . The sum of the digits is $\boxed{\textbf{(D)}\; 13}$ .

Solution 2

We know from the problem that Dash goes 3 steps further than Cozy per jump (assuming they aren't within 4 steps from the top). That means that if Dash takes 19 fewer jumps than Cozy to get to the top of the staircase, the staircase must be at least 57 steps high (3*19=57). We then start using guess-and-check:

$57$ steps: $\left \lceil {57/2} \right \rceil = 29$ jumps for Cozy, and $\left \lceil {57/5} \right \rceil = 12$ jumps for Dash, giving a difference of $17$ jumps.

$58$ steps: $\left \lceil {58/2} \right \rceil = 29$ jumps for Cozy, and $\left \lceil {57/5} \right \rceil = 12$ jumps for Dash, giving a difference of $17$ jumps.

$59$ steps: $\left \lceil {59/2} \right \rceil = 30$ jumps for Cozy, and $\left \lceil {59/5} \right \rceil = 12$ jumps for Dash, giving a difference of $18$ jumps.

By the time we test $61$ steps, we notice that when the number of steps exceeds a multiple of $2$ , the difference in jumps increases. So, we have to find the next number that will increase the difference. $62$ doesn't because both both Cozy's and Dash's number of jumps increases, but $63$ does, and $64$ . $65$ actually gives a difference of 20 jumps, but $66$ goes back down to 19 (because Dash had to take another jump when Cozy didn't). We don't need to go any further because the difference will stay above 19 onward.

Therefore, the possible numbers of steps in the staircase are $63$ , $64$ , and $66$ , giving a sum of $193$ . The sum of those digits is $13$ , so the answer is $\boxed{D}$

Solution 3

We're looking for natural numbers $x$ such that $\left \lceil{\frac{x}{5}}\right \rceil + 19 = \left \lceil{\frac{x}{2}}\right \rceil$ .

Let's call $x = 10a + b$ . We now have $考题21 2015 AMC 10B$ , or

$19 - 3a = \left \lceil{\frac{b}{2}}\right \rceil - \left \lceil{\frac{b}{5}}\right \rceil$ .

Obviously, since $b \le 10$ , this will not work for any value under 6. In addition, since obviously $\frac{b}{2} \ge \frac{b}{5}$ , this will not work for any value over six, so we have $a = 6$ and $\left \lceil{\frac{b}{2}}\right \rceil - \left \lceil{\frac{b}{5}}\right \rceil = 1.$

This can be achieved when $\left \lceil{\frac{b}{5}}\right \rceil = 1$ and $考题21 2015 AMC 10B$ , or when $\left \lceil{\frac{b}{5}}\right \rceil = 2$ and $\left \lceil{\frac{b}{2}}\right \rceil = 3$ .

Case One:

We have $b \le 5$ and $3 \le b \le 4$ , so $b = 3, 4$ .

Case Two:

We have $6 \le b \le 9$ and $5 \le b \le 6$ , so $b = 6$ .

We then have $63 + 64 + 66 = 193$ , which has a digit sum of $\boxed{13}$ .

Solution 4

Translate the problem into following equation:

$n = 5D - \{0 \sim 4\} = 2C - \{0 \sim 1\}$

Since $C = D + 19$ , we have

$考题21 2015 AMC 10B$

i.e.,

$考题21 2015 AMC 10B$

We then have $D = 13$ when $\{1\} - \{0\}$ or $\{2\} - \{1\}$ (dog's last jump has 2 steps and cat's last jump has 1 step), which yields $n = 64$ and $n = 63$ respectively.