2015 AMC 10B Problems/Problem 21

Problem

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 $s$ ?

$\textbf{(A) }9\qquad\textbf{(B) }11\qquad\textbf{(C) }12\qquad\textbf{(D) }13\qquad\textbf{(E) }15$

Solutions

Solution 1

We can translate this wordy problem into this simple equation:

$\left\lceil \frac{s}{2} \right\rceil - 19 = \left\lceil \frac{s}{5} \right\rceil$

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

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'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 $2a + \left \lceil{\frac{b}{5}}\right \rceil + 19 = 5a + \left \lceil{\frac{b}{2}}\right \rceil$ , 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 $\left \lceil{\frac{b}{2}}\right \rceil = 2$ , 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 3

We know from the problem that Dash goes up 3 steps more 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:

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

$\vdots$

We notice a pattern here: the number of jumps Dash takes increases when the number of steps exceeds a multiple of 5, and the number of jumps Cozy takes increases when the number of steps exceeds a multiple of 2. Also, if the number of steps exceeds a multiple of both 2 and 5, the difference in jumps doesn't change (because both numbers increase).

So, we have to find the next number that will increase the difference. $61$ doesn't because it's a multiple of 2 and 5, and $62$ doesn't because it doesn't exceed any larger multiples of 2 or 5. $63$ does, however, and $64$ and $66$ . We don't need to go any further because the difference will stay above 19.

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 $\boxed{D}$

2015 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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