# Difference between revisions of "2015 AMC 12B 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 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 $s$?

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

## 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

It can easily be seen that the problem can be expressed by the equation: $$\left\lceil \frac{s}{2} \right\rceil - \left\lceil \frac{s}{5} \right\rceil = 19$$

However, because the ceiling function is difficult to work with, we can rewrite the previous equation as:

$$\frac{s+a}{2} - \frac{s+b}{5} = 19$$ Where $a \in \{0,1\}$ and $b \in \{0,1,2,3,4\}$ Multiplying both sides by ten and simplifying, we get: $$5s+5a-2s-2b=190$$ $$3s = 190+2b-5a$$ $$s = 63 + \frac{1+2b-5a}{3}$$

Because s must be an integer, we need to find the values of $a$ and $b$ such that $2b-5a \equiv 2 \mod 3$. We solve using casework.

Case 1: $a = 0$

If $a = 0$, we have $2b \equiv 2 \mod 3$. We can easily see that $b = 1$ or $b = 4$, which when plugged into our original equation lead to $s = 64$ and $s=66$ respectively.

Case 2: $a = 1$

If $a = 1$, we have $2b-5 \equiv 2 \mod 3$, which can be rewritten as $2b \equiv 1 \mod 3$. We can again easily see that $b = 2$ is the only solution, which when plugged into our original equation lead to $s = 63$.

Adding these together we get $64+66+63=193$. The sum of the digits is $\boxed{\textbf{(D)}\; 13}$.

## Solution 3

As before, we write the equation:

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

To get a ballpark estimate of where $s$ might lie, we remove the ceiling functions to find:

$$\frac{s}{2} - 19 = \frac{s}{5}.$$

This gives $\frac{3s}{10} = 19$, and thus values for $s$ will be around $\frac{190}{3} = 63.\overline3$.

Now, to establish some bounds around this estimated working value, we note that if $s=60$, Cozy takes 30 steps while Dash takes 12, a difference of 18. If $s=70$, Cozy takes 35 steps while Dash takes 14, a difference of 21. When $s$ increases from a multiple of ten, the difference will never decrease beyond what it is at the multiple of ten, and likewise, when it decreases, it never becomes greater than at the multiple of ten, so any working values of $s$ will be between $60$ and $70$.

Then, by inspection, $s=63, 64,$ or $66$, so $\sum s = 193 \implies \boxed{\textbf{(D)}\; 13}.$

## Solution 4

Notice that the possible number of steps in the staircase is around 60 to 70. By testing all of the values between 60 and 70, we see that 63, 64 and 66 work. Adding those up gives 193, so the answer is $1+9+3=\boxed{\textbf{(D)}\; 13}.$

~IceMatrix

## See Also

 2015 AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 20 Followed byProblem 22 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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