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Difference between revisions of "2015 AMC 12B Problems/Problem 21"

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<math>\textbf{(A)}\; 9 \qquad\textbf{(B)}\; 11 \qquad\textbf{(C)}\; 12 \qquad\textbf{(D)}\; 13 \qquad\textbf{(E)}\; 15</math>
 
<math>\textbf{(A)}\; 9 \qquad\textbf{(B)}\; 11 \qquad\textbf{(C)}\; 12 \qquad\textbf{(D)}\; 13 \qquad\textbf{(E)}\; 15</math>
  
−
==Solution==
+
==Solution 1==
 
We can translate this wordy problem into this simple equation:
 
We can translate this wordy problem into this simple equation:
  
Line 22: Line 22:
  
 
Summing up we get <math>63+64+66=193</math>. The sum of the digits is <math>\boxed{\textbf{(D)}\; 13}</math>.
 
Summing up we get <math>63+64+66=193</math>. The sum of the digits is <math>\boxed{\textbf{(D)}\; 13}</math>.
 +
 +
==Solution 2==
 +
It can easily be seen that the problem can be expressed by the equation:
 +
<cmath>\left\lceil \frac{s}{2} \right\rceil - \left\lceil \frac{s}{5} \right\rceil = 19</cmath>
 +
 +
However, because the ceiling function is difficult to work with, we can rewrite the previous equation as:
 +
 +
<cmath>\frac{s+a}{2} - \frac{s+b}{5} = 19</cmath> Where <math>a \in \{0,1\}</math> and <math>b \in \{0,1,2,3,4\}</math> Multiplying both sides by ten and simplifying, we get:
 +
<cmath>5s+5a-2s-2b=190</cmath>
 +
<cmath>3s = 190+2b-5a</cmath>
 +
<cmath>s = 63 + \frac{1+2b-5a}{3}</cmath>
 +
 +
Because s must be an integer, we need to find the values of <math>a</math> and <math>b</math> such that <math>2b-5a \equiv 2 \mod 3</math>. We solve using casework.
 +
 +
Case 1: <math>a = 0</math>
 +
 +
If <math>a = 0</math>, we have <math>2b \equiv 2 \mod 3</math>. We can easily see that <math>b = 1</math> or <math>b = 4</math>, which when plugged into our original equation lead to <math>s = 64</math> and <math>s=66</math> respectively.
 +
 +
<br>
 +
 +
Case 2: <math>a = 1</math>
 +
 +
If <math>a = 1</math>, we have <math>2b-5 \equiv 2 \mod 3</math>, which can be rewritten as <math>2b \equiv 1 \mod 3 </math>. We can again easily see that <math>b = 2</math> is the only solution, which when plugged into our original equation lead to <math>s = 63</math>.
 +
 +
Adding these together we get <math>64+66+63=193</math>. The sum of the digits is <math>\boxed{\textbf{(D)}\; 13}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2015|ab=B|num-a=22|num-b=20}}
 
{{AMC12 box|year=2015|ab=B|num-a=22|num-b=20}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 19:10, 5 September 2015

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 $\boxed{\textbf{(D)}\; 13}$.

See Also

2015 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 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

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