Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2015 AMC 10B Problems/Problem 21"

(→‎Solution 3)
m (→‎Solution 2)
(35 intermediate revisions by 9 users not shown)
Line 1: Line 1:
 
==Problem==
 
==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 <math>5</math> steps left).  Suppose the Dash takes <math>19</math> fewer jumps than Cozy to reach the top of the staircase.  Let <math>s</math> denote the sum of all possible numbers of steps this staircase can have.  What is the sum of the digits of <math>s</math>?
+
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 <math>5</math> steps left).  Suppose Dash takes <math>19</math> fewer jumps than Cozy to reach the top of the staircase.  Let <math>s</math> denote the sum of all possible numbers of steps this staircase can have.  What is the sum of the digits of <math>s</math>?
  
 
<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>
  
−
==Solutions==
+
==Solution 1==
  
−
===Solution 1===
+
Let <math>n</math> be the number of steps. We have
−
We can translate this wordy problem into this simple equation:
 
  
−
<cmath>\left\lceil \frac{s}{2} \right\rceil - 19 = \left\lceil \frac{s}{5} \right\rceil</cmath>
+
<cmath>\left\lceil \frac{n}{2} \right\rceil - 19 = \left\lceil \frac{n}{5} \right\rceil</cmath>
  
 
We will proceed to solve this equation via casework.
 
We will proceed to solve this equation via casework.
  
−
Case 1: <math>\left\lceil \frac{s}{2} \right\rceil = \frac{s}{2}</math>
+
Case <math>1</math>: <math>\left\lceil \frac{n}{2} \right\rceil = \frac{n}{2}</math>
  
−
Our equation becomes <math>\frac{s}{2} - 19 = \frac{s}{5} + \frac{j}{5}</math>, where <math>j \in \{0,1,2,3,4\}</math> Using the fact that <math>s</math> is an integer, we quickly find that <math>j=1</math> and <math>j=4</math> yield <math>s=64</math> and <math>s=66</math>, respectively.
+
Our equation becomes <math>\frac{n}{2} - 19 = \frac{n}{5} + \frac{j}{5}</math>, where <math>j \in \{0,1,2,3,4\}</math> Using the fact that <math>n</math> is an integer, we quickly find that <math>j=1</math> and <math>j=4</math> yield <math>n=64</math> and <math>n=66</math>, respectively.
  
 
<br>
 
<br>
  
−
Case 2: <math>\left\lceil \frac{s}{2} \right\rceil = \frac{s}{2}+\frac{1}{2}</math>
+
Case <math>2</math>: <math>\left\lceil \frac{n}{2} \right\rceil = \frac{n}{2}+\frac{1}{2}</math>
  
−
Our equation becomes <math>\frac{s}{2} +\frac{1}{2} - 19 = \frac{s}{5} + \frac{j}{5}</math>, where <math>j \in \{0,1,2,3,4\}</math> Using the fact that <math>s</math> is an integer, we quickly find that <math>j=2</math> yields <math>s=63</math>. Summing up we get <math>63+64+66=193</math>. The sum of the digits is <math>\boxed{\textbf{(D)}\; 13}</math>.
+
Our equation becomes <math>\frac{n}{2} +\frac{1}{2} - 19 = \frac{n}{5} + \frac{j}{5}</math>, where <math>j \in \{0,1,2,3,4\}</math> Using the fact that <math>n</math> is an integer, we quickly find that <math>j=2</math> yields <math>n=63</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===
+
==Solution 2==
 +
 
 +
We know from the problem that Dash goes <math>3</math> steps further than Cozy per jump (assuming they aren't within <math>4</math> steps from the top). That means that if Dash takes <math>19</math> fewer jumps than Cozy to get to the top of the staircase, the staircase must be at least <math>3 \cdot 19=57</math> steps high. We then start using guess-and-check:
 +
 
 +
<math>57</math> steps: <math>\left \lceil {57/2} \right \rceil = 29</math> jumps for Cozy, and <math>\left \lceil {57/5} \right \rceil = 12</math> jumps for Dash, giving a difference of <math>17</math> jumps.
 +
 
 +
<math>58</math> steps: <math>\left \lceil {58/2} \right \rceil = 29</math> jumps for Cozy, and <math>\left \lceil {58/5} \right \rceil = 12</math> jumps for Dash, giving a difference of <math>17</math> jumps.
 +
 
 +
<math>59</math> steps: <math>\left \lceil {59/2} \right \rceil = 30</math> jumps for Cozy, and <math>\left \lceil {59/5} \right \rceil = 12</math> jumps for Dash, giving a difference of <math>18</math> jumps.
 +
 
 +
<math>60</math> steps: <math>\left \lceil {60/2} \right \rceil = 30</math> jumps for Cozy, and <math>\left \lceil {60/5} \right \rceil = 12</math> jumps for Dash, giving a difference of <math>18</math> jumps.
 +
 
 +
<math>\vdots</math>
 +
 
 +
By the time we test <math>61</math> steps, we notice that when the number of steps exceeds a multiple of <math>2</math>, the difference in jumps increases. So, we have to find the next number that will increase the difference. <math>62</math> doesn't because both both Cozy's and Dash's number of jumps increases, but <math>63</math> does, and <math>64</math>. <math>65</math> actually gives a difference of <math>20</math> jumps, but <math>66</math> goes back down to <math>19</math> (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 <math>19</math> onward.
 +
 
 +
Therefore, the possible numbers of steps in the staircase are <math>63</math>, <math>64</math>, and <math>66</math>, giving a sum of <math>193</math>. The sum of those digits is <math>13</math>, so the answer is <math>\boxed{D}</math>
 +
 
 +
*Minor Edits by Lolgod2
 +
 
 +
==Solution 3==
  
 
We're looking for natural numbers <math>x</math> such that <math>\left \lceil{\frac{x}{5}}\right \rceil + 19 = \left \lceil{\frac{x}{2}}\right \rceil</math>.
 
We're looking for natural numbers <math>x</math> such that <math>\left \lceil{\frac{x}{5}}\right \rceil + 19 = \left \lceil{\frac{x}{2}}\right \rceil</math>.
Line 31: Line 50:
 
<math>19 - 3a = \left \lceil{\frac{b}{2}}\right \rceil - \left \lceil{\frac{b}{5}}\right \rceil</math>.
 
<math>19 - 3a = \left \lceil{\frac{b}{2}}\right \rceil - \left \lceil{\frac{b}{5}}\right \rceil</math>.
  
−
Obviously, since <math>b \le 10</math>, this will not work for any value under 6. In addition, since obviously <math>\frac{b}{2} \ge \frac{b}{5}</math>, this will not work for any value over six, so we have <math>a = 6</math> and <math>\left \lceil{\frac{b}{2}}\right \rceil - \left \lceil{\frac{b}{5}}\right \rceil = 1.</math>
+
Obviously, since <math>b \le 10</math>, this will not work for any value of <math>a</math> under <math>6</math>. In addition, since obviously <math>\frac{b}{2} \ge \frac{b}{5}</math>, this will not work for any value over six, so we have <math>a = 6</math> and <math>\left \lceil{\frac{b}{2}}\right \rceil - \left \lceil{\frac{b}{5}}\right \rceil = 1.</math>
  
 
This can be achieved when <math>\left \lceil{\frac{b}{5}}\right \rceil = 1</math> and <math>\left \lceil{\frac{b}{2}}\right \rceil = 2</math>, or when <math>\left \lceil{\frac{b}{5}}\right \rceil = 2</math> and <math>\left \lceil{\frac{b}{2}}\right \rceil = 3</math>.
 
This can be achieved when <math>\left \lceil{\frac{b}{5}}\right \rceil = 1</math> and <math>\left \lceil{\frac{b}{2}}\right \rceil = 2</math>, or when <math>\left \lceil{\frac{b}{5}}\right \rceil = 2</math> and <math>\left \lceil{\frac{b}{2}}\right \rceil = 3</math>.
Line 45: Line 64:
 
We then have <math>63 + 64 + 66 = 193</math>, which has a digit sum of <math>\boxed{13}</math>.
 
We then have <math>63 + 64 + 66 = 193</math>, which has a digit sum of <math>\boxed{13}</math>.
  
−
===Solution 3===
+
==Solution 4==
−
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:
+
Translate the problem into following equation:
  
−
<math>57</math> steps: <math>\left \lceil {57/2} \right \rceil = 29</math> steps for Cozy, and <math>\left \lceil {57/5} \right \rceil = 12</math> steps for Dash, giving a difference of <math>17</math> jumps.
+
<math>n = 5D - \{0 \sim 4\} = 2C - \{0 \sim 1\}</math>
  
−
<math>\vdots</math>
+
Since <math>C = D + 19</math>, we have
 +
 
 +
<math>5D - \{0 \sim 4\} = 2D + 38 - \{0 \sim 1\}</math>
 +
 
 +
i.e.,
 +
 
 +
<math>3D = 38 + \{0 \sim 4\} - \{0 \sim 1\}</math>
 +
 
 +
We then have <math>D = 13</math> when <math>\{1\} - \{0\}</math> or <math>\{2\} - \{1\}</math> (the dog's last jump has <math>2</math> steps and the cat's last jump has <math>1</math> step), which yields <math>n = 64</math> and <math>n = 63</math> respectively.
  
−
By the time we test <math>61</math> steps (which gives a difference of <math>18</math> jumps), we notice a pattern: 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).  
+
Another solution is <math>D = 14</math> when <math>\{4\} - \{0\}</math>, which yields <math>n = 66</math>.
  
−
So, we have to find the next number that will increase the difference. <math>62</math> doesn't because it doesn't exceed any larger multiples of 2 or 5, but <math>63</math> does, and <math>64</math> and <math>66</math>. We don't need to go any further because the difference will stay above 19.
+
Therefore, with <math>63 + 64 + 66 = 193</math>, the digit sum is <math>\boxed{13}</math>.
  
−
Therefore, the possible numbers of steps in the staircase are <math>63</math>, <math>64</math>, and <math>66</math>, giving a sum of <math>193</math>. The sum of those digits is <math>13</math>, so the answer is <math>\boxed{D}</math>
+
==Video Solution==
 +
https://youtu.be/TpHZVbBGmVQ
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2015|ab=B|num-b=20|num-a=22}}
 
{{AMC10 box|year=2015|ab=B|num-b=20|num-a=22}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 00:51, 16 June 2022

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 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

Let $n$ be the number of steps. We have

\[\left\lceil \frac{n}{2} \right\rceil - 19 = \left\lceil \frac{n}{5} \right\rceil\]

We will proceed to solve this equation via casework.

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

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


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

Our equation becomes $\frac{n}{2} +\frac{1}{2} - 19 = \frac{n}{5} + \frac{j}{5}$, where $j \in \{0,1,2,3,4\}$ Using the fact that $n$ is an integer, we quickly find that $j=2$ yields $n=63$. 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 $3 \cdot 19=57$ steps high. 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 $\left \lceil {58/5} \right \rceil = 12$ 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.

$60$ steps: $\left \lceil {60/2} \right \rceil = 30$ jumps for Cozy, and $\left \lceil {60/5} \right \rceil = 12$ jumps for Dash, giving a difference of $18$ jumps.

$\vdots$

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 $\boxed{D}$

  • Minor Edits by Lolgod2

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 $2a + \left \lceil{\frac{b}{5}}\right \rceil + 19 = 5a + \left \lceil{\frac{b}{2}}\right \rceil$, 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 of $a$ 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 4

Translate the problem into following equation:

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

Since $C = D + 19$, we have

$5D - \{0 \sim 4\} = 2D + 38 - \{0 \sim 1\}$

i.e.,

$3D = 38 + \{0 \sim 4\} - \{0 \sim 1\}$

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

Another solution is $D = 14$ when $\{4\} - \{0\}$, which yields $n = 66$.

Therefore, with $63 + 64 + 66 = 193$, the digit sum is $\boxed{13}$.

Video Solution

https://youtu.be/TpHZVbBGmVQ

See Also

2015 AMC 10B (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 10 Problems and Solutions

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10B_Problems/Problem_21&oldid=174917"