Difference between revisions of "2015 AMC 10B Problems/Problem 21"
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We will proceed to solve this equation via casework. | We will proceed to solve this equation via casework. | ||
− | Case <math>1</math>: <math>\left\lceil \frac{s}{2} \right\rceil = \frac{s}{2}</math> | + | Case <math>1</math> (<math>s</math> is an even number): <math>\left\lceil \frac{s}{2} \right\rceil = \frac{s}{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{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. | ||
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<br> | <br> | ||
− | Case <math>2</math>: <math>\left\lceil \frac{s}{2} \right\rceil = \frac{s}{2}+\frac{1}{2}</math> | + | Case <math>2</math> (<math>s</math> is an odd number): <math>\left\lceil \frac{s}{2} \right\rceil = \frac{s}{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{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>. |
Revision as of 16:59, 20 September 2020
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 steps left). Suppose the Dash takes fewer jumps than Cozy to reach the top of the staircase. Let denote the sum of all possible numbers of steps this staircase can have. What is the sum of the digits of ?
Solution 1
We can translate this wordy problem into this simple equation:
We will proceed to solve this equation via casework.
Case ( is an even number):
Our equation becomes , where Using the fact that is an integer, we quickly find that and yield and , respectively.
Case ( is an odd number):
Our equation becomes , where Using the fact that is an integer, we quickly find that yields . Summing up we get . The sum of the digits is .
Solution 2
We know from the problem that Dash goes steps further than Cozy per jump (assuming they aren't within steps from the top). That means that if Dash takes fewer jumps than Cozy to get to the top of the staircase, the staircase must be at least steps high (3*19=57). We then start using guess-and-check:
steps: jumps for Cozy, and jumps for Dash, giving a difference of jumps.
steps: jumps for Cozy, and jumps for Dash, giving a difference of jumps.
steps: jumps for Cozy, and jumps for Dash, giving a difference of jumps.
By the time we test steps, we notice that when the number of steps exceeds a multiple of , the difference in jumps increases. So, we have to find the next number that will increase the difference. doesn't because both both Cozy's and Dash's number of jumps increases, but does, and . actually gives a difference of jumps, but goes back down to (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 onward.
Therefore, the possible numbers of steps in the staircase are , , and , giving a sum of . The sum of those digits is , so the answer is
Solution 3
We're looking for natural numbers such that .
Let's call . We now have , or
.
Obviously, since , this will not work for any value under . In addition, since obviously , this will not work for any value over six, so we have and
This can be achieved when and , or when and .
Case One:
We have and , so .
Case Two:
We have and , so .
We then have , which has a digit sum of .
Solution 4
Translate the problem into following equation:
Since , we have
i.e.,
We then have when or (the dog's last jump has steps and the cat's last jump has step), which yields and respectively.
Another solution is when , which yields .
Therefore, with , the digit sum is .
Video Solution
~IceMatrix
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.