Difference between revisions of "2015 AMC 12B Problems/Problem 21"
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Adding these together we get <math>64+66+63=193</math>. The sum of the digits is <math>\boxed{\textbf{(D)}\; 13}</math>. | Adding these together we get <math>64+66+63=193</math>. The sum of the digits is <math>\boxed{\textbf{(D)}\; 13}</math>. | ||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | As before, we write the equation: | ||
+ | |||
+ | <cmath>\left\lceil \frac{s}{2} \right\rceil - 19 = \left\lceil \frac{s}{5} \right\rceil.</cmath> | ||
+ | |||
+ | To get a ballpark estimate of where <math>s</math> might lie, we remove the ceiling functions to find: | ||
+ | |||
+ | <cmath>\frac{s}{2} - 19 = \frac{s}{5}.</cmath> | ||
+ | |||
+ | This gives <math>\frac{3s}{10} = 19</math>, and thus values for <math>s</math> will be around <math>\frac{190}{3} = 63.\overline3</math>. | ||
+ | |||
+ | Now, to establish some bounds around this estimated working value, we note that if <math>s=60</math>, Cozy takes 30 steps while Dash takes 12, a difference of 18. If <math>s=70</math>, Cozy takes 35 steps while Dash takes 14, a difference of 21. When <math>s</math> 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 <math>s</math> will be between <math>60</math> and <math>70</math>. | ||
+ | |||
+ | Then, by inspection, <math>s=63, 64,</math> or <math>66</math>, so <math>\sum s = 193 \implies \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 20:37, 11 February 2018
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 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 1:
Our equation becomes , where Using the fact that is an integer, we quickly find that and yield and , respectively.
Case 2:
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
It can easily be seen that the problem can be expressed by the equation:
However, because the ceiling function is difficult to work with, we can rewrite the previous equation as:
Where and Multiplying both sides by ten and simplifying, we get:
Because s must be an integer, we need to find the values of and such that . We solve using casework.
Case 1:
If , we have . We can easily see that or , which when plugged into our original equation lead to and respectively.
Case 2:
If , we have , which can be rewritten as . We can again easily see that is the only solution, which when plugged into our original equation lead to .
Adding these together we get . The sum of the digits is .
Solution 3
As before, we write the equation:
To get a ballpark estimate of where might lie, we remove the ceiling functions to find:
This gives , and thus values for will be around .
Now, to establish some bounds around this estimated working value, we note that if , Cozy takes 30 steps while Dash takes 12, a difference of 18. If , Cozy takes 35 steps while Dash takes 14, a difference of 21. When 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 will be between and .
Then, by inspection, or , so
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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