Difference between revisions of "2015 AMC 10B Problems/Problem 21"
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So, we have to find the next number that will increase the difference. <math>61</math> doesn't because it's a multiple of 2 and 5, and <math>62</math> doesn't because it doesn't exceed any larger multiples of 2 or 5. <math>63</math> does, however, and <math>64</math> and <math>66</math>. We don't need to go any further because the difference will stay above 20. | So, we have to find the next number that will increase the difference. <math>61</math> doesn't because it's a multiple of 2 and 5, and <math>62</math> doesn't because it doesn't exceed any larger multiples of 2 or 5. <math>63</math> does, however, and <math>64</math> and <math>66</math>. We don't need to go any further because the difference will stay above 20. | ||
+ | |||
+ | 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> | ||
==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 23:00, 21 December 2017
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
?
Solutions
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
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 6. 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 3
We know from the problem that Dash goes up 3 steps more 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:
steps for Cozy, and
steps for Dash, giving a difference of
steps.
steps for Cozy, and
steps for Dash, giving a difference of
steps.
steps for Cozy, and
steps for Dash, giving a difference of
steps.
steps for Cozy, and
steps for Dash, giving a difference of
steps.
We notice a pattern here: 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).
So, we have to find the next number that will increase the difference. doesn't because it's a multiple of 2 and 5, and
doesn't because it doesn't exceed any larger multiples of 2 or 5.
does, however, and
and
. We don't need to go any further because the difference will stay above 20.
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
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 |
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