Difference between revisions of "2015 AMC 10B Problems/Problem 21"
Mr doudou14 (talk | contribs) (→Solution 3) |
Mr doudou14 (talk | contribs) (→Solution 3) |
||
Line 53: | Line 53: | ||
By the time we test <math>61</math> steps (which gives a difference of <math>18</math> jumps), we notice 3 things: | By the time we test <math>61</math> steps (which gives a difference of <math>18</math> jumps), we notice 3 things: | ||
+ | |||
-When the number of steps exceeds a multiple of 5, the difference in jumps increases (because the number of jumps Dash takes increases) | -When the number of steps exceeds a multiple of 5, the difference in jumps increases (because the number of jumps Dash takes increases) | ||
Revision as of 23:19, 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 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:
steps:
jumps for Cozy, and
jumps for Dash, giving a difference of
jumps.
By the time we test steps (which gives a difference of
jumps), we notice 3 things:
-When the number of steps exceeds a multiple of 5, the difference in jumps increases (because the number of jumps Dash takes increases)
-When the number of steps exceeds a multiple of 2, the difference in jumps decreases (because the number of jumps Cozy takes increases)
-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 doesn't exceed any larger multiples of 2 or 5, but
does, and
.
actually gives a difference of 20 jumps, but
goes back down to 19. We don't need to go any further because the difference will stay above 19.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.