Difference between revisions of "2019 AMC 10B Problems/Problem 25"
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~Solution by mxnxn | ~Solution by mxnxn | ||
| + | ==Solution3== | ||
| + | That problem have 3 different cases like 0 or 0->1 or 1->1 | ||
| + | If one sequence last two is 1->1 than next well be 0 | ||
| + | If one sequense last two is 0->1 then next can be 0 or 1 | ||
| + | If one sequense last one is 0 then next well be 1 | ||
| + | 1 2 3 4 5 6 7 8 9 | ||
| + | 0 1 0 1 1 1 2 2 3 4 | ||
| + | 0->1 0 1 0 1 1 1 2 2 3 | ||
| + | 1->1 0 0 1 0 1 1 1 2 2 | ||
| + | 10 11 12 13 14 15 16 17 18 | ||
| + | 0 5 7 9 12 16 21 28 37 49 | ||
| + | 0->1 4 5 7 9 12 16 21 28 37 | ||
| + | 1->1 3 4 5 7 9 12 16 21 28 | ||
| + | But last one is 0 so just 1->1->0 or 0->1->1 well be right->answer is 37+28=65(C) | ||
| + | |||
| + | ~Solution by rayfunmath | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2019|ab=B|num-b=24|after=Last Problem}} | {{AMC10 box|year=2019|ab=B|num-b=24|after=Last Problem}} | ||
{{AMC12 box|year=2019|ab=B|num-b=22|num-a=24}} | {{AMC12 box|year=2019|ab=B|num-b=22|num-a=24}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 04:17, 15 February 2019
- The following problem is from both the 2019 AMC 10B #25 and 2019 AMC 12B #23, so both problems redirect to this page.
Problem
How many sequences of 
s and 
s of length 
are there that begin with a , end with a , contain no two consecutive s, and contain no three consecutive s?
Solution 1 (Recursion)
We can deduce that any valid sequence of length 
will start with a 0 followed by either "10" or "110".
Because of this, we can define a recursive function:
This is because for any valid sequence of length , you can append either "10" or "110" and the resulting sequence would still satisfy the given conditions.
and 
, so you follow the recursion up until 
~Solution by MagentaCobra
Solution 2 (Casework)
After any given zero, the next zero must appear exactly two or three spots down the line. And we started at position 1 and ended at position 19, so we moved over 18. Therefore, we must add a series of 2's and 3's to get 18. How can we do this?
Option 1: nine 2's (there is only 1 way to arrange this).
Option 2: two 3's and six 2's (
ways to arrange this).
Option 3: four 3's and three 2's (
ways to arrange this).
Option 4: six 3's (there is only 1 way to arrange this).
Sum the four numbers given above: 1+28+35+1=65
~Solution by mxnxn
Solution3
That problem have 3 different cases like 0 or 0->1 or 1->1 If one sequence last two is 1->1 than next well be 0 If one sequense last two is 0->1 then next can be 0 or 1 If one sequense last one is 0 then next well be 1 1 2 3 4 5 6 7 8 9 0 1 0 1 1 1 2 2 3 4 0->1 0 1 0 1 1 1 2 2 3 1->1 0 0 1 0 1 1 1 2 2
10 11 12 13 14 15 16 17 18 0 5 7 9 12 16 21 28 37 49 0->1 4 5 7 9 12 16 21 28 37 1->1 3 4 5 7 9 12 16 21 28 But last one is 0 so just 1->1->0 or 0->1->1 well be right->answer is 37+28=65(C)
~Solution by rayfunmath
See Also
| 2019 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 24 |
Followed by Last Problem | |
| 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 | ||
| 2019 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 22 |
Followed by Problem 24 |
| 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
