Difference between revisions of "2018 AMC 8 Problems/Problem 19"
(Improved clarity of Solution 2) |
Virjoy2001 (talk | contribs) m (→Problem 19) |
||
Line 1: | Line 1: | ||
− | + | =Problem 19= | |
In a sign pyramid a cell gets a "+" if the two cells below it have the same sign, and it gets a "-" if the two cells below it have different signs. The diagram below illustrates a sign pyramid with four levels. How many possible ways are there to fill the four cells in the bottom row to produce a "+" at the top of the pyramid? | In a sign pyramid a cell gets a "+" if the two cells below it have the same sign, and it gets a "-" if the two cells below it have different signs. The diagram below illustrates a sign pyramid with four levels. How many possible ways are there to fill the four cells in the bottom row to produce a "+" at the top of the pyramid? | ||
Revision as of 13:35, 20 August 2019
Problem 19
In a sign pyramid a cell gets a "+" if the two cells below it have the same sign, and it gets a "-" if the two cells below it have different signs. The diagram below illustrates a sign pyramid with four levels. How many possible ways are there to fill the four cells in the bottom row to produce a "+" at the top of the pyramid?
Solution 1
Instead of + and -, let us use 1 and 0, respectively. If we let , , , and be the values of the four cells on the bottom row, then the three cells on the next row are equal to , , and taken modulo 2 (this is exactly the same as finding , and so on). The two cells on the next row are and taken modulo 2, and lastly, the cell on the top row gets .
Thus, we are looking for the number of assignments of 0's and 1's for , , , such that , or in other words, is odd. As , this is the same as finding the number of assignments such that . Notice that, no matter what , , and are, this uniquely determines . There are ways to assign 0's and 1's arbitrarily to , , and , so the answer is .
Solution 2
Each row is fully determined by its leftmost cell (the other cells in the row are restricted from the cells above). Since we are given the first row already, we still need to decide the other rows. The first cell in each row has only possibilities (+ and -), so we have ways.
See Also
2018 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.