Difference between revisions of "2018 AMC 8 Problems/Problem 19"
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− | Instead of + and -, let us use 1 and 0, respectively. If we let <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> be the values of the four cells on the bottom row, then the three cells on the next row are equal to <math>a+b</math>, <math>b+c</math>, and <math>c+d</math> taken modulo 2 (this is exactly the same as finding <math>a \text{ XOR } b</math>, and so on). The two cells on the next row are <math>a+2b+c</math> and <math>b+2c+d</math> taken modulo 2, and lastly, the cell on the top row gets <math>a+3b+3c+d \pmod{2}</math>. | + | Instead of + and -, let us use 1 and 0, respectively. If we let <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> be the values of the four cells on the bottom row, then the three cells on the next row are equal to <math>a+b</math>, <math>b+c</math>, and <math>c+d</math> taken modulo (mod) 2 (this is exactly the same as finding <math>a \text{ XOR } b</math>, and so on). The two cells on the next row are <math>a+2b+c</math> and <math>b+2c+d</math> taken modulo (mod) 2, and lastly, the cell on the top row gets <math>a+3b+3c+d \pmod{2}</math>. |
Thus, we are looking for the number of assignments of 0's and 1's for <math>a</math>, <math>b</math>, <math>c</math>, <math>d</math> such that <math>a+3b+3c+d \equiv 1 \pmod{2}</math>, or in other words, is odd. As <math>3 \equiv 1 \pmod{2}</math>, this is the same as finding the number of assignments such that <math>a+b+c+d \equiv 1 \pmod{2}</math>. Notice that, no matter what <math>a</math>, <math>b</math>, and <math>c</math> are, this uniquely determines <math>d</math>. There are <math>2^3 = 8</math> ways to assign 0's and 1's arbitrarily to <math>a</math>, <math>b</math>, and <math>c</math>, so the answer is <math>\boxed{\textbf{(C) } 8}</math>. | Thus, we are looking for the number of assignments of 0's and 1's for <math>a</math>, <math>b</math>, <math>c</math>, <math>d</math> such that <math>a+3b+3c+d \equiv 1 \pmod{2}</math>, or in other words, is odd. As <math>3 \equiv 1 \pmod{2}</math>, this is the same as finding the number of assignments such that <math>a+b+c+d \equiv 1 \pmod{2}</math>. Notice that, no matter what <math>a</math>, <math>b</math>, and <math>c</math> are, this uniquely determines <math>d</math>. There are <math>2^3 = 8</math> ways to assign 0's and 1's arbitrarily to <math>a</math>, <math>b</math>, and <math>c</math>, so the answer is <math>\boxed{\textbf{(C) } 8}</math>. |
Revision as of 19:58, 8 November 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 (mod) 2 (this is exactly the same as finding , and so on). The two cells on the next row are and taken modulo (mod) 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
The sign of the next row on the pyramid depends on previous row. There are two options for the previous row, - or +. There are three rows to the pyramid that depend on what the top row is. Therefore, the ways you can make a + on the top is , which is C. -totoro.selsel
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 |
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