Difference between revisions of "2018 AMC 8 Problems/Problem 19"
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Scrabbler94 (talk | contribs) (→Solution: previous solution was quite poor as it's not immediately obvious why half of all assignments of +/- give a + on top.) |
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==Solution== | ==Solution== | ||
| + | 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>. | ||
| − | + | 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>. | |
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{{AMC8 box|year=2018|num-b=18|num-a=20}} | {{AMC8 box|year=2018|num-b=18|num-a=20}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 17:33, 21 November 2018
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?
![[asy] unitsize(2cm); path box = (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--cycle; draw(box); label("$+$",(0,0)); draw(shift(1,0)*box); label("$-$",(1,0)); draw(shift(2,0)*box); label("$+$",(2,0)); draw(shift(3,0)*box); label("$-$",(3,0)); draw(shift(0.5,0.4)*box); label("$-$",(0.5,0.4)); draw(shift(1.5,0.4)*box); label("$-$",(1.5,0.4)); draw(shift(2.5,0.4)*box); label("$-$",(2.5,0.4)); draw(shift(1,0.8)*box); label("$+$",(1,0.8)); draw(shift(2,0.8)*box); label("$+$",(2,0.8)); draw(shift(1.5,1.2)*box); label("$+$",(1.5,1.2)); [/asy]](http://latex.artofproblemsolving.com/f/5/9/f59b1af815a0ba23baf250a48b1029ee9d1bd5e4.png)
Solution
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 
.
| 2018 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 18 |
Followed by Problem 20 | |
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| All AJHSME/AMC 8 Problems and Solutions | ||
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