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Difference between revisions of "2013 AMC 8 Problems/Problem 6"
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<math>\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 6</math>
 
<math>\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 6</math>
  
−
==Solution==
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==Solution 1 (Working Backwards)==
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===Solution 1: Working Backwards===
 
 
Let the value in the empty box in the middle row be <math>x</math>, and the value in the empty box in the top row be <math>y</math>. <math>y</math> is the answer we're looking for.
 
Let the value in the empty box in the middle row be <math>x</math>, and the value in the empty box in the top row be <math>y</math>. <math>y</math> is the answer we're looking for.
  
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</asy>
 
</asy>
  
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We see that <math>600 = 30x</math>, making <math>x = 20</math>.
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From the diagram, <math>600 = 30x</math>, making <math>x = 20</math>.
  
 
<asy>
 
<asy>
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It follows that <math>20 = 5y</math>, so <math>y = \boxed{\textbf{(C)}\ 4}</math>.
 
It follows that <math>20 = 5y</math>, so <math>y = \boxed{\textbf{(C)}\ 4}</math>.
  
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===Solution 2: Jumping Back to the Start===
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==Solution 2 (Jumping to the Start)==
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Another way to do this problem is to realize what makes up the bottommost number. This method doesn't work quite as well for this problem, but in a larger tree, it might be faster. (In this case, Solution 1 would be faster since there's only two missing numbers.)
+
Another way to do this problem is to realize what makes up the bottommost number. This method doesn't work quite as well for this problem, but in a larger tree, it might be faster. Again, let the value in the empty box in the middle row be <math>x</math>, and the value in the empty box in the top row be <math>y</math>. <math>y</math> is the answer we're looking for.
−
 
 
−
Again, let the value in the empty box in the middle row be <math>x</math>, and the value in the empty box in the top row be <math>y</math>. <math>y</math> is the answer we're looking for.
 
  
 
<asy>
 
<asy>
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draw((1,2)--(1,4)--(3,4)--(3,2)--cycle);
 
draw((1,2)--(1,4)--(3,4)--(3,2)--cycle);
 
label("600",(0,-1));
 
label("600",(0,-1));
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label("$z$",(-1,1));
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label("30",(-1,1));
 
label("6",(-2,3));
 
label("6",(-2,3));
 
label("5",(0,3));
 
label("5",(0,3));
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600 = 150y\\
 
600 = 150y\\
 
\boxed{\textbf{(C)}\ 4} = y</math>
 
\boxed{\textbf{(C)}\ 4} = y</math>
 +
 +
==Video Solution==
 +
https://youtu.be/4tSvBcnv6dA ~savannahsolver
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2013|num-b=5|num-a=7}}
 
{{AMC8 box|year=2013|num-b=5|num-a=7}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 23:05, 1 February 2023

Problem

The number in each box below is the product of the numbers in the two boxes that touch it in the row above. For example, $30 = 6\times5$. What is the missing number in the top row?

[asy] unitsize(0.8cm); draw((-1,0)--(1,0)--(1,-2)--(-1,-2)--cycle); draw((-2,0)--(0,0)--(0,2)--(-2,2)--cycle); draw((0,0)--(2,0)--(2,2)--(0,2)--cycle); draw((-3,2)--(-1,2)--(-1,4)--(-3,4)--cycle); draw((-1,2)--(1,2)--(1,4)--(-1,4)--cycle); draw((1,2)--(1,4)--(3,4)--(3,2)--cycle); label("600",(0,-1)); label("30",(-1,1)); label("6",(-2,3)); label("5",(0,3)); [/asy]

$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 6$

Solution 1 (Working Backwards)

Let the value in the empty box in the middle row be $x$, and the value in the empty box in the top row be $y$. $y$ is the answer we're looking for.

[asy] unitsize(0.8cm); draw((-1,0)--(1,0)--(1,-2)--(-1,-2)--cycle); draw((-2,0)--(0,0)--(0,2)--(-2,2)--cycle); draw((0,0)--(2,0)--(2,2)--(0,2)--cycle); draw((-3,2)--(-1,2)--(-1,4)--(-3,4)--cycle); draw((-1,2)--(1,2)--(1,4)--(-1,4)--cycle); draw((1,2)--(1,4)--(3,4)--(3,2)--cycle); label("600",(0,-1)); label("30",(-1,1)); label("6",(-2,3)); label("5",(0,3)); label("$x$",(1,1)); label("$y$",(2,3)); [/asy]

From the diagram, $600 = 30x$, making $x = 20$.

[asy] unitsize(0.8cm); draw((-1,0)--(1,0)--(1,-2)--(-1,-2)--cycle); draw((-2,0)--(0,0)--(0,2)--(-2,2)--cycle); draw((0,0)--(2,0)--(2,2)--(0,2)--cycle); draw((-3,2)--(-1,2)--(-1,4)--(-3,4)--cycle); draw((-1,2)--(1,2)--(1,4)--(-1,4)--cycle); draw((1,2)--(1,4)--(3,4)--(3,2)--cycle); label("600",(0,-1)); label("30",(-1,1)); label("6",(-2,3)); label("5",(0,3)); label("20",(1,1)); label("$y$",(2,3)); [/asy]

It follows that $20 = 5y$, so $y = \boxed{\textbf{(C)}\ 4}$.

Solution 2 (Jumping to the Start)

Another way to do this problem is to realize what makes up the bottommost number. This method doesn't work quite as well for this problem, but in a larger tree, it might be faster. Again, let the value in the empty box in the middle row be $x$, and the value in the empty box in the top row be $y$. $y$ is the answer we're looking for.

[asy] unitsize(0.8cm); draw((-1,0)--(1,0)--(1,-2)--(-1,-2)--cycle); draw((-2,0)--(0,0)--(0,2)--(-2,2)--cycle); draw((0,0)--(2,0)--(2,2)--(0,2)--cycle); draw((-3,2)--(-1,2)--(-1,4)--(-3,4)--cycle); draw((-1,2)--(1,2)--(1,4)--(-1,4)--cycle); draw((1,2)--(1,4)--(3,4)--(3,2)--cycle); label("600",(0,-1)); label("30",(-1,1)); label("6",(-2,3)); label("5",(0,3)); label("$x$",(1,1)); label("$y$",(2,3)); [/asy]

We can write some equations:

$600 = 30x\\ 30 = 6\cdot 5\\ x = 5y$

Now we can substitute into the first equation using the two others:

$600 = (6\cdot5)(5y)\\ 600= 6\cdot5\cdot5\cdot y\\ 600=6\cdot25\cdot y\\ 600 = 150y\\ \boxed{\textbf{(C)}\ 4} = y$

Video Solution

https://youtu.be/4tSvBcnv6dA ~savannahsolver

See Also

2013 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
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All AJHSME/AMC 8 Problems and Solutions

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