Difference between revisions of "2001 AMC 10 Problems/Problem 22"

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Line 150: Line 150:
 
==Systems of Equations==
 
==Systems of Equations==
 
Create an equation for every row, column, and diagonal. Let <math>e</math> be the sum of the rows, columns, and diagonals.  
 
Create an equation for every row, column, and diagonal. Let <math>e</math> be the sum of the rows, columns, and diagonals.  
<math>w+v+24=e</math>
+
<cmath>w+v+24=e</cmath>
<math>x+y+18=e</math>
+
<cmath>x+y+18=e</cmath>
<math>z+46=e</math>
+
<cmath>z+46=e</cmath>
<math>v+43=e</math>
+
<cmath>v+43=e</cmath>
<math>x+z+24=e</math>
+
<cmath>x+z+24=e</cmath>
<math>w+y+21=e</math>
+
<cmath>w+y+21=e</cmath>
<math>x+w+25=e</math>
+
<cmath>x+w+25=e</cmath>
<math>x+v+21=e</math>.
+
<cmath>x+v+21=e</cmath>.
  
 
Notice that <math>z+46=e</math> and <math>x+z+24=e</math> both have <math>z</math>. Equate them and you get that <math>x=22</math>.
 
Notice that <math>z+46=e</math> and <math>x+z+24=e</math> both have <math>z</math>. Equate them and you get that <math>x=22</math>.

Revision as of 02:38, 17 June 2021

Problem

In the magic square shown, the sums of the numbers in each row, column, and diagonal are the same. Five of these numbers are represented by $v$, $w$, $x$, $y$, and $z$. Find $y + z$.

[asy] unitsize(10mm); defaultpen(linewidth(1pt)); for(int i=0; i<=3; ++i) { draw((0,i)--(3,i)); draw((i,0)--(i,3)); } label("$25$",(0.5,0.5)); label("$z$",(1.5,0.5)); label("$21$",(2.5,0.5)); label("$18$",(0.5,1.5)); label("$x$",(1.5,1.5)); label("$y$",(2.5,1.5)); label("$v$",(0.5,2.5)); label("$24$",(1.5,2.5)); label("$w$",(2.5,2.5));[/asy]

$\textbf{(A)}\ 43 \qquad \textbf{(B)}\ 44 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 46 \qquad \textbf{(E)}\ 47$


Solutions

Solution 1

We know that $y+z=2v$, so we could find one variable rather than two.

$v+24+w=43+v$

$24+w=43$

$w=19$

[asy] unitsize(10mm); defaultpen(linewidth(1pt)); for(int i=0; i<=3; ++i) { draw((0,i)--(3,i)); draw((i,0)--(i,3)); } label("$25$",(0.5,0.5)); label("$z$",(1.5,0.5)); label("$21$",(2.5,0.5)); label("$18$",(0.5,1.5)); label("$x$",(1.5,1.5)); label("$y$",(2.5,1.5)); label("$v$",(0.5,2.5)); label("$24$",(1.5,2.5)); label("$19$",(2.5,2.5));[/asy]

$44+x=24+x+z \implies z=20$

[asy] unitsize(10mm); defaultpen(linewidth(1pt)); for(int i=0; i<=3; ++i) { draw((0,i)--(3,i)); draw((i,0)--(i,3)); } label("$25$",(0.5,0.5)); label("$20$",(1.5,0.5)); label("$21$",(2.5,0.5)); label("$18$",(0.5,1.5)); label("$x$",(1.5,1.5)); label("$y$",(2.5,1.5)); label("$v$",(0.5,2.5)); label("$24$",(1.5,2.5)); label("$19$",(2.5,2.5));[/asy]

The sum per row is $25+21+20=66$.

Thus $66-18-25=66-43=v=23$.

Since we needed $2v$ and we know $v=23$, $23 \times 2 = \boxed{\textbf{(D)}\ 46}$.

Solution 2

$v+24+w=43+v$

$24+w=43$

$w=19$

[asy] unitsize(10mm); defaultpen(linewidth(1pt)); for(int i=0; i<=3; ++i) { draw((0,i)--(3,i)); draw((i,0)--(i,3)); } label("$25$",(0.5,0.5)); label("$z$",(1.5,0.5)); label("$21$",(2.5,0.5)); label("$18$",(0.5,1.5)); label("$x$",(1.5,1.5)); label("$y$",(2.5,1.5)); label("$v$",(0.5,2.5)); label("$24$",(1.5,2.5)); label("$19$",(2.5,2.5));[/asy]

$44+x=24+x+z \implies z=20$

[asy] unitsize(10mm); defaultpen(linewidth(1pt)); for(int i=0; i<=3; ++i) { draw((0,i)--(3,i)); draw((i,0)--(i,3)); } label("$25$",(0.5,0.5)); label("$20$",(1.5,0.5)); label("$21$",(2.5,0.5)); label("$18$",(0.5,1.5)); label("$x$",(1.5,1.5)); label("$y$",(2.5,1.5)); label("$v$",(0.5,2.5)); label("$24$",(1.5,2.5)); label("$19$",(2.5,2.5));[/asy]

The magic sum is determined by the bottom row. $25+20+21=66$.

Solving for $y$:

$y=66-19-21=66-40=26$.

To find our answer, we need to find $y+z$. $y+z=20+26 = \boxed{\textbf{(D)}\ 46}$.

Really Easy Solution

A nice thing to know is that any $3$ numbers that go through the middle form an arithmetic sequence.

Using this, we know that $x=(24+z)/2$, or $2x=24+z$ because $x$ would be the average.

We also know that because $x$ is the average the magic sum would be $3x$, so we can also write the equation $3x-46=z$ using the bottom row.

Solving for x in this system we get $x=22$, so now using the arithmetic sequence knowledge we find that $y=26$ and $z=20$.

Adding these we get $\boxed{\textbf{(D)}\ 46}$.


-harsha12345

Systems of Equations

Create an equation for every row, column, and diagonal. Let $e$ be the sum of the rows, columns, and diagonals. \[w+v+24=e\] \[x+y+18=e\] \[z+46=e\] \[v+43=e\] \[x+z+24=e\] \[w+y+21=e\] \[x+w+25=e\] \[x+v+21=e\].

Notice that $z+46=e$ and $x+z+24=e$ both have $z$. Equate them and you get that $x=22$. Using that same strategy, we use $v+43=e$ instead. $w+v+24=e$ is good for our purposes. It turns out that $w=19$. Since we already know those numbers, and $x+w+25=e$, We can say that $e$ will be $66$. We are now able to solve: $x+z+24=e$, $w+y+21=e$, $x+v+21=e$, and $x+y+18=e$. Respectively, $v=23$, $w=19$, $x=22$, $y=26$, and $z=20$. We only require The sum of $y+z$, which is $26+20=46$. We get that the sum of $y$ and $z$ respectively is $\boxed{\textbf{(D)}\ 4}$

-OofPirate

Video Solution

https://youtu.be/9guPi81LgfM

~savannahsolver

See Also

2001 AMC 10 (ProblemsAnswer KeyResources)
Preceded by
Problem 21
Followed by
Problem 23
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

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