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

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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 <math> v </math>, <math> w </math>, <math> x </math>, <math> y </math>, and <math> z </math>. Find <math> y + z </math>.
 
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 <math> v </math>, <math> w </math>, <math> x </math>, <math> y </math>, and <math> z </math>. Find <math> y + z </math>.
−
 
−
<math> \textbf{(A)}\ 43 \qquad \textbf{(B)}\ 44 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 46 \qquad \textbf{(E)}\ 47 </math>
 
  
 
<asy>
 
<asy>
Line 22: Line 20:
 
label("$24$",(1.5,2.5));
 
label("$24$",(1.5,2.5));
 
label("$w$",(2.5,2.5));</asy>
 
label("$w$",(2.5,2.5));</asy>
 +
 +
<math> \textbf{(A)}\ 43 \qquad \textbf{(B)}\ 44 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 46 \qquad \textbf{(E)}\ 47 </math>
 +
  
 
==Solutions==
 
==Solutions==
 +
 +
==Video solution 1==
 +
 +
https://www.youtube.com/watch?v=-v6vCwJAGtI
 +
 +
-DaBob
  
 
===Solution 1===
 
===Solution 1===
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=== Really Easy Solution ===
 
=== Really Easy Solution ===
−
A nice thing to know is that any <math>3</math> numbers that goes through the middle forms an arithmetic sequence.
+
A nice thing to know is that any <math>3</math> numbers that go through the middle form an arithmetic sequence.
  
−
Using this, we know that <math>x=24+z/2</math>, or <math>2x=24+z</math> because <math>x</math> would be the average.
+
Using this, we know that <math>x=(24+z)/2</math>, or <math>2x=24+z</math> because <math>x</math> would be the average.
  
 
We also know that because <math>x</math> is the average the magic sum would be <math>3x</math>, so we can also write the equation <math>3x-46=z</math> using the bottom row.
 
We also know that because <math>x</math> is the average the magic sum would be <math>3x</math>, so we can also write the equation <math>3x-46=z</math> using the bottom row.
Line 142: Line 149:
 
Solving for x in this system we get <math>x=22</math>, so now using the arithmetic sequence knowledge we find that <math>y=26</math> and <math>z=20</math>.
 
Solving for x in this system we get <math>x=22</math>, so now using the arithmetic sequence knowledge we find that <math>y=26</math> and <math>z=20</math>.
  
−
Adding these we get <math>/boxed{46}</math>.
+
Adding these we get <math>\boxed{\textbf{(D)}\ 46}</math>.
  
  
 
-harsha12345
 
-harsha12345
 +
 +
==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.
 +
<cmath>w+v+24=e</cmath>
 +
<cmath>x+y+18=e</cmath>
 +
<cmath>z+46=e</cmath>
 +
<cmath>v+43=e</cmath>
 +
<cmath>x+z+24=e</cmath>
 +
<cmath>w+y+21=e</cmath>
 +
<cmath>x+w+25=e</cmath>
 +
<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>.
 +
Using that same strategy, we use <math>v+43=e</math> instead. <math>w+v+24=e</math> is good for our purposes. It turns out that <math>w=19</math>. Since we already know those numbers, and <math>x+w+25=e</math>, We can say that <math>e</math> will be <math>66</math>. We are now able to solve: <math>x+z+24=e</math>, <math>w+y+21=e</math>, <math>x+v+21=e</math>, and <math>x+y+18=e</math>. Respectively, <math>v=23</math>, <math>w=19</math>, <math>x=22</math>, <math>y=26</math>, and <math>z=20</math>. We only require The sum of <math>y+z</math>, which is <math>26+20=46</math>.
 +
We get that the sum of <math>y</math> and <math>z</math> respectively is <math>\boxed{\textbf{(D)}\ 46}</math>
 +
 +
-OofPirate
 +
 +
==Video Solution 2==
 +
https://youtu.be/9guPi81LgfM
 +
 +
~savannahsolver
  
 
==See Also==
 
==See Also==

Latest revision as of 15:44, 9 August 2022

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

Video solution 1

https://www.youtube.com/watch?v=-v6vCwJAGtI

-DaBob

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 $\boxed{\textbf{(D)}\ 46}$

-OofPirate

Video Solution 2

https://youtu.be/9guPi81LgfM

~savannahsolver

See Also

2001 AMC 10 (Problems • Answer Key • Resources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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