2001 AMC 10 Problems/Problem 22

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

Video Solution

https://youtu.be/9guPi81LgfM

~savannahsolver

2001 AMC 10 (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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