Difference between revisions of "2001 AMC 10 Problems/Problem 22"
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To find our answer, we need to find <math> y+z </math>. <math> y+z=20+26 = \boxed{\textbf{(D)}\ 46} </math>. | To find our answer, we need to find <math> y+z </math>. <math> y+z=20+26 = \boxed{\textbf{(D)}\ 46} </math>. | ||
| − | === | + | === Really Easy Solution === |
| − | + | A nice thing to know is that any <math>3</math> numbers that goes through the middle forms 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. | |
| − | + | 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. | |
| − | + | 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>. | |
Revision as of 07:58, 2 April 2019
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 
, 
, 
, 
, and 
. Find 
.
![[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]](http://latex.artofproblemsolving.com/4/e/2/4e24d4a82884d1873aa08b1502e6996d8292e7ed.png)
Solutions
Solution 1
We know that 
, so we could find one variable rather than two.
![[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]](http://latex.artofproblemsolving.com/a/c/9/ac93cbfe3fce005ca5ee9103f9f203476e189db6.png)
![[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]](http://latex.artofproblemsolving.com/1/2/6/1261087702926461e26052453481b55bb6c07a10.png)
The sum per row is 
.
Thus 
.
Since we needed 
and we know 
, 
.
Solution 2
The magic sum is determined by the bottom row. 
.
Solving for :
.
To find our answer, we need to find 
. 
.
Really Easy Solution
A nice thing to know is that any 
numbers that goes through the middle forms an arithmetic sequence.
Using this, we know that 
, or 
because would be the average.
We also know that because is the average the magic sum would be 
, so we can also write the equation 
using the bottom row.
Solving for x in this system we get 
, so now using the arithmetic sequence knowledge we find that 
and 
.
Adding these we get 
.
-harsha12345
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. 
