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 === | ||
| + | any 3 numbers that goes through the middle forms an arithmetic sequence, | ||
| + | through this we know that x=24+z/2 or 2x=24+z because x would be the average | ||
| + | we also know that the because x is the average the magic sum would be 3x, so we can also write the equation 3x-46=z using hte bottom row | ||
| + | solving for x in this system we get 22, so know using the arithmetic sequence knowledge we find that y=26 and z=20. | ||
==See Also== | ==See Also== | ||
Revision as of 22:58, 29 December 2018
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
any 3 numbers that goes through the middle forms an arithmetic sequence, through this we know that x=24+z/2 or 2x=24+z because x would be the average we also know that the because x is the average the magic sum would be 3x, so we can also write the equation 3x-46=z using hte bottom row solving for x in this system we get 22, so know using the arithmetic sequence knowledge we find that y=26 and z=20.
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. 
