Difference between revisions of "2001 AMC 10 Problems/Problem 22"
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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>. | ||
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>. | 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)}\ | + | We get that the sum of <math>y</math> and <math>z</math> respectively is <math>\boxed{\textbf{(D)}\ 46}</math> |
-OofPirate | -OofPirate |
Revision as of 12:44, 17 June 2021
Contents
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 .
Solutions
Solution 1
We know that , so we could find one variable rather than two.
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 go through the middle form 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
Systems of Equations
Create an equation for every row, column, and diagonal. Let be the sum of the rows, columns, and diagonals. .
Notice that and both have . Equate them and you get that . Using that same strategy, we use instead. is good for our purposes. It turns out that . Since we already know those numbers, and , We can say that will be . We are now able to solve: , , , and . Respectively, , , , , and . We only require The sum of , which is . We get that the sum of and respectively is
-OofPirate
Video Solution
~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.