2001 AMC 10 Problems/Problem 22: Difference between revisions
Pidigits125 (talk | contribs) |
Pidigits125 (talk | contribs) |
||
| Line 134: | Line 134: | ||
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>. | ||
==See Also== | |||
{{AMC10 box|year=2001|num-b=21|num-a=23}} | |||
Revision as of 21:05, 16 March 2011
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
.
.
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 | ||