Difference between revisions of "2004 AMC 10A Problems/Problem 16"
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==Solution 2== | ==Solution 2== | ||
− | We use complementary counting. | + | We use complementary counting. There are only <math>2\times2</math> and <math>1\times1</math> squares that do not contain the black square. Counting, there are <math>12</math> <math>2\times2</math>, and <math>25-1 = 24</math> <math>1\times1</math> squares that do not contain the black square. That gives <math>12+24=36</math> squares that don't contain it. There are a total of <math>25+16+9+4+2+1 = 55</math> squares possible, therefore there are <math>55-36 = 19</math> squares that contains the black square, which is <math>\boxed{\mathrm{(D)}\ 19}</math>. |
− | There are only | ||
− | Counting, there are <math>12</math> | ||
− | That gives <math>12+24=36</math> squares that don't contain it. | ||
− | There are a total of <math>25+16+9+4+2+1 = 55</math> squares possible, therefore there are <math>55-36 = 19</math> squares that | ||
==See also== | ==See also== |
Revision as of 11:28, 21 July 2014
Contents
Problem
The grid shown contains a collection of squares with sizes from
to
. How many of these squares contain the black center square?
Solution 1
There are:
of the
squares containing the black square,
of the
squares containing the black square,
of the
squares containing the black square,
of the
squares containing the black square,
of the
squares containing the black square.
Thus, the answer is .
Solution 2
We use complementary counting. There are only and
squares that do not contain the black square. Counting, there are
, and
squares that do not contain the black square. That gives
squares that don't contain it. There are a total of
squares possible, therefore there are
squares that contains the black square, which is
.
See also
2004 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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.