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Difference between revisions of "2004 AMC 10A Problems/Problem 16"

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==Solution 2==
 
==Solution 2==
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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>.
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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+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>.
  
 
==See also==
 
==See also==

Revision as of 13:14, 27 December 2015

Problem

The $5\times 5$ grid shown contains a collection of squares with sizes from $1\times 1$ to $5\times 5$. How many of these squares contain the black center square?

2004 AMC 10A problem 16.png

$\mathrm{(A) \ } 12 \qquad \mathrm{(B) \ } 15 \qquad \mathrm{(C) \ } 17 \qquad \mathrm{(D) \ } 19\qquad \mathrm{(E) \ } 20$

Solution 1

There are:

  • $1$ of the $1\times 1$ squares containing the black square,
  • $4$ of the $2\times 2$ squares containing the black square,
  • $9$ of the $3\times 3$ squares containing the black square,
  • $4$ of the $4\times 4$ squares containing the black square,
  • $1$ of the $5\times 5$ squares containing the black square.

Thus, the answer is $1+4+9+4+1=19\Rightarrow\boxed{\mathrm{(D)}\ 19}$.

Solution 2

We use complementary counting. There are only $2\times2$ and $1\times1$ squares that do not contain the black square. Counting, there are $12$ $2\times2$, and $25-1 = 24$ $1\times1$ squares that do not contain the black square. That gives $12+24=36$ squares that don't contain it. There are a total of $25+16+9+4+1 = 55$ squares possible, therefore there are $55-36 = 19$ squares that contains the black square, which is $\boxed{\mathrm{(D)}\ 19}$.

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

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