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

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<math> \mathrm{(A) \ } 12 \qquad \mathrm{(B) \ } 15 \qquad \mathrm{(C) \ } 17 \qquad \mathrm{(D) \ }  19\qquad \mathrm{(E) \ } 20  </math>
 
<math> \mathrm{(A) \ } 12 \qquad \mathrm{(B) \ } 15 \qquad \mathrm{(C) \ } 17 \qquad \mathrm{(D) \ }  19\qquad \mathrm{(E) \ } 20  </math>
  
==Solution==
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==Solution 1==
 
There are:
 
There are:
 
*<math>1</math> of the <math>1\times 1</math> squares containing the black square,
 
*<math>1</math> of the <math>1\times 1</math> squares containing the black square,
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Thus, the answer is <math>1+4+9+4+1=19\Rightarrow \boxed{\mathrm{(D)}}</math>.
 
Thus, the answer is <math>1+4+9+4+1=19\Rightarrow \boxed{\mathrm{(D)}}</math>.
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==Solution 2==
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We use complementary counting.
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There are only 2x2 and 1x1 squares that do not contain the black square.
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Counting, there are <math>12</math> 2x2, and <math>25-1 = 24</math> 1x1 squares that do not contain the black square.
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That gives <math>12+24=36</math> squares that don't contain it.
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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 do not contain the black square, which is <math>\boxed{\mathrm{(D)}}</math>.  .
  
 
==See also==
 
==See also==

Revision as of 17:52, 30 December 2010

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 $1+4+9+4+1=19\Rightarrow \boxed{\mathrm{(D)}}$.

Solution 2

We use complementary counting. There are only 2x2 and 1x1 squares that do not contain the black square. Counting, there are $12$ 2x2, and $25-1 = 24$ 1x1 squares that do not contain the black square. That gives $12+24=36$ squares that don't contain it. There are a total of $25+16+9+4+2+1 = 55$ squares possible, therefore there are $55-36 = 19$ squares that do not contain the black square, which is $\boxed{\mathrm{(D)}}$. .

See also

2004 AMC 10A (ProblemsAnswer KeyResources)
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