2004 AMC 10A Problems/Problem 16

Revision as of 10:28, 21 July 2014 by Hesa57 (talk | contribs) (Solution 2)

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 $25+16+9+4+2+1 = 55$ 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 (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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All AMC 10 Problems and Solutions

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