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

(New page: The answer is 19 or (D) since the number of ways of arranging squares 1x1 through 3x3 are squares( as in power of degree) of their sides. As for the 4x4 and 5x5, it's easy to find the few ...)
 
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The answer is 19 or (D) since the number of ways of arranging squares 1x1 through 3x3 are squares( as in power of degree) of their sides. As for the 4x4 and 5x5, it's easy to find the few ways visually.
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==Problem==
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The <math>5\times 5</math> grid shown contains a collection of squares with sizes from <math>1\times 1</math> to <math>5\times 5</math>. How many of these squares contain the black center square?
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[[Image:2004 AMC 10A problem 16.png]]
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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>
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==Solution==
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===Solution 1===
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Since there are five types of squares: <math>1 \times 1, 2 \times 2, 3 \times 3, 4 \times 4,</math> and <math>5 \times 5.</math> We must find how many of each square contain the black shaded square in the center.
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If we list them, we get that
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*There is <math>1</math> of all <math>1\times 1</math> squares, containing the black square
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*There are <math>4</math> of all <math>2\times 2</math> squares, containing the black square
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*There are <math>9</math> of all <math>3\times 3</math> squares, containing the black square
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*There are <math>4</math> of all <math>4\times 4</math> squares, containing the black square
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*There is <math>1</math> of all <math>5\times 5</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)}\ 19}</math>.
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===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> squares, 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 <math>(25</math> - <math>1\times1</math> squares <math>16</math> - <math>2\times2</math> squares <math>9</math> - <math>3\times3</math> squares <math>4</math> - <math>4\times4</math> squares and <math>1</math> - <math>5\times5</math> square), therefore there are <math>55-36 = 19</math> squares that contain the black square, which is <math>\boxed{\mathrm{(D)}\ 19}</math>.
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== Video Solutions ==
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*https://youtu.be/0W3VmFp55cM?t=4697
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*https://youtu.be/aMmF6jz6xA4
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==See also==
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{{AMC10 box|year=2004|ab=A|num-b=15|num-a=17}}
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[[Category:Introductory Combinatorics Problems]]
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{{MAA Notice}}

Revision as of 23:44, 16 January 2021

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

Solution 1

Since there are five types of squares: $1 \times 1, 2 \times 2, 3 \times 3, 4 \times 4,$ and $5 \times 5.$ We must find how many of each square contain the black shaded square in the center.

If we list them, we get that

  • There is $1$ of all $1\times 1$ squares, containing the black square
  • There are $4$ of all $2\times 2$ squares, containing the black square
  • There are $9$ of all $3\times 3$ squares, containing the black square
  • There are $4$ of all $4\times 4$ squares, containing the black square
  • There is $1$ of all $5\times 5$ 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 $12$-$2\times2$ squares, 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 $(25$ - $1\times1$ squares $16$ - $2\times2$ squares $9$ - $3\times3$ squares $4$ - $4\times4$ squares and $1$ - $5\times5$ square), therefore there are $55-36 = 19$ squares that contain the black square, which is $\boxed{\mathrm{(D)}\ 19}$.

Video Solutions

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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