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

Revision as of 04:31, 4 November 2022

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?

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

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 Solution by OmegaLearn

https://youtu.be/HhdpuJt78Hg?t=168

~ pi_is_3.14

@@ Line 22: / Line 22: @@
 ===Solution 2===
 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>.
+== Video Solution by OmegaLearn ==
+https://youtu.be/HhdpuJt78Hg?t=168
+~ pi_is_3.14
 == Video Solutions ==

2004 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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