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

Latest revision as of 00:37, 14 August 2024

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?

$[asy] for (int i=0; i<5; ++i) { for (int j=0; j<5; ++j) { draw((-2.5+i, -2.5+j)--(-1.5+i, -2.5+j) -- (-1.5+i, -1.5+j) -- (-2.5+i, -1.5+j)--cycle); } fill((-0.5,-0.5)--(-0.5, 0.5)--(0.5,0.5) -- (0.5,-0.5)--cycle, black); } [/asy]$

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

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 ~VictorZhang

@@ Line 1: / Line 1: @@
 ==Problem==
 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?
-[[Image:2004 AMC 10A problem 16.png]]
+<asy>
+for (int i=0; i<5; ++i) {
+ for (int j=0; j<5; ++j) {
+   draw((-2.5+i, -2.5+j)--(-1.5+i, -2.5+j) -- (-1.5+i, -1.5+j) -- (-2.5+i, -1.5+j)--cycle);
+ }
+ fill((-0.5,-0.5)--(-0.5, 0.5)--(0.5,0.5) -- (0.5,-0.5)--cycle, black);
+}
+</asy>
 <math> \mathrm{(A) \ } 12 \qquad \mathrm{(B) \ } 15 \qquad \mathrm{(C) \ } 17 \qquad \mathrm{(D) \ }  19\qquad \mathrm{(E) \ } 20  </math>
-==Solution 1==
+==Solution==
-There are:
+===Solution 1===
-*<math>1</math> of the <math>1\times 1</math> squares containing the black square,
+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.
-*<math>4</math> of the <math>2\times 2</math> squares containing the black square,
-*<math>9</math> of the <math>3\times 3</math> squares containing the black square,
+If we list them, we get that
-*<math>4</math> of the <math>4\times 4</math> squares containing the black square,
+*There is <math>1</math> of all <math>1\times 1</math> squares, containing the black square
-*<math>1</math> of the <math>5\times 5</math> squares containing the black square.
+*There are <math>4</math> of all <math>2\times 2</math> squares, containing the black square
+*There are <math>9</math> of all <math>3\times 3</math> squares, containing the black square
+*There are <math>4</math> of all <math>4\times 4</math> squares, containing the black square
+*There is <math>1</math> of all <math>5\times 5</math> squares, containing the black square
+Thus, the answer is <math>1+4+9+4+1=19\Rightarrow\boxed{\mathrm{(D)}\ 19}</math>.
+===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
-Thus, the answer is <math>1+4+9+4+1=19\Rightarrow \boxed{\mathrm{(D)}}</math>.
+~ pi_is_3.14
+~VictorZhang
-==Solution 2==
+== Video Solutions ==
-We use complementary counting.
+*https://youtu.be/0W3VmFp55cM?t=4697
-There are only 2x2 and 1x1 squares that do not contain the black square.
+*https://youtu.be/aMmF6jz6xA4
-Counting, there are <math>12</math> 2x2, and <math>25-1 = 24</math> 1x1 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 do not contain the black square, which is <math>\boxed{\mathrm{(D)}}</math>.   .
 ==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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