Difference between revisions of "2012 AMC 10A Problems/Problem 14"

(Solution)
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<math> \textbf{(A)}\ 480 \qquad\textbf{(B)}\ 481 \qquad\textbf{(C)}\ 482 \qquad\textbf{(D)}\ 483 \qquad\textbf{(E)}\ 484</math>
 
<math> \textbf{(A)}\ 480 \qquad\textbf{(B)}\ 481 \qquad\textbf{(C)}\ 482 \qquad\textbf{(D)}\ 483 \qquad\textbf{(E)}\ 484</math>
  
== Solution ==
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== Solution 1==
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There are 15 rows with 15 black tiles, and 16 rows with 16 black tiles, so the answer is <math>15^2+16^2 =225+256= \boxed{\textbf{(B)}\ 481}</math>
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==Solution 2==
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We build the <math>31 \times 31</math> checkerboard starting with a board of <math>30 \times 30</math> that is exactly half black.  There are <math>15 \cdot 30</math> black tiles in this region.
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Add to this <math>30 \times 30</math> checkerboard a <math>1 \times 30</math> strip on the bottom that has <math>15</math> black tiles.
  
There are 15 rows with 15 black tiles, and 16 rows with 16 black tiles, so the answer is <math>15^2+16^2 =225+256= \boxed{\textbf{(B)}\ 481}</math>
+
Add to this <math>31 \times 30</math> checkerboard a <math>31 \times 1</math> strip on the right that has <math>15 + 1</math> black tiles.
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In total, there are <math>15 \cdot 30 + 15 + 15 + 1 = 481</math> tiles, giving an answer of <math>\boxed{\textbf{(B)}\ 481}</math>
  
 
== See Also ==
 
== See Also ==
  
 
{{AMC10 box|year=2012|ab=A|num-b=13|num-a=15}}
 
{{AMC10 box|year=2012|ab=A|num-b=13|num-a=15}}

Revision as of 00:31, 9 February 2012

Problem

Chubby makes nonstandard checkerboards that have $31$ squares on each side. The checkerboards have a black square in every corner and alternate red and black squares along every row and column. How many black squares are there on such a checkerboard?

$\textbf{(A)}\ 480 \qquad\textbf{(B)}\ 481 \qquad\textbf{(C)}\ 482 \qquad\textbf{(D)}\ 483 \qquad\textbf{(E)}\ 484$

Solution 1

There are 15 rows with 15 black tiles, and 16 rows with 16 black tiles, so the answer is $15^2+16^2 =225+256= \boxed{\textbf{(B)}\ 481}$

Solution 2

We build the $31 \times 31$ checkerboard starting with a board of $30 \times 30$ that is exactly half black. There are $15 \cdot 30$ black tiles in this region.

Add to this $30 \times 30$ checkerboard a $1 \times 30$ strip on the bottom that has $15$ black tiles.

Add to this $31 \times 30$ checkerboard a $31 \times 1$ strip on the right that has $15 + 1$ black tiles.

In total, there are $15 \cdot 30 + 15 + 15 + 1 = 481$ tiles, giving an answer of $\boxed{\textbf{(B)}\ 481}$

See Also

2012 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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All AMC 10 Problems and Solutions
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