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Difference between revisions of "2012 AMC 10A Problems/Problem 14"
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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>
  
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== 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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Note: On getting <math>16^2</math>+<math>15^2</math> you only need to calculate the units digit
 +
 
 +
==Solution 2==
  
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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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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.
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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}}
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{{MAA Notice}}

Revision as of 22:52, 21 November 2020

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 $15^2+16^2 =225+256= \boxed{\textbf{(B)}\ 481}$

Note: On getting $16^2$+$15^2$ you only need to calculate the units digit

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 (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
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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