Difference between revisions of "2012 AMC 10A Problems/Problem 14"
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== Solution 1== | == Solution 1== | ||
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> | 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> | ||
| + | |||
| + | Note: When solving <math>16^2</math>+<math>15^2</math>, you only need to calculate the units digit. | ||
==Solution 2== | ==Solution 2== | ||
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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> | In total, there are <math>15 \cdot 30 + 15 + 15 + 1 = 481</math> tiles, giving an answer of <math>\boxed{\textbf{(B)}\ 481}</math> | ||
| + | |||
| + | ==Solution 3== | ||
| + | |||
| + | For every <math>2 \times 31</math> strip, there are 31 black tiles and 31 white tiles. There are 15 such strips in total and another <math>1 \times 31</math> strip having 16 black titles. | ||
| + | |||
| + | In total, there are <math>31\times 15+16=481</math> black tiles. | ||
| + | |||
| + | ~Bran_Quin | ||
| + | |||
| + | ==Solution 4== | ||
| + | |||
| + | Drawing smaller scale sketches, we notice that the odd columns of an <math>n \times n</math> (where <math>n</math> is odd) board have <math>\left \lceil \frac{n}{2} \right \rceil</math> black tiles, while the even columns have <math>\left \lfloor \frac{n}{2} \right \rfloor</math> black tiles. In our case, we have a <math>31 \times 31</math> board. We have <math>16</math> odd columns, and <math>15</math> even columns, so the number of black tiles in total is <math>16^2 + 15^2 = \boxed{\textbf{(B)}\ 481}</math>. | ||
| + | |||
| + | ~kn07 | ||
== 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}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 17:40, 27 November 2022
Problem
Chubby makes nonstandard checkerboards that have 
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?
Solution 1
There are 15 rows with 15 black tiles, and 16 rows with 16 black tiles, so the answer is 
Note: When solving 
+
, you only need to calculate the units digit.
Solution 2
We build the 
checkerboard starting with a board of 
that is exactly half black. There are 
black tiles in this region.
Add to this checkerboard a 
strip on the bottom that has 
black tiles.
Add to this 
checkerboard a 
strip on the right that has 
black tiles.
In total, there are 
tiles, giving an answer of 
Solution 3
For every 
strip, there are 31 black tiles and 31 white tiles. There are 15 such strips in total and another 
strip having 16 black titles.
In total, there are 
black tiles.
~Bran_Quin
Solution 4
Drawing smaller scale sketches, we notice that the odd columns of an 
(where 
is odd) board have 
black tiles, while the even columns have 
black tiles. In our case, we have a board. We have 
odd columns, and even columns, so the number of black tiles in total is 
.
~kn07
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
