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

(Solution 1)
(Solution 3)
(7 intermediate revisions by 6 users not shown)
Line 7: Line 7:
 
== 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: One getting <math>16^2</math>+<math>15^2</math> you only need to calculate the units digit
+
 
 +
Note: When solving <math>16^2</math>+<math>15^2</math>, you only need to calculate the units digit.
  
 
==Solution 2==
 
==Solution 2==
Line 18: Line 19:
  
 
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_Qin
  
 
== See Also ==
 
== See Also ==

Revision as of 23:25, 7 August 2021

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: When solving $16^2$+$15^2$, 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 $\boxed{\textbf{(B)}\ 481}$

Solution 3

For every $2 \times 31$ strip, there are 31 black tiles and 31 white tiles. There are 15 such strips in total and another $1 \times 31$ strip having 16 black titles.

In total, there are $31\times 15+16=481$ black tiles.

~Bran_Qin

See Also

2012 AMC 10A (ProblemsAnswer KeyResources)
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. AMC logo.png