Difference between revisions of "2012 AMC 10A Problems/Problem 14"
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==Solution 3== | ==Solution 3== | ||
− | For every <math>2 \times 31</math> strip, there are 31 black tiles and 31 white tiles. There are 15 | + | 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. |
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In total, there are <math>31\times 15+16=481</math> black tiles. | In total, there are <math>31\times 15+16=481</math> black tiles. |
Revision as of 22:25, 7 August 2021
Contents
[hide]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_Qin
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.