Difference between revisions of "2004 AMC 10A Problems/Problem 16"
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==Solution== | ==Solution== | ||
− | There | + | There are: |
+ | *<math>1</math> of the <math>1\times 1</math> squares containing the black square, | ||
+ | *<math>4</math> of the <math>2\times 2</math> squares containing the black square, | ||
+ | *<math>9</math> of the <math>3\times 3</math> squares containing the black square, | ||
+ | *<math>4</math> of the <math>4\times 4</math> squares containing the black square, | ||
+ | *<math>1</math> of the <math>5\times 5</math> squares containing the black square. | ||
− | + | Thus, the answer is <math>1+4+9+4+1=19\Rightarrow \boxed{\mathrm{(D)}}</math>. | |
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− | <math>1+4+9+4+1=19\Rightarrow \boxed{\mathrm{(D) | ||
==See also== | ==See also== | ||
{{AMC10 box|year=2004|ab=A|num-b=15|num-a=17}} | {{AMC10 box|year=2004|ab=A|num-b=15|num-a=17}} | ||
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+ | [[Category:Introductory Combinatorics Problems]] |
Revision as of 16:43, 15 April 2008
Problem
The grid shown contains a collection of squares with sizes from to . How many of these squares contain the black center square?
Solution
There are:
- of the squares containing the black square,
- of the squares containing the black square,
- of the squares containing the black square,
- of the squares containing the black square,
- of the squares containing the black square.
Thus, the answer is .
See also
2004 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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 |