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 |