Difference between revisions of "2004 AMC 10A Problems/Problem 16"

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==Solution==
 
==Solution==
There is one 1-1 square containing the black square
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There are:
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*<math>1</math> of the <math>1\times 1</math> squares containing the black square,
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*<math>4</math> of the <math>2\times 2</math> squares containing the black square,
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*<math>9</math> of the <math>3\times 3</math> squares containing the black square,
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*<math>4</math> of the <math>4\times 4</math> squares containing the black square,
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*<math>1</math> of the <math>5\times 5</math> squares containing the black square.
  
There are 4 2-2 squares containing the black square
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Thus, the answer is <math>1+4+9+4+1=19\Rightarrow \boxed{\mathrm{(D)}}</math>.
 
 
There are 9 3-3 squares containing the black square
 
 
 
There are 4 4-4's.
 
 
 
There is 1 5-5.
 
 
 
<math>1+4+9+4+1=19\Rightarrow \boxed{\mathrm{(D) \ }}</math>
 
  
 
==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 $5\times 5$ grid shown contains a collection of squares with sizes from $1\times 1$ to $5\times 5$. How many of these squares contain the black center square?

2004 AMC 10A problem 16.png

$\mathrm{(A) \ } 12 \qquad \mathrm{(B) \ } 15 \qquad \mathrm{(C) \ } 17 \qquad \mathrm{(D) \ }  19\qquad \mathrm{(E) \ } 20$

Solution

There are:

  • $1$ of the $1\times 1$ squares containing the black square,
  • $4$ of the $2\times 2$ squares containing the black square,
  • $9$ of the $3\times 3$ squares containing the black square,
  • $4$ of the $4\times 4$ squares containing the black square,
  • $1$ of the $5\times 5$ squares containing the black square.

Thus, the answer is $1+4+9+4+1=19\Rightarrow \boxed{\mathrm{(D)}}$.

See also

2004 AMC 10A (ProblemsAnswer KeyResources)
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