Difference between revisions of "2012 AMC 8 Problems/Problem 17"

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A square with integer side length is cut into 10 squares, all of which have integer side length and at least 8 of which have area 1. What is the smallest possible value of the length of the side of the original square?
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==Problem==A square with integer side length is cut into 10 squares, all of which have integer side length and at least 8 of which have area 1. What is the smallest possible value of the length of the side of the original square?
  
 
<math> \textbf{(A)}\hspace{.05in}3\qquad\textbf{(B)}\hspace{.05in}4\qquad\textbf{(C)}\hspace{.05in}5\qquad\textbf{(D)}\hspace{.05in}6\qquad\textbf{(E)}\hspace{.05in}7 </math>
 
<math> \textbf{(A)}\hspace{.05in}3\qquad\textbf{(B)}\hspace{.05in}4\qquad\textbf{(C)}\hspace{.05in}5\qquad\textbf{(D)}\hspace{.05in}6\qquad\textbf{(E)}\hspace{.05in}7 </math>
  
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==Solution==
 
The first answer choice <math> {\textbf{(A)}\ 3} </math>, can be eliminated since there must be <math> 10 </math> squares with integer side lengths. We then test the next largest sidelength which is <math> 4 </math>. The square with area <math> 16 </math> can be partitioned into <math> 8 </math> squares with area <math> 1 </math> and two squares with area <math> 4 </math>, which satisfies all the conditions of the problem. Therefore, the smallest possible value of the length of the side of the original square is <math> \boxed{\textbf{(B)}\ 4} </math>.
 
The first answer choice <math> {\textbf{(A)}\ 3} </math>, can be eliminated since there must be <math> 10 </math> squares with integer side lengths. We then test the next largest sidelength which is <math> 4 </math>. The square with area <math> 16 </math> can be partitioned into <math> 8 </math> squares with area <math> 1 </math> and two squares with area <math> 4 </math>, which satisfies all the conditions of the problem. Therefore, the smallest possible value of the length of the side of the original square is <math> \boxed{\textbf{(B)}\ 4} </math>.
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2012|num-b=16|num-a=18}}
 
{{AMC8 box|year=2012|num-b=16|num-a=18}}

Revision as of 11:57, 9 December 2012

==Problem==A square with integer side length is cut into 10 squares, all of which have integer side length and at least 8 of which have area 1. What is the smallest possible value of the length of the side of the original square?

$\textbf{(A)}\hspace{.05in}3\qquad\textbf{(B)}\hspace{.05in}4\qquad\textbf{(C)}\hspace{.05in}5\qquad\textbf{(D)}\hspace{.05in}6\qquad\textbf{(E)}\hspace{.05in}7$

Solution

The first answer choice ${\textbf{(A)}\ 3}$, can be eliminated since there must be $10$ squares with integer side lengths. We then test the next largest sidelength which is $4$. The square with area $16$ can be partitioned into $8$ squares with area $1$ and two squares with area $4$, which satisfies all the conditions of the problem. Therefore, the smallest possible value of the length of the side of the original square is $\boxed{\textbf{(B)}\ 4}$.

See Also

2012 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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 AJHSME/AMC 8 Problems and Solutions
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