Difference between revisions of "2012 AMC 8 Problems/Problem 17"
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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? | + | ==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> | ||
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?
Solution
The first answer choice 
, can be eliminated since there must be 
squares with integer side lengths. We then test the next largest sidelength which is 
. The square with area 
can be partitioned into 
squares with area 
and two squares with area , which satisfies all the conditions of the problem. Therefore, the smallest possible value of the length of the side of the original square is 
.
See Also
| 2012 AMC 8 (Problems • Answer Key • Resources) | ||
| 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 | ||
