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? | 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> \ | + | <math> \text{(A)}\hspace{.05in}3\qquad\text{(B)}\hspace{.05in}4\qquad\text{(C)}\hspace{.05in}5\qquad\text{(D)}\hspace{.05in}6\qquad\text{(E)}\hspace{.05in}7 </math> |
+ | |||
+ | ==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 smallest 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== | ||
+ | {{AMC8 box|year=2012|num-b=16|num-a=18}} | ||
+ | {{MAA Notice}} |
Latest revision as of 17:46, 16 January 2021
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 smallest 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.