Difference between revisions of "1998 AHSME Problems/Problem 1"
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==Solution== | ==Solution== | ||
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| + | Looking at the list of <math>w</math> and <math>y</math> values that must match up left-to-right, we have <math>(w,y) = A(4,6), B(1,3), C(3,5), D(7,4), E(9,7)</math>. Looking for digits that only appear once, we see that <math>9</math>, <math>6</math>, <math>1</math>, and <math>5</math> cannot match up to other digits, and thus must appear on the ends. <math>1</math> and <math>9</math> only appear on the left, and thus their respective blocks <math>B</math> and <math>E</math> must appear on the left. Similarly, <math>6</math> and <math>5</math> only appear on the right, and thus their blocks <math>A</math> and <math>C</math> must appear on the right-most block of their row. Therefore, <math>D</math>, the only block without a unique digit, must be the top-center block. | ||
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| + | Now that we have placed one block, <math>D</math>, we can only place block <math>E</math> to the left of <math>D</math>, and block <math>A</math> to the right of <math>D</math>. Thus, <math>\boxed{E}</math> is the right answer. | ||
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| + | Completing the puzzle, the top boxes read <math>EDA</math>, while the bottom two boxes read <math>BC</math>. | ||
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==See Also== | ==See Also== | ||
| − | {{AHSME box|year=1998|before=First Problem|num-a= | + | {{AHSME box|year=1998|before=First Problem|num-a=2}} |
| + | {{MAA Notice}} | ||
Latest revision as of 13:28, 5 July 2013
Problem 1
Each of the sides of five congruent rectangles is labeled with an integer. In rectangle A, 
. In rectangle B, 
. In rectangle C, 
. In rectangle D, 
. In rectangle E, 
. These five rectangles are placed, without rotating or reflecting, in position as below. Which of the rectangle is the top leftmost one?
![[asy] draw((0,5)--(0,7)--(3,7)--(3,5)--cycle); draw((0,4)--(9,4)--(9,2)--(6,2)--(6,0)--(0,0)--cycle); draw((3,0)--(3,4));draw((6,2)--(6,4));draw((0,2)--(6,2)); label("$w$",(0,6),(-1,0));label("$x$",(1.5,7),(0,1));label("$y$",(3,6),(1,0));label("$z$",(1.5,5),(0,-1)); [/asy]](http://latex.artofproblemsolving.com/f/d/4/fd4ce8e9f69eccb7bc297cde34a60f44244789af.png)
Solution
Looking at the list of 
and 
values that must match up left-to-right, we have 
. Looking for digits that only appear once, we see that 
, 
, 
, and 
cannot match up to other digits, and thus must appear on the ends. and only appear on the left, and thus their respective blocks 
and 
must appear on the left. Similarly, and only appear on the right, and thus their blocks 
and 
must appear on the right-most block of their row. Therefore, 
, the only block without a unique digit, must be the top-center block.
Now that we have placed one block, , we can only place block to the left of , and block to the right of . Thus, 
is the right answer.
Completing the puzzle, the top boxes read 
, while the bottom two boxes read 
.
See Also
| 1998 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by First Problem |
Followed by Problem 2 | |
| 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 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
