Difference between revisions of "1994 AJHSME Problems/Problem 20"
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To minimize the numerator, let <math>W=1</math> and <math>Y=2</math>. | To minimize the numerator, let <math>W=1</math> and <math>Y=2</math>. | ||
| − | <cmath>\frac{9+16}{72} = \boxed{\text{(D)}\ \frac{25}{72}}</cmath> | + | <cmath>\frac{9+16}{72} = \boxed{\text{(D)}\rightarrow \frac{25}{72}}</cmath> |
==See Also== | ==See Also== | ||
{{AJHSME box|year=1994|num-b=19|num-a=21}} | {{AJHSME box|year=1994|num-b=19|num-a=21}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 12:37, 30 October 2016
Problem
Let 
and 
be four different digits selected from the set
If the sum 
is to be as small as possible, then must equal
Solution
![\[\frac{W}{X} + \frac{Y}{Z} = \frac{WZ+XY}{XZ}\]](http://latex.artofproblemsolving.com/9/1/4/914fbb8336720d6bceff189dfc81868bda5f761e.png)
Small fractions have small numerators and large denominators. To maximize the denominator, let 
and 
.
![\[\frac{9W+8Y}{72}\]](http://latex.artofproblemsolving.com/a/c/4/ac476641187669bdc5c0780909ff690c015b388f.png)
To minimize the numerator, let 
and 
.
![\[\frac{9+16}{72} = \boxed{\text{(D)}\rightarrow \frac{25}{72}}\]](http://latex.artofproblemsolving.com/b/7/1/b71f5d688316ac9410c0586159d59f4e33942c57.png)
See Also
| 1994 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 19 |
Followed by Problem 21 | |
| 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. 
