Difference between revisions of "1994 AJHSME Problems/Problem 20"
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<cmath>\frac{9+16}{72} = \boxed{\text{(D)}\rightarrow \frac{25}{72}}</cmath> | <cmath>\frac{9+16}{72} = \boxed{\text{(D)}\rightarrow \frac{25}{72}}</cmath> | ||
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− | + | ==Solution 2== | |
− | + | to make the smallest fraction, you need the lowest numerator and the highest denominator. So, take the first 2 and last 2 digits of the set: 1,2 and 8,9. Balance the equations to be "even". Since 1 is smaller than 2, put it over 8. You get 1/8 + 2/9 = 25/72, or D | |
− | + | -goldenn | |
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==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}} |
Latest revision as of 23:17, 10 September 2019
Contents
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
Small fractions have small numerators and large denominators. To maximize the denominator, let and .
To minimize the numerator, let and .
Solution 2
to make the smallest fraction, you need the lowest numerator and the highest denominator. So, take the first 2 and last 2 digits of the set: 1,2 and 8,9. Balance the equations to be "even". Since 1 is smaller than 2, put it over 8. You get 1/8 + 2/9 = 25/72, or D
-goldenn
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.