Difference between revisions of "1986 AJHSME Problems/Problem 2"
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We see that <math>\frac{1}{3} = \frac{5}{15}</math>, and <math>\frac{2}{5} = \frac{6}{15}</math>, so obviously <math>\frac{1}{3}</math> is smaller. | We see that <math>\frac{1}{3} = \frac{5}{15}</math>, and <math>\frac{2}{5} = \frac{6}{15}</math>, so obviously <math>\frac{1}{3}</math> is smaller. | ||
| − | <math>\boxed{\text{A}}</math> | + | <math>\boxed{\text{A) } \frac{1}{3} }</math> |
| + | |||
| + | ==Alternate Solution== | ||
| + | |||
| + | Find the reciprocals of all numbers. | ||
| + | |||
| + | |||
| + | \( \frac{1}{3} = 3 \) | ||
| + | |||
| + | |||
| + | \( \frac{2}{5} = \frac{5}{2} = 2.5 \) | ||
| + | |||
| + | |||
| + | \( 1 = 1 \) | ||
| + | |||
| + | |||
| + | \( 5 = \frac{1}{5} = 0.2 \) | ||
| + | |||
| + | |||
| + | \( 1986 = \frac{1}{1986} = 0.0005035246727 \) | ||
| + | |||
| + | |||
| + | Therefore, just by comparing the options, we get to know that \( \frac{1}{3} \) has the largest reciprocal. | ||
| + | |||
| + | Thus, the answer is | ||
| + | |||
| + | <math>\boxed{\text{A) } \frac{1}{3} }</math> | ||
| + | |||
| + | ~ Alternate solution by abcde26 | ||
==See Also== | ==See Also== | ||
Latest revision as of 02:16, 23 October 2025
Problem
Which of the following numbers has the largest reciprocal?
Solution
For positive numbers, the larger the number, the smaller its reciprocal. Likewise, smaller numbers have larger reciprocals.
Thus, all we have to do is find the smallest number.
But which one is it? 
? or 
?
We see that 
, and 
, so obviously is smaller.
Alternate Solution
Find the reciprocals of all numbers.
\( \frac{1}{3} = 3 \)
\( \frac{2}{5} = \frac{5}{2} = 2.5 \)
\( 1 = 1 \)
\( 5 = \frac{1}{5} = 0.2 \)
\( 1986 = \frac{1}{1986} = 0.0005035246727 \)
Therefore, just by comparing the options, we get to know that \( \frac{1}{3} \) has the largest reciprocal.
Thus, the answer is
~ Alternate solution by abcde26
See Also
| 1986 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
