2018 AMC 12B Problems/Problem 17
Contents
Problem
Let and be positive integers such that and is as small as possible. What is ?
Solution 1
We claim that, between any two fractions and , if , the fraction with smallest denominator between them is . To prove this, we see that
which reduces to . We can easily find that , giving an answer of .
Solution 2 (requires justification)
Assume that the difference results in a fraction of the form . Then,
Also assume that the difference results in a fraction of the form . Then,
Solving the system of equations yields and . Therefore, the answer is
Solution 3
Cross-multiply the inequality to get
Then,
Since , are integers, is an integer. To minimize , start from , which gives . This limits to be greater than , so test values of starting from . However, to do not give integer values of .
Once , it is possible for to be equal to , so could also be equal to The next value, , is not a solution, but gives . Thus, the smallest possible value of is , and the answer is .
Solution 4
Graph the regions and . Note that the lattice point (9,16) is the smallest magnitude one which appears within the region bounded by the two graphs. Thus, our fraction is and the answer is .
Remark: This also gives an intuitive geometric proof of the mediant.
Solution 5 (Using answer choices to prove mediant)
As the other solutions do, the mediant is between the two fractions, with a difference of ABCp\frac{k-11}{k}k=12,13,14,15\frac{k-13}{k}k=14,15,\frac{4}{7}\approx 0.6$
See Also
2018 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 16 |
Followed by Problem 18 |
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