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 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 using vectors.
Solution 5 (Using answer choices to prove mediant)
As the other solutions do, the mediant is between the two fractions, with a difference of . Suppose that the answer was not , then the answer must be or as otherwise would be negative. Then, the possible fractions with lower denominator would be for and for which are clearly not anywhere close to
Solution 6
Inverting the given inequality we get
which simplifies to
We can now substitute . Note we need to find .
which simplifies to
Cleary is greater than . We will now substitute to get
The inequality simplifies to . The inequality simplifies to . Combining the two we get
Since and are integers, the smallest values of and that satisfy the above equation are and respectively. Substituting these back in, we arrive with an answer of .
Solution 7
Start with . Repeat the following process until you arrive at the answer: if the fraction is less than or equal to , add to the numerator; otherwise, if it is greater than or equal to , add one to the denominator. We have:
.
Solution 8
Because q and p are positive integers with , we can let where . Now, the problem condition reduces to
Our first inequality is which gives us .
Our second inequality is which gives us .
Hence, .
It is clear that we are aiming to find the least positive integer value of k such that there is at least one value of p that satisfies the inequality.
Now, simple casework through the answer choices of the problem reveals that .
Solution 9 (Quick inspection)
Checking possible fractions within the interval can get us to the answer, but only if we do it with more skill. The interval can also be written as . This represents fraction with the numerator a little bit more than half the denominator. Every fraction we consider must not exceed this range.
The denominators to be considered are . We check . At this point we know that we've got our fraction and our answer is
The inspection was made faster by considering the fact that .
So, once a fraction was gotten which was greater than we jump to the next denominator.
We then make sure we consider fractions with higher positive difference between the denominator and numerator. And we also do not forget that the numerator must be greater than half of the denominator.
( was obviously skipped because it is equal to .)
~OlutosinNGA
See Also
2018 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 16 |
Followed by Problem 18 |
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