2018 AMC 12B Problems/Problem 17
Contents
- 1 Problem
- 2 Solution 1 (Generalization)
- 3 Solution 2 (Generalization)
- 4 Solution 3 (Substitution)
- 5 Solution 4 (Substitution)
- 6 Solution 5 (Manipulation)
- 7 Solution 6 (Solves for p)
- 8 Solution 7 (Inspection)
- 9 Solution 8 (Inspection)
- 10 Solution 9 (Graphing)
- 11 Solution 10 (Answer Choices)
- 12 Solution 11 (Answer Choices)
- 13 Solution 12 (Educated Guess)
- 14 See Also
Problem
Let and be positive integers such that and is as small as possible. What is ?
Solution 1 (Generalization)
More generally, let and be positive integers such that and From we have or From we have or Since note that:
- Multiplying by multiplying by and adding the results, we get
- Multiplying by multiplying by and adding the results, we get
To minimize we set from which Together, we can prove that For this problem, we have and so and The answer is
Remark
We will prove each part of the compound inequality in
- and
Let so The precondition becomes so
It follows that Moreover, the equality case occurs if and only if
We can prove by the same process. Similarly, the equality case occurs if and only if
Let and so and
It follows that Moreover, this part of is independent of the precondition
~MRENTHUSIASM
Solution 2 (Generalization)
Define variables and as Solution 1 does. Moreover, this solution refers to inequalities and in Solution 1.
Note that Multiplying both sides of by we get
For this problem, we have and so At we have from which Therefore, the answer is
Alternatively, refer to the Remark section in Solution 1 for further generalizations.
~pieater314159 ~MRENTHUSIASM
Solution 3 (Substitution)
Inverting the given inequality we get which simplifies to We can now substitute Note we need to find which simplifies to Clearly We will now substitute to get The inequality simplifies to
The inequality simplifies to
Combining the two inequalities, 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 4 (Substitution)
Because and are positive integers with we can let where Now, the problem condition reduces to
Our first inequality is which gives us or
Our second inequality is which gives us or
Hence, we have or
It is clear that we are aiming to find the least positive integer value of such that there is at least one value of that satisfies the inequality.
Now, simple casework through the answer choices of the problem reveals that
Solution 5 (Manipulation)
We subtract from both sides of the equation, so Then for to be as small as possible, has to be so is and is
~purplepenguin2
Solution 6 (Solves for p)
Cross-multiply the inequality to get Then, we have or Since and 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 7 (Inspection)
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
Therefore, the answer is
Solution 8 (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
Solution 9 (Graphing)
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 10 (Answer Choices)
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 11 (Answer Choices)
In ascending order, we can use answer choices, values for as a method of figuring out our answer through the means of substitution. Let the assumed difference be Then, We thus have two inequalities: and
Solving for in these equalities, we get So, is between and making it as is a positive integer (again, at this point, this is still an assumption). This would set
Since the minimum difference is
~mesmore
Solution 12 (Educated Guess)
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
Refer to Solution 1 for the full justification.
See Also
2018 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 16 |
Followed by Problem 18 |
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