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
A
B
C
p
\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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