Difference between revisions of "2018 AMC 12B Problems/Problem 17"
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Our first inequality is <math>\frac{5}{9}<\frac{p}{p+k}</math> which gives us <math>5p+5k<9p\implies \frac{5}{4}k<p</math>. | Our first inequality is <math>\frac{5}{9}<\frac{p}{p+k}</math> which gives us <math>5p+5k<9p\implies \frac{5}{4}k<p</math>. | ||
− | Our second inequality is <math>\frac{p}{p+k}<\frac{4}{7} which gives us < | + | Our second inequality is <math>\frac{p}{p+k}<\frac{4}{7}</math> which gives us <math>7p<4p+4k\implies p<\frac{4}{3}k</math>. |
− | Hence, < | + | Hence, <math>\frac{5}{4}k<p<\frac{4}{3}k\implies 15k<12p<16k</math>. |
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. | 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 q-p=p+k-p=k\implies k\ge{\boxed{7}} | + | Now, simple casework through the answer choices of the problem reveals that q-p=p+k-p=k\implies k\ge{\boxed{7}}$. |
+ | |||
==See Also== | ==See Also== | ||
Revision as of 15:43, 11 July 2018
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 q-p=p+k-p=k\implies k\ge{\boxed{7}}$.
See Also
2018 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 16 |
Followed by Problem 18 |
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 AMC 12 Problems and Solutions |
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