Difference between revisions of "2018 AMC 12B Problems/Problem 17"
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<cmath>\frac{1}{bd}=\frac{c}{d}-\frac{a}{b}=\left(\frac{c}{d}-\frac{p}{q}\right)+\left(\frac{p}{q}-\frac{a}{b}\right) \geq \frac{1}{dq}+\frac{1}{bq},</cmath> | <cmath>\frac{1}{bd}=\frac{c}{d}-\frac{a}{b}=\left(\frac{c}{d}-\frac{p}{q}\right)+\left(\frac{p}{q}-\frac{a}{b}\right) \geq \frac{1}{dq}+\frac{1}{bq},</cmath> | ||
− | which reduces to <math>q\geq b+d</math>. We can easily find that <math>p=a+c</math>, giving an answer of <math> | + | which reduces to <math>q\geq b+d</math>. We can easily find that <math>p=a+c</math>, giving an answer of <math>\boxed{\textbf{(A)}\ 7}</math>. (pieater314159) |
==Solution 2 (requires justification)== | ==Solution 2 (requires justification)== | ||
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<math>4q - 7p = 1</math> | <math>4q - 7p = 1</math> | ||
− | Solving the system of equations yields <math>q=16</math> and <math>p=9</math>. Therefore, the answer is <math>16-9=\boxed{7}</math> | + | Solving the system of equations yields <math>q=16</math> and <math>p=9</math>. Therefore, the answer is <math>\boxed{\textbf{(A)}\ 7}</math> |
+ | |||
+ | ==Solution 3== | ||
+ | Cross-multiply the inequality to get <cmath>35q < 63p < 36q.</cmath> | ||
+ | |||
+ | Then, | ||
+ | <cmath>0 < 63p-35q < q,</cmath> | ||
+ | <cmath>0 < 7(9p-5q) < q.</cmath> | ||
+ | |||
+ | Since <math>p</math>, <math>q</math> are integers, <math>9p-5q</math> is an integer. To minimize <math>q</math>, start from <math>9p-5q=1</math>, which gives <math>p=\frac{5q+1}{9}</math>. This limits <math>q</math> to be greater than <math>7</math>, so test values of <math>q</math> starting from <math>q=8</math>. However, <math>q=8</math> to <math>q=14</math> do not give integer values of <math>p</math>. | ||
+ | |||
+ | Once <math>q>14</math>, it is possible for <math>9p-5q</math> to be equal to <math>2</math>, so <math>p</math> could also be equal to <math>\frac{5q+2}{9}.</math> The next value, <math>q=15</math>, is not a solution, but <math>q=16</math> gives <math>p=\frac{5\cdot 16 + 1}{9} = 9</math>. Thus, the smallest possible value of <math>q</math> is <math>16</math>, and the answer is <math>16-9= \boxed{\textbf{(A)}\ 7}</math>. | ||
==See Also== | ==See Also== |
Revision as of 18:28, 16 February 2018
Problem
Let and
be positive integers such that
and
isi 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
. (pieater314159)
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
.
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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