Difference between revisions of "2022 AIME I Problems/Problem 7"
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==Solution 2== | ==Solution 2== | ||
Since we are trying to minimize <cmath>\dfrac{a \cdot b \cdot c - d \cdot e \cdot f}{g \cdot h \cdot i},</cmath> we want to minimize its numerator and maximize its denominator. One way to do this is to make the numerator <math>1</math> and the denominator as large as possible. This means that <math>a\cdot b\cdot c</math> has to be a different parity than <math>d\cdot e\cdot f.</math> Using this and reserving <math>8</math> and <math>9</math> for the denominator, we notice that <cmath>\dfrac{2 \cdot 3\cdot 6 - 1 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 9}=\frac{1}{288}.</cmath> | Since we are trying to minimize <cmath>\dfrac{a \cdot b \cdot c - d \cdot e \cdot f}{g \cdot h \cdot i},</cmath> we want to minimize its numerator and maximize its denominator. One way to do this is to make the numerator <math>1</math> and the denominator as large as possible. This means that <math>a\cdot b\cdot c</math> has to be a different parity than <math>d\cdot e\cdot f.</math> Using this and reserving <math>8</math> and <math>9</math> for the denominator, we notice that <cmath>\dfrac{2 \cdot 3\cdot 6 - 1 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 9}=\frac{1}{288}.</cmath> | ||
− | Since the maximum denominator is <math> | + | Since the maximum denominator is <math>7\cdot 8\cdot 9 < 2\cdot 288,</math> we conclude that <math>\frac{1}{288}</math> will be less than any other fraction we can come up with with a numerator greater than <math>1.</math> This means that all we need to check is fractions with numerator <math>1</math> and denominator greater than <math>288.</math> The only alternatives we need to consider are <math>5\cdot 8\cdot 9</math> and <math>6\cdot 8\cdot 9</math> in the denominator. The parity restriction allows us to focus on numerators where either <math>a,b,c</math> are all odd or <math>d,e,f</math> are all odd, so our choices are <math>1\cdot 3\cdot 7</math> (paired with either <math>2\cdot 4\cdot 5</math> or <math>2\cdot 4\cdot 6</math>) and <math>1\cdot 3\cdot 5</math> (paired with <math>2\cdot4\cdot7</math>), respectively. Neither gives us a numerator of <math>1,</math> so the minimum fraction is <math>\frac{1}{288}</math> and thus the answer is <math>1+288=\boxed{289}.</math> |
~jgplay | ~jgplay |
Revision as of 17:51, 21 February 2022
Contents
Problem
Let be distinct integers from
to
The minimum possible positive value of
can be written as
where
and
are relatively prime positive integers. Find
Solution 1
To minimize a positive fraction, we minimize its numerator and maximize its denominator. It is clear that
If we minimize the numerator, then Note that
so
It follows that
and
are consecutive composites with prime factors no other than
and
The smallest values for
and
are
and
respectively. So, we have
and
from which
If we do not minimize the numerator, then Note that
Together, we conclude that the minimum possible positive value of is
Therefore, the answer is
~MRENTHUSIASM
Solution 2
Since we are trying to minimize we want to minimize its numerator and maximize its denominator. One way to do this is to make the numerator
and the denominator as large as possible. This means that
has to be a different parity than
Using this and reserving
and
for the denominator, we notice that
Since the maximum denominator is
we conclude that
will be less than any other fraction we can come up with with a numerator greater than
This means that all we need to check is fractions with numerator
and denominator greater than
The only alternatives we need to consider are
and
in the denominator. The parity restriction allows us to focus on numerators where either
are all odd or
are all odd, so our choices are
(paired with either
or
) and
(paired with
), respectively. Neither gives us a numerator of
so the minimum fraction is
and thus the answer is
~jgplay
See Also
2022 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.