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Difference between revisions of "2022 AIME I Problems/Problem 7"

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==Solution 2==
 
==Solution 2==
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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 the numerator to be as small as possible and the denominator as big as possible. One way to do this is to make the numerator one 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>9</math> and <math>8</math> for the denominator, we notice that <cmath>\dfrac{2 \cdot  3\cdot 6 - 7 \cdot 1 \cdot 5}{9 \cdot 8 \cdot 4}=\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>9\cdot 8\cdot 7=504,</math> which is less than <math>2\cdot 288,</math> <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 numerator greater than <math>288</math>. The only alternatives we need to consider are <math>9\cdot 8\cdot 6</math> and <math>9\cdot 8\cdot 5</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>7\cdot 3\cdot 1</math> (paired with either <math>2\cdot 4\cdot 5</math> or <math>2\cdot 4\cdot 6</math>) or <math>5\cdot 3\cdot 1</math>. Neither gives us a numerator of <math>1,</math> so we can conclude that the minimum fraction is <math>\frac{1}{288}</math> and thus the answer is <math>1+288=\boxed{289}</math>.
+
Since the maximum denominator is <math>97\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>) or <math>1\cdot 3\cdot 5</math> (paired with <math>2\cdot4\cdot7</math>). 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 04:49, 21 February 2022

Problem

Let $a,b,c,d,e,f,g,h,i$ be distinct integers from $1$ to $9.$ The minimum possible positive value of \[\dfrac{a \cdot b \cdot c - d \cdot e \cdot f}{g \cdot h \cdot i}\] can be written as $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Solution 1

To minimize a positive fraction, we minimize its numerator and maximize its denominator. It is clear that $\frac{a \cdot b \cdot c - d \cdot e \cdot f}{g \cdot h \cdot i} \geq \frac{1}{7\cdot8\cdot9}.$

If we minimize the numerator, then $a \cdot b \cdot c - d \cdot e \cdot f = 1.$ Note that $a \cdot b \cdot c \cdot d \cdot e \cdot f = (a \cdot b \cdot c) \cdot (a \cdot b \cdot c - 1) \geq 6! = 720,$ so $a \cdot b \cdot c \geq 28.$ It follows that $a \cdot b \cdot c$ and $d \cdot e \cdot f$ are consecutive composites with prime factors no other than $2,3,5,$ and $7.$ The smallest values for $a \cdot b \cdot c$ and $d \cdot e \cdot f$ are $36$ and $35,$ respectively. So, we have $\{a,b,c\} = \{2,3,6\}, \{d,e,f\} = \{1,5,7\},$ and $\{g,h,i\} = \{4,8,9\},$ from which $\frac{a \cdot b \cdot c - d \cdot e \cdot f}{g \cdot h \cdot i} = \frac{1}{288}.$

If we do not minimize the numerator, then $a \cdot b \cdot c - d \cdot e \cdot f > 1.$ Note that $\frac{a \cdot b \cdot c - d \cdot e \cdot f}{g \cdot h \cdot i} \geq \frac{2}{7\cdot8\cdot9} > \frac{1}{288}.$

Together, we conclude that the minimum possible positive value of $\frac{a \cdot b \cdot c - d \cdot e \cdot f}{g \cdot h \cdot i}$ is $\frac{1}{288}.$ Therefore, the answer is $1+288=\boxed{289}.$

~MRENTHUSIASM

Solution 2

Since we are trying to minimize \[\dfrac{a \cdot b \cdot c - d \cdot e \cdot f}{g \cdot h \cdot i},\] we want to minimize its numerator and maximize its denominator. One way to do this is to make the numerator $1$ and the denominator as large as possible. This means that $a\cdot b\cdot c$ has to be a different parity than $d\cdot e\cdot f.$ Using this and reserving $8$ and $9$ for the denominator, we notice that \[\dfrac{2 \cdot 3\cdot 6 - 1 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 9}=\frac{1}{288}.\] Since the maximum denominator is $97\cdot 8\cdot 9 < 2\cdot 288,$ we conclude that $\frac{1}{288}$ will be less than any other fraction we can come up with with a numerator greater than $1.$ This means that all we need to check is fractions with numerator $1$ and denominator greater than $288.$ The only alternatives we need to consider are $5\cdot 8\cdot 9$ and $6\cdot 8\cdot 9$ in the denominator. The parity restriction allows us to focus on numerators where either $a,b,c$ are all odd or $d,e,f$ are all odd, so our choices are $1\cdot 3\cdot 7$ (paired with either $2\cdot 4\cdot 5$ or $2\cdot 4\cdot 6$) or $1\cdot 3\cdot 5$ (paired with $2\cdot4\cdot7$). Neither gives us a numerator of $1,$ so the minimum fraction is $\frac{1}{288}$ and thus the answer is $1+288=\boxed{289}.$

~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

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