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

Revision as of 17:33, 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

@@ Line 13: / Line 13: @@
 ~MRENTHUSIASM
-==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 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 choice is one of
-* <math>1\cdot 3\cdot 5</math> (paired with <math>2\cdot4\cdot7</math>)
-* <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>)
-* <math>1\cdot 5\cdot 7</math> (paired with <math>2\cdot3\cdot4</math>)
-* <math>3\cdot 5\cdot 7</math> (paired with <math>1\cdot2\cdot4</math>)
-None of them 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
 ==See Also==
 {{AIME box|year=2022|n=I|num-b=6|num-a=8}}
 {{MAA Notice}}

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

Revision as of 17:33, 21 February 2022

Problem

Solution 1

See Also