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2022 AIME I Problems/Problem 7

Revision as of 17:51, 21 February 2022 by MRENTHUSIASM (talk | contribs) (→‎Solution 2)

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 $7\cdot 8\cdot 9 < 2\cdot 288,$ 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 $2\cdot 4\cdot 6$) and $1\cdot 3\cdot 5$ (paired with $2\cdot4\cdot7$), respectively. 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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