Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2022 AIME I Problems/Problem 7"
(→‎Solution 3 (Educated Trial & Error))
 
(29 intermediate revisions by 5 users not shown)
Line 3: Line 3:
 
Let <math>a,b,c,d,e,f,g,h,i</math> be distinct integers from <math>1</math> to <math>9.</math> The minimum possible positive value of <cmath>\dfrac{a \cdot b \cdot c - d \cdot e \cdot f}{g \cdot h \cdot i}</cmath> can be written as <math>\frac{m}{n},</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n.</math>
 
Let <math>a,b,c,d,e,f,g,h,i</math> be distinct integers from <math>1</math> to <math>9.</math> The minimum possible positive value of <cmath>\dfrac{a \cdot b \cdot c - d \cdot e \cdot f}{g \cdot h \cdot i}</cmath> can be written as <math>\frac{m}{n},</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n.</math>
  
−
==Solution 1 (sophisticated bash)==
+
==Solution 1 (Optimization)==
−
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>
+
To minimize a positive fraction, we minimize its numerator and maximize its denominator. It is clear that <math>\frac{a \cdot b \cdot c - d \cdot e \cdot f}{g \cdot h \cdot i} \geq \frac{1}{7\cdot8\cdot9}.</math>
−
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>.
 
  
−
-jgplay
+
If we minimize the numerator, then <math>a \cdot b \cdot c - d \cdot e \cdot f = 1.</math> Note that <math>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,</math> so <math>a \cdot b \cdot c \geq 28.</math> It follows that <math>a \cdot b \cdot c</math> and <math>d \cdot e \cdot f</math> are consecutive composites with prime factors no other than <math>2,3,5,</math> and <math>7.</math> The smallest values for <math>a \cdot b \cdot c</math> and <math>d \cdot e \cdot f</math> are <math>36</math> and <math>35,</math> respectively. So, we have <math>\{a,b,c\} = \{2,3,6\}, \{d,e,f\} = \{1,5,7\},</math> and <math>\{g,h,i\} = \{4,8,9\},</math> from which <math>\frac{a \cdot b \cdot c - d \cdot e \cdot f}{g \cdot h \cdot i} = \frac{1}{288}.</math>
 +
 
 +
If we do not minimize the numerator, then <math>a \cdot b \cdot c - d \cdot e \cdot f > 1.</math> Note that <math>\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}.</math>
 +
 
 +
Together, we conclude that the minimum possible positive value of <math>\frac{a \cdot b \cdot c - d \cdot e \cdot f}{g \cdot h \cdot i}</math> is <math>\frac{1}{288}.</math> Therefore, the answer is <math>1+288=\boxed{289}.</math>
 +
 
 +
~MRENTHUSIASM ~jgplay
 +
 
 +
==Solution 2 (Bash)==
 +
Obviously, to find the correct answer, we need to get the largest denominator with the smallest numerator.
 +
 
 +
To bash efficiently, we can start out with <math>7\cdot8\cdot9</math> as our denominator. This, however, leaves us with the numbers <math>1, 2, 3, 4, 5,</math> and <math>6</math> left. The smallest we can make out of this is <math>1\cdot5\cdot6 - 2\cdot3\cdot4 = 30 - 24 = 6</math>. When simplified, it gives us <math>\frac{1}{84}</math>, which gives a small answer of <math>85</math>. Obviously there are larger answers than this.
 +
 
 +
After the first bash, we learn to bash even more efficiently, we can consider <u>both</u> the numerator and the denominator when guessing. We know the numerator has to be extremely small while still having a large denominator. When bashing, we soon find out the couple <math>(a,b,c)=(2,3,6)</math> and <math>(d,e,f)=(1,5,7)</math>.
 +
 
 +
This gives us a numerator of <math>36-35=1</math>, which is by far the smallest yet. With the remaining numbers <math>4, 8,</math> and <math>9</math>, we get <math>\frac{36-35}{4\cdot8\cdot9}=\frac{1}{288}</math>.
 +
 
 +
Finally, we add up our numerator and denominator: The answer is <math>1+288=\boxed{289}</math>.
 +
 
 +
~[[OrenSH|orenbad]]
 +
 
 +
==Solution 3 (Educated Trial and Error)==
 +
To minimize the numerator, we must have <math>abc - def = 1</math>. Thus, one of these products must be odd and the other must be even. The odd product must consist of only odd numbers. The smallest such value <math>(d, e, f) = (1, 3, 5)</math> cannot result in a difference of <math>1</math>, and the next smallest product, <math>(d, e, f) = (1, 3, 7)</math> cannot either, but <math>(d, e, f) = (1, 5, 7)</math> can if <math>(a, b, c) = (2, 3, 6)</math>. Thus, the denominator must be <math>(g, h, i) = (4, 8, 9)</math>, and the smallest fraction possible is <math>\dfrac{36 - 35}{288} = \dfrac{1}{288}</math>, making the answer <math>1 + 288 = \boxed{289}</math>.
 +
 
 +
~[https://artofproblemsolving.com/wiki/index.php/User:A_mathemagician A_MatheMagician]
 +
 
 +
==See Also==
 +
{{AIME box|year=2022|n=I|num-b=6|num-a=8}}
 +
{{MAA Notice}}

Latest revision as of 12:04, 16 January 2023

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 (Optimization)

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 ~jgplay

Solution 2 (Bash)

Obviously, to find the correct answer, we need to get the largest denominator with the smallest numerator.

To bash efficiently, we can start out with $7\cdot8\cdot9$ as our denominator. This, however, leaves us with the numbers $1, 2, 3, 4, 5,$ and $6$ left. The smallest we can make out of this is $1\cdot5\cdot6 - 2\cdot3\cdot4 = 30 - 24 = 6$. When simplified, it gives us $\frac{1}{84}$, which gives a small answer of $85$. Obviously there are larger answers than this.

After the first bash, we learn to bash even more efficiently, we can consider both the numerator and the denominator when guessing. We know the numerator has to be extremely small while still having a large denominator. When bashing, we soon find out the couple $(a,b,c)=(2,3,6)$ and $(d,e,f)=(1,5,7)$.

This gives us a numerator of $36-35=1$, which is by far the smallest yet. With the remaining numbers $4, 8,$ and $9$, we get $\frac{36-35}{4\cdot8\cdot9}=\frac{1}{288}$.

Finally, we add up our numerator and denominator: The answer is $1+288=\boxed{289}$.

~orenbad

Solution 3 (Educated Trial and Error)

To minimize the numerator, we must have $abc - def = 1$. Thus, one of these products must be odd and the other must be even. The odd product must consist of only odd numbers. The smallest such value $(d, e, f) = (1, 3, 5)$ cannot result in a difference of $1$, and the next smallest product, $(d, e, f) = (1, 3, 7)$ cannot either, but $(d, e, f) = (1, 5, 7)$ can if $(a, b, c) = (2, 3, 6)$. Thus, the denominator must be $(g, h, i) = (4, 8, 9)$, and the smallest fraction possible is $\dfrac{36 - 35}{288} = \dfrac{1}{288}$, making the answer $1 + 288 = \boxed{289}$.

~A_MatheMagician

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2022_AIME_I_Problems/Problem_7&oldid=186799"