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

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~MRENTHUSIASM ~jgplay
 
~MRENTHUSIASM ~jgplay
  
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==Solution 2 (bash)==
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==Solution 2 (Bash)==
 
Obviously, to find the correct answer, we need to get the largest denominator with the smallest numerator.  
 
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 the answer of <math>\frac{1}{84}</math> which gives a small answer of <math>85</math>. Obviously there are larger answers than this.
 
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 the answer of <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 BOTH the numerator and the denominator when guessing. We know the numerator has to be extremely small while still having a large denominator. When basing, we soon find out the couple <math>(1, 5, 7)</math> and <math>(2, 3, 6)</math>
+
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 basing, we soon find out the couple <math>(1, 5, 7)</math> and <math>(2, 3, 6)</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 our answer of <math>\frac{36-35}{4\cdot8\cdot9}=\frac{1}{288}</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 our answer of <math>\frac{36-35}{4\cdot8\cdot9}=\frac{1}{288}</math>.

Revision as of 19:23, 31 December 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

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 the answer of $\frac{1}{84}$ 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 basing, we soon find out the couple $(1, 5, 7)$ and $(2, 3, 6)$

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 our answer of $\frac{36-35}{4\cdot8\cdot9}=\frac{1}{288}$.

To finalize, we add up our numerator and denominator which gives us $1+288=\boxed{289}$ as our answer.

~orenbad

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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