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Difference between revisions of "2022 AIME I Problems/Problem 7"
(→‎Solution 1 (sophisticated bash))
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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 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 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>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>.
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-jgplay

Revision as of 20:11, 17 February 2022

Solution 1 (sophisticated bash)

Since we are trying to minimize \[\dfrac{a \cdot b \cdot c - d \cdot e \cdot f}{g \cdot h \cdot i},\] 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 $a\cdot b\cdot c$ has to be a different parity than $d\cdot e\cdot f.$ Using this, and reserving $9$ and $8$ for the denominator, we notice that \[\dfrac{2 \cdot 3\cdot 6 - 7 \cdot 1 \cdot 5}{9 \cdot 8 \cdot 4}=\frac{1}{288}.\] Since the maximum denominator is $9\cdot 8\cdot 7=504,$ which is less than $2\cdot 288,$ $\frac{1}{288}$ 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 numerator greater than $288$. The only alternatives we need to consider are $9\cdot 8\cdot 6$ and $9\cdot 8\cdot 5$ 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 $7\cdot 3\cdot 1$ (paired with either $2\cdot 4\cdot 5$ or $2\cdot 4\cdot 6$) or $5\cdot 3\cdot 1$. Neither gives us a numerator of $1,$ so we can conclude that the minimum fraction is $\frac{1}{288}$ and thus the answer is $1+288=\boxed{289}$.

-jgplay

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