Difference between revisions of "2022 AIME I Problems/Problem 7"
MRENTHUSIASM (talk | contribs) (→Solution 2: Tried my very best making this solution more rigorous, but I realized that there are many more cases for the denominator: 5*7*9, 5*8*9, 6*7*9, 6*8*9, 6*7*8, 7*8*9. So, I decide to remove this sol and combine credits in Sol 1.) |
MRENTHUSIASM (talk | contribs) m (→Solution 1) |
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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 | + | ==Solution== |
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> | 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> | ||
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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> | 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 | + | ~MRENTHUSIASM ~Jgplay |
==See Also== | ==See Also== | ||
{{AIME box|year=2022|n=I|num-b=6|num-a=8}} | {{AIME box|year=2022|n=I|num-b=6|num-a=8}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 17:37, 21 February 2022
Problem
Let 
be distinct integers from 
to 
The minimum possible positive value of ![\[\dfrac{a \cdot b \cdot c - d \cdot e \cdot f}{g \cdot h \cdot i}\]](http://latex.artofproblemsolving.com/c/6/5/c653d108c03120ef617df93994006dd9a956c44c.png)
can be written as 
where 
and 
are relatively prime positive integers. Find 
Solution
To minimize a positive fraction, we minimize its numerator and maximize its denominator. It is clear that 
If we minimize the numerator, then 
Note that 
so 
It follows that 
and 
are consecutive composites with prime factors no other than 
and 
The smallest values for and are 
and 
respectively. So, we have 
and 
from which 
If we do not minimize the numerator, then 
Note that 
Together, we conclude that the minimum possible positive value of 
is 
Therefore, the answer is 
~MRENTHUSIASM ~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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
