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Difference between revisions of "2000 PMWC Problems/Problem T8"

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==Solution==
 
==Solution==
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Given that <math>k</math>, <math>n</math>, and <math>m</math> are positive integers, it is obvious that none of them can be 1, because that would immediately make the quantity <math>\frac{1}{k}+\frac{1}{n}+\frac{1}{m}</math> too large.
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What about having one of them equal <math>2</math>? We can actually prove through casework that at least one of the integers must be equal to <math>2</math> by considering what would happen if none of <math>k</math>, <math>n</math>, or <math>m</math> were equal to <math>2</math>. If this was true, then the smallest <math>k</math>, <math>n</math>, or <math>m</math> could be would be <math>3</math>. If all three were equal to <math>3</math>, then we would have <math>\frac{1}{3}+\frac{1}{3}+\frac{1}{3} = 1</math>, which is too large. However, the next largest value for the sum <math>\frac{1}{k}+\frac{1}{n}+\frac{1}{m}</math> would be <math>\frac{1}{3}+\frac{1}{3}+\frac{1}{4} = \frac{11}{12}</math>, which is too small. Thus there are no possibilities that have no <math>2</math>s, meaning we can set any one of the variables <math>k</math>, <math>n</math>, or <math>m</math> equal to <math>2</math> and continue solving.
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Let's say we choose <math>k</math> to equal <math>2</math>. Then we have <math>\frac{9}{20}<\frac{1}{n}+\frac{1}{m}<\frac{1}{2}</math>. Obviously we can't have another <math>2</math>, but what if we had a <math>3</math>? It turns out that we can prove that we must have a <math>3</math> in much the same way we proved that there must be a <math>2</math>. If we assume that there is no <math>3</math>, the largest value of <math>\frac{1}{n}+\frac{1}{m}</math> would be <math>\frac{1}{4}+\frac{1}{4} = \frac{1}{2}</math>, which is too large. Just as last time, the next largest possibility, <math>\frac{1}{4}+\frac{1}{5} = \frac{9}{20}</math>, is too small. Thus, there are no possibilities that have no <math>3</math>s, meaning one of the remaining variables must be <math>3</math>. If we randomly select <math>n</math> to equal <math>3</math>, we are left with the equation <math>\frac{9}{20} - \frac{1}{3} < \frac{1}{m} < \frac{1}{2} - \frac{1}{3}</math>, which means <math>\frac{7}{60} < \frac{1}{m} < \frac{1}{6}</math>. This inequality shows that <math>m</math> can be equal to <math>7</math> or <math>8</math>. The problem asks for the least possible value of <math>k+n+m</math>, so we say that <math>m = 7</math>.
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To conclude, we have <math>\frac{1}{k}+\frac{1}{n}+\frac{1}{m} = \frac{1}{2}+\frac{1}{3}+\frac{1}{7} = \frac{21}{42}+\frac{14}{42}+\frac{6}{42} = \frac{41}{42}</math>, which is indeed greater than <math>\frac{19}{20}</math> and less than <math>1</math>. This configuration, <math>(2,3,7)</math> is one of two that satisfies the given conditions, the other being <math>(2,3,8)</math>. Of these two solutions, <math>(2,3,7)</math> has the lower sum, meaning our answer is <math>2 + 3 + 7 = \boxed{12}</math>
  
 
==See Also==
 
==See Also==
 
Back to test: https://artofproblemsolving.com/wiki/index.php/2000_PMWC_Problems
 
Back to test: https://artofproblemsolving.com/wiki/index.php/2000_PMWC_Problems

Latest revision as of 10:54, 24 December 2019

Problem

There are positive integers $k$, $n$, $m$ such that $\frac{19}{20}<\frac{1}{k}+\frac{1}{n}+\frac{1}{m}<1$. What is the smallest possible value of $k+n+m$?

Solution

Given that $k$, $n$, and $m$ are positive integers, it is obvious that none of them can be 1, because that would immediately make the quantity $\frac{1}{k}+\frac{1}{n}+\frac{1}{m}$ too large.


What about having one of them equal $2$? We can actually prove through casework that at least one of the integers must be equal to $2$ by considering what would happen if none of $k$, $n$, or $m$ were equal to $2$. If this was true, then the smallest $k$, $n$, or $m$ could be would be $3$. If all three were equal to $3$, then we would have $\frac{1}{3}+\frac{1}{3}+\frac{1}{3} = 1$, which is too large. However, the next largest value for the sum $\frac{1}{k}+\frac{1}{n}+\frac{1}{m}$ would be $\frac{1}{3}+\frac{1}{3}+\frac{1}{4} = \frac{11}{12}$, which is too small. Thus there are no possibilities that have no $2$s, meaning we can set any one of the variables $k$, $n$, or $m$ equal to $2$ and continue solving.


Let's say we choose $k$ to equal $2$. Then we have $\frac{9}{20}<\frac{1}{n}+\frac{1}{m}<\frac{1}{2}$. Obviously we can't have another $2$, but what if we had a $3$? It turns out that we can prove that we must have a $3$ in much the same way we proved that there must be a $2$. If we assume that there is no $3$, the largest value of $\frac{1}{n}+\frac{1}{m}$ would be $\frac{1}{4}+\frac{1}{4} = \frac{1}{2}$, which is too large. Just as last time, the next largest possibility, $\frac{1}{4}+\frac{1}{5} = \frac{9}{20}$, is too small. Thus, there are no possibilities that have no $3$s, meaning one of the remaining variables must be $3$. If we randomly select $n$ to equal $3$, we are left with the equation $\frac{9}{20} - \frac{1}{3} < \frac{1}{m} < \frac{1}{2} - \frac{1}{3}$, which means $\frac{7}{60} < \frac{1}{m} < \frac{1}{6}$. This inequality shows that $m$ can be equal to $7$ or $8$. The problem asks for the least possible value of $k+n+m$, so we say that $m = 7$.


To conclude, we have $\frac{1}{k}+\frac{1}{n}+\frac{1}{m} = \frac{1}{2}+\frac{1}{3}+\frac{1}{7} = \frac{21}{42}+\frac{14}{42}+\frac{6}{42} = \frac{41}{42}$, which is indeed greater than $\frac{19}{20}$ and less than $1$. This configuration, $(2,3,7)$ is one of two that satisfies the given conditions, the other being $(2,3,8)$. Of these two solutions, $(2,3,7)$ has the lower sum, meaning our answer is $2 + 3 + 7 = \boxed{12}$

See Also

Back to test: https://artofproblemsolving.com/wiki/index.php/2000_PMWC_Problems

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