Find Smallest Positive Integer

[anemone]

Find the smallest positive integer $n$ such that for every integer $m$ with $0<m<1993$, there exists an integer $k$ for which \(\displaystyle \frac{m}{1993}<\frac{k}{n}<\frac{m+1}{1994}\).

 

[Wilmer]

Find the smallest positive integer $n$ such that for every integer $m$ with $0<m<1993$, there exists an integer $k$ for which \(\displaystyle \frac{m}{1993}<\frac{k}{n}<\frac{m+1}{1994}\).

Terrific puzzle, Anemone!

Smallest n = 3987

Example using low m (1) and high m (1992):

(m + 1) / 1994 > k / n > m / 1993

m = 1:
2 / 1994 > 3 / 3987 > 1 / 1993

m = 1992:
1993 / 1994 > 3985 / 3987 > 1992 / 1993

Unfortunately, got no exotic formula for you (Bandit)

 

[anemone]

Terrific puzzle, Anemone!

Smallest n = 3987

Example using low m (1) and high m (1992):

(m + 1) / 1994 > k / n > m / 1993

m = 1:
2 / 1994 > 3 / 3987 > 1 / 1993

m = 1992:
1993 / 1994 > 3985 / 3987 > 1992 / 1993

Hi Wilmer,

Thanks for participating and thanks for the compliment to this problem as well.:)

Yes, 3987 is the answer to this problem but...

Unfortunately, got no exotic formula for you(Bandit)

30 minutes in the corner, please...(Tongueout)

 

[Wilmer]

30 minutes in the corner, please...(Tongueout)

No fair! You only asked: "Find the smallest positive integer [FONT=MathJax_Math-italic]n" (Punch)[/FONT]

 

[anemone]

I'll post the solution in full later because I think there might be others who still want to attempt it!

 

[Albert1]

Find the smallest positive integer $n$ such that for every integer $m$ with $0<m<1993$, there exists an integer $k$ for which \(\displaystyle \frac{m}{1993}<\frac{k}{n}<\frac{m+1}{1994}\).

we know $0<\dfrac {k}{n}<1$
using $0<\dfrac {1}{2}<\dfrac {2}{3}<\dfrac {3}{4}<--<\dfrac {3985}{3986}<\dfrac {3986}{3987}<\dfrac {3987}{3988}<1$
$2\times \dfrac {m}{3986}<\dfrac {k_0}{n_0}<2\times \dfrac{m+1}{3988}<1$
take $n=n_0 ,\,\, and \,\, k=k_0$
now it is easy to see that the smallese value of n =3987

 

[anemone]

we know $0<\dfrac {k}{n}<1$
using $0<\dfrac {1}{2}<\dfrac {2}{3}<\dfrac {3}{4}<--<\dfrac {3995}{3996}<\dfrac {3996}{3997}<\dfrac {3997}{3998}<1$
$2\times \dfrac {m}{3996}<\dfrac {k_0}{n_0}<2\times \dfrac{m+1}{3998}<1$
take $n=n_0 ,\,\, and \,\, k=k_0$
now it is easy to see that the smallese value of n =3997

Hi Albert,

Thanks for participating and I assume you meant \(\displaystyle \dfrac {3985}{3986}<\dfrac {3986}{3987}<\dfrac {3987}{3988}\) rather than \(\displaystyle \dfrac {3995}{3996}<\dfrac {3996}{3997}<\dfrac {3997}{3998}\), is that true?If that's true, according to your reasoning we would arrive to \(\displaystyle \dfrac {3985}{1993}<\dfrac {2(3986)}{3987}<\dfrac {3987}{1994}\) and I don't think this helps much as what the original question stated is it wants the difference between the two numerators on both fractions to be 1 but $3987-3985=2$. Hmm...What do you think, Albert?

 

[Opalg]

Find the smallest positive integer $n$ such that for every integer $m$ with $0<m<1993$, there exists an integer $k$ for which \(\displaystyle \frac{m}{1993}<\frac{k}{n}<\frac{m+1}{1994}\).

If \(\displaystyle \frac ab < \frac cd\) then \(\displaystyle \frac ab < \frac{a+c}{b+d} < \frac cd\) (assuming that all those numbers are positive). Therefore \(\displaystyle \frac m{1993} < \frac {2m+1}{3987} < \frac{m+1}{1994}.\)

(But that doesn't prove that $3987$ is the smallest such number.)

 

[anemone]

Solution:

Spoiler

If we have \(\displaystyle \frac{1992}{1993}<\frac{k}{n}<\frac{1993}{1994}\), then \(\displaystyle \frac{1993-1992}{1993}>\frac{n-k}{n}>\frac{1994-1993}{1994}\).

Simplifying we get

\(\displaystyle \frac{1}{1993}>\frac{n-k}{n}>\frac{1}{1994}\)

\(\displaystyle 1993<\frac{n}{n-k}<1994\)

Clearly \(\displaystyle n-k \ne 1\), so \(\displaystyle n-k \ge 2\).

Hence, \(\displaystyle n>1993(n-k) \ge 3986\) and so $n \ge 3987$.

Remark: This method of solving isn't mine, I saw it somewhere and wanted so much to share it here with MHB.

 

[Albert1]

Hi Albert,

Thanks for participating and I assume you meant \(\displaystyle \dfrac {3985}{3986}<\dfrac {3986}{3987}<\dfrac {3987}{3988}\) rather than \(\displaystyle \dfrac {3995}{3996}<\dfrac {3996}{3997}<\dfrac {3997}{3998}\), is that true?If that's true, according to your reasoning we would arrive to \(\displaystyle \dfrac {3985}{1993}<\dfrac {2(3986)}{3987}<\dfrac {3987}{1994}\) and I don't think this helps much as what the original question stated is it wants the difference between the two numerators on both fractions to be 1 but $3987-3985=2$. Hmm...What do you think, Albert?

sorry a mistake in calculation , now I will do this way :
$\dfrac {2m}{3986}\leq \dfrac{3984}{3986}\approx 0.999498243$
$\dfrac {2m+2}{3988}\leq \dfrac{3986}{3988}\approx 0.999498495$
for any 0<m<1993 ($m\in N$)
and we will find min(n) satisfying :
$\dfrac {3984}{3986}<\dfrac{k}{n}<\dfrac {3986}{3988}<1-----(1)$
$\dfrac {-3986}{3988}<\dfrac{-k}{n}<\dfrac {-3984}{3986}$

$\dfrac {2}{3988}<\dfrac{n-k}{n}<\dfrac {2}{3986}$

the rest is the same as the solution from anemone
in fact from (1) it is clear to see min(n)=3987

 

[Albert1]

we know $0<\dfrac {k}{n}<1$
using $0<\dfrac {1}{2}<\dfrac {2}{3}<\dfrac {3}{4}<--<\dfrac {3985}{3986}<\dfrac {3986}{3987}<\dfrac {3987}{3988}<1$
$2\times \dfrac {m}{3986}<\dfrac {k_0}{n_0}<2\times \dfrac{(m+1)}{3988}<1$
take $n=n_0 ,\,\, and \,\, k=k_0$
now it is easy to see that the smallese value of n =3987

the previous post (a mistake in calculation) has been changed above