- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,937

- Thread starter anemone
- Start date

- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,937

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

- Thread starter
- Admin
- #3

- Feb 14, 2012

- 3,937

HiTerrific 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

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

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

30 minutes in the corner, please...Unfortunately, got no exotic formula for you

No fair! You only asked: "Find the smallest positive integer [FONT=MathJax_Math-italic]n" [/FONT]30 minutes in the corner, please...

- Thread starter
- Admin
- #5

- Feb 14, 2012

- 3,937

- Jan 25, 2013

- 1,225

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

Last edited:

- Thread starter
- Admin
- #7

- Feb 14, 2012

- 3,937

Hiwe 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

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,

- Moderator
- #8

- Feb 7, 2012

- 2,799

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.)

- Thread starter
- Admin
- #9

- Feb 14, 2012

- 3,937

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.

- Jan 25, 2013

- 1,225

sorry a mistake in calculation , now I will do this way :HiAlbert,

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?

$\dfrac {2m}{3986}\leq \dfrac{3984}{3986}\approx 0.999498243$

$\dfrac {2m+2}{3988}\leq \dfrac{3986}{3988}\approx 0.999498495$

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

Last edited:

- Jan 25, 2013

- 1,225

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