Difference between revisions of "1988 OIM Problems/Problem 2"

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== Problem ==
 
== Problem ==
Let <math>a</math>, <math>b</math>, <math>c</math>, <math>d</math>, <math>p</math>, and <math>q</math>, be non-zero natural numbers that verify <math>ad-bc=1</math>, and <math>\frac{a}{b} >\frac{p}{q}>\frac{c}{d}.
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Let <math>a</math>, <math>b</math>, <math>c</math>, <math>d</math>, <math>p</math>, and <math>q</math>, be non-zero natural numbers that verify <math>ad-bc=1</math>, and <math>\frac{a}{b} >\frac{p}{q}>\frac{c}{d}</math>.
  
 
Prove:
 
Prove:
  
(i) </math>q \ge b+d<math>
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(i) <math>q \ge b+d</math>
  
(ii) If </math>q=b+d<math>, then </math>p=a+c$.
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(ii) If <math>q=b+d</math>, then <math>p=a+c</math>.
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 +
~translated into English by Tomas Diaz. ~orders@tomasdiaz.com
  
 
== Solution ==
 
== Solution ==
 
{{solution}}
 
{{solution}}
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 +
== See also ==
 +
https://www.oma.org.ar/enunciados/ibe3.htm

Latest revision as of 12:28, 13 December 2023

Problem

Let $a$, $b$, $c$, $d$, $p$, and $q$, be non-zero natural numbers that verify $ad-bc=1$, and $\frac{a}{b} >\frac{p}{q}>\frac{c}{d}$.

Prove:

(i) $q \ge b+d$

(ii) If $q=b+d$, then $p=a+c$.

~translated into English by Tomas Diaz. ~orders@tomasdiaz.com

Solution

This problem needs a solution. If you have a solution for it, please help us out by adding it.

See also

https://www.oma.org.ar/enunciados/ibe3.htm