Question 1205868
<pre>

Let n be the smallest possible answer.
Then there are positive integers p,q,r such that

{{{system(4p+2=n,9q+5=n,7r+2=n)}}}

{{{4p+2=7r+2}}}

{{{4p=7r}}}

{{{p/r=7/4}}}

{{{system(p=7k,r=4k)}}} with k positive, and small as possible

{{{system(4(7k)+2=n,9q+5=n,7(4k)+2=n)}}}

{{{system(28k+2=n,9q+5=n,28k+2=n)}}}

{{{system(28k+2=n,9q+5=n)}}}

{{{28k+2=9q+5}}}

{{{28k=9q+3}}} 

{{{(27+1)k=9q+3}}}  Writing all integers in terms of their nearest multiple of
              the smallest coefficient of a variable in absolute value. A
              trick used in solving Diophantine equations.  

{{{27k+k=9q+3}}} 

{{{3k+k/9=q+3/9}}}  dividing through by that coefficient 9.

{{{3k-q=3/9-k/9=A}}} Getting fractions together and setting equal to some
                     integer A

{{{27k-9q=3-k=9A}}}

{{{3-k=9A}}}

{{{k=3-9A}}}

{{{A=0}}} the only value that makes k positive and small as possible

{{{k=3}}}

{{{system(p=7k,r=4k)}}}

{{{system(p=7(3),r=4(3))}}}

{{{system(p=21,r=12)}}}

{{{28(3)=9q+3}}}

{{{84=9q+3}}}

{{{81=9q}}}

{{{9=q}}}

{{{system(4(21)+2=n,9(9)+5=n,7(12)+2=n)}}}

{{{system(84+2=n,81+5=n,84+2=n)}}}

{{{system(86=n,86=n,86=n)}}}

The answer is 86.

Edwin</pre>