[tex]\overline{abc}=5k\\
5k:n=36\ rest\ 1=>5k=36n+1=>n= \frac{5k-1}{36} = \frac{40k-8}{288}[1]\\
5k:p=32\ rest\ 5=>5k=32p+5=>p= \frac{5k-5}{32} = \frac{45k-45}{288}[2]\\
p-n= \frac{5k-37}{288} = v\in N\\
5k-37=288v\\
k= \frac{288v+37}{5} = \frac{57\cdot5v+3v+5\cdot7+2}{5} =57v+7+ \frac{3v+2}{5}\in N \\
\frac{3v+2}{5}\in N=>v=1,6,...\\
Pentru\ v=1=>k=65\\
\overline{abc}=65\cdot5=325\\
325:9=36\ rest\ 1\\
325:10=32 \ rest\ 5\\
n=9\ si \ p=10 \ sunt \ numere \ consecutive.\\
[/tex]
Obs.Pentru v=6=>
1765=353*5
1765:49=36 rest 1
1765:55=32 rest 5
p-n=55-49=6 , adica daca numarul era de forma abcd, p era cu 6 mai mare decat n.
