Răspunsuri

2014-02-14T18:38:43+02:00
2013=3*11*61; 1*2*3*...*70=(3*11*61)*1*2*4*5*...*70 =2013*1*2*4*5*...*10*12*13*...*60*62*...*70 
Aplicam Teorema impartirii cu rest:
2013*1*2*4*5*...*70+1234=2013*c+r , r<2013
Observam ca avem 2013 in ambii membri=> r=1234;  c=1*2*4*5*...*9*10*12*13*...*60*62*...*70 

in loc de prima relatie:2013=3*11*61 obs ca in 1*2*...*70 avem 3 11 si 61 deci putem scrie ca 3*11*61*1*2*4*...*70
de fapt am editat
e corect ce am scris