Question

Sa se determine restul impartirii numarului 1x2x3x4x...x69x70 +1234 la 2013

Answer 1

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