Question

determinati restul impartirii numarului N = 1 x 2 x 3 x ..... x 100 + 1001 la 120

Answer 1

[tex]120=5*3*8\\1*2*3*4*5*6*7*8...*100=\\(5*3*8)*(1*2*4*6*7*9*...*100)=M_1_2_0\\ 1001:120=8\ rest\ 41\\Restul=41 [/tex]