A=1+3+5+...+41
Folosim 1+3+5+...+2n-1=n^2, n>0 si nr. nat.
2n-1=41
n=21
A=21^2=(7^2)(3^2)
B=10(1+3+...+41)=10A=10x(7^2)(3^2)
C=100(1+3+...+41)=100A=100x(7^2)(3^2)
A+B+C=A+10A+100A=111A=111x(7^2)(3^2)=111x7x7x3x3=777x63
A+B+C este multiplu de 63, deci se divide cu el.
