Question

aflati ca numarul A+B+C este divizibil cu 63,unde A=1+3+...+41,B=10+30+...+410 si C=100+300+...+4100.

Answer 1

A=suma Gauss=[n(n+1)]:2=(41×42):2=861=A
B=10+30+...+410 progresie aritmetica cu an=410, a1=10 si r=20
an=a1+(n-1)×r
410=10+(n-1)×20
410=10+20n-20
420=20n
n=21
S=[(10+410)×21]:2=8610,B=4410
C=100+300+...+4100
an=4100, a1=100, r=200
an=a1+(n-1)×r
4100=100+(n-1)×200
4100=100+200n-200
4200=200n
n=4200:200
n=21
S=[(a1+an)×n]:2=[(4100+100)×21]:2=44100=C
A+B+C=861+4410+44100=49371