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roblema : patru numere consecutive , cu suma 12518 ;

Răspuns :

a+b+c+d=12518    b=a+1   c=b+1=a+2    d=c+1=b+2=a+3
a+a+1+a+2+a+3=12518
4a+6=12518
4a=12518-6
4a=12512
a=12512/4
a=3128
b=a+1=3128+1=3129
c=b+1=3129+1=3130
d=c+1=3130+1=3131