Question

suma a opt numere naturale impare consecutiveeste egala cu 192.determinati cele opt numere.

Answer 1

a+b+c+d+e+f+g+h=192
a+(a+2)+(a+4)+(a+6)+(a+8)+(a+10)+(a+12)+(a+14)=192
8a+56=192
8a=192-56
8a=136
a=136:8
a=17
b=17+2
b=19
c=17+4
c=21
d=17+6
d=23
e=17+8
e=25
f=17+10
f=27
g=17+12
g=29
h=17+14
h=31
proba:17+19+21+23+25+27+29+31=
           =36+21+23+25+27+29+31=
           =57+23+25+27+29+31=
           =80+25+27+29+31=
           =105+27+29+31=
            =132+29+31=
            =161+31=192

Answer 2

a+1+a+3+a+5+a+7+a+9+a+11+a+13+a+15=192
8a=192-64
a=128:8
a=16
a+1=17
a+3=19
a+5=21
a+7=23
a+9=25
a+11=27
a+13=29
a+15=31
17+19+21+23+25+27+29+31=192(verificare)