Question

suma a trei nr consecutive se imparte la 7 si obtinem catul 27 si restul 6. determinati numerele

Answer 1

$(a+a+1+a+2):7=27r6 \\ a+a+1+a+2=7*27+6 \\ a+a+1+a+2=195 \\ 3a+3=195 \\ 3a=195-3 \\ a=192:3 \\ a=64 \\ nr.consecutive.sunt.64,65,66$

Answer 2

(x+x+1+x+2):7-24 rest 6
3x+3=24x7+6
3x+3=195
3x=195-3
3x=192
x=192:3
x=64
x+1=64+1=65
x+2=64+2=66