Question

suma a trei numere naturale consecutive se imparte la 7 si obtinem catul 24 si restul 6.Determinati numerele. MULTUMESC!!!

Answer 1

a+(a+1)+(a+2)=3(a+1)
3(a+1):7=24 si rest 6
3(a+1)=24×7+6

3a+3=168+6

3a=174-3

3a=171, a=171:3=57

a=57,
b=a+1=58
c=a+2=59

deci a=57, b=58  si c=59
verificare

Suma  se imparte la 7 si obtinem catul 24 si restul 6
 57+58+59=174
174:7=24  rest 6

Answer 2

[a+(a+1)+(a+2)]:7=24 rest 6
a+a+1+a+2=24*7+6
3a+3=168+6
3a+3=174  | - 3
3a=171
a=171:3
a=57 
a+1=57+1
a+1=58
a+2=57+2
a+2=59
Numerele consecutive sunt:
57 ; 58 ; 59
verificare
(57+58+59):7=24 rest 6
174:7=24 rest 6
174=24*7+6
174=168+6
174=174