Question

determinati cel mai mic numar natural care impartit pe rand la 12 ,15su 27 da de fiecare data restul 3 si caturi nenule

Answer 1

x:12=c1 rest 3
x=c1*12+3
x-3=c1*12
x-3=M12
x:15=c2 rest 3
x=c2*15+3
x-3=c2*15
x-3=M15
x:27=c3 rest 3
x=c3*27+3
x-3=c3*27
x-3=M27
Deci:
x-3=M12 intersectat cu M15 intersectat cu M27
x-3=M(12,15,27)
x-3=cmmmc(12,15,27)
Descompunem numerele 12,15,27 in factori primi
12/3
4/2
2/2
1/
12=2^2*3^1
15/5
3/3
1/
15=5^1*3^1
27/3
9/3
3/3
1/
27=3^3
Se iau toti factorii primi la puterile cele mai mari
cmmmc(12,15,27)=3^3*5^1*2^2=27*5*4=27*20=540
x-3=540
x=540+3
x=543
Proba:
543:12=45 rest 3
543:15=36 rest 3
543:27=20 rest 3