n:24=c1r14
n:35=c2r25
n:45=c3 r 35
scriu teorema impartirii cu rest pt fiecare situatie
n=24c1+14=24c1+24-10=24(c1+1)-10
n=35c2+25=35c2+35-10=35(c2+1)-10
n=45c3+35=45c3+45-10=45(c3+1)-10
adun 10 la fiecare ecuatie
n+10=24(c1+1)=35(c2+1)=45(c3+1)
rezulta ca n+10 e multiplu comun al nr 24,35 si 45
aflu c.m.m.m.c. al nr
24=2^3*3
35=5*7
45=3^2*5
[24,35,45]=2^3*3^2*5*7=2520
n+10=2520, n=2520-10=2510
d. Se rezolva observand ca diferenta intre impartitor si rest e aceeasi adica 7
n:12=c1r5⇒n=12c1+12-7=12(c1+1)-7
n:15=c2r8 ⇒n=15c2+15-7=15(c2+1)-7
n:18=c3 r 11 ⇒ n=18c3+11=18c3+18-7=18(c3+1)-7
adun 7 si obtin ca n+7=12(c1+1)=15(c2+1)=18(c3+1)
inseamna ca n+7 e c.m.m.m.c. al nr 12,15 si 18
12=2^2*3
15=3*5
18=2*3^2
[12,15,18]=2^2*3^2*5=180
n=180+7=187
Sper ca ai inteles!
