1)Scrieti numerele naturale cel mult egale cu 130, care au exact opt divizori.
8=2*2*2 sau 2*4 sau 4*2
Asadar: n+1=2 => n=1
sau n+1=4 => n=3
deci puterile pot fi :
a¹*b¹*c¹ , sau a¹*b³ , sau a³*b¹
cei mai mici a,b,c sunt 2,3,5,7,11,13,17,19,23...
adica:
2*3*5=30=1,2,3,5,6,10,30
2*3*7=42=1,2,3,6,7,14,21,42
2*3*11=66= 1,2,3,6,11,22,33,66
2*3*17=102=1,2,3,6,17,34,51,102
2*3*19=114=1,2,3,6,19,38,57,114
2*3*23=nu=138>130
2*5*7=70=1,2,5,7,10,14,35,70
2*5*11=110=1,2,5,10,11,22,55,110
2*5*13=130=1,2,5,10,13,26,65,130
2*5*17=170>130=nu
2*7*11=154>130=nu
2*11*13=286>130=nu
3*5*7=105=1,3,5,7,15,21,35,105
3*5*11=165>130=nu
3*7*11=231>130=nu
5*7*11=285>130=nu
2*3³=2*27=54=1,2,3,6,9,18,27,54
2*5³=2*125=250>130-nu
2³*3=24=1,2,3,4,6,8,12,24
2³*5=40=1,2,4,5,8,10,20,40
2³*7=8*7=56=1,2,4,7,8,14,28,54
2³*11=88=1,2,4,8,11,22,44,88
2³*13=8*13=104=1,2,4,8,13,26,52,104
2³*17 =136>130 nu
Asadar,
numerele naturale cel mult egale cu 130, care au exact 8 divizori sunt:
24,30,40,42,54,56,66,70,88,102,104,105,110,114,130.
2)Determinati nt natural n , stiind ca:
a) n+4 este divizor al nr. 18
n+4=18 =>n=14
n+4=9 =>n=5
n+4=6 =>n=2
=>n∈{2,5,4}
b) 18 supra n+2 apartine numerelor naturale.
18/(n+2) =1 =>n=18-2=16
18/(n+2) =2 =>2n=18-4=14 => n=7
18/(n+2) =3 =>3=18-6=12 =>n=4
18/(n+2) =4 sau 5 nu se poate deoarece 4 si 5 nu sunt divizori ai lui 18
18/(n+2) =6 =>6n=18-12=6 =>n=1
18/(n+2) =18 =>18n=18-36=-18 => n= -1 nu este natural
n∈{1,4,7,16}
