1.
A=2⁹×3⁷×5¹³
B=2²⁵×3⁸×5¹¹
A×B=2⁹×3⁷×5¹³ × 2²⁵×3⁸×5¹¹=2³⁴×3¹⁵×5²⁴=2¹⁰×3⁵×(2×5)²⁴=2¹⁰×3⁵×10²⁴
deci vom avea 24 de zerouri
2.
x=24a+14
24a divizibil cu 4
restul va fi dat de 14:4 =3 si rest 2 deci restul este 2
3.3ˣ+3ˣ⁺¹+3ˣ⁺²=351
3ˣ(3⁰+3¹+3²)=351
3ˣ × (1+3+9)=351
3ˣ × 13=351
3ˣ =27
3ˣ =3³
x=3
4.
restul poate fi={1,2,3,4,5,6}
7×3=21
7×3+1=22
7×3+2=23
7×3+3=24
7×3+4=25
7×3+5=26
7×3+6=27
nr sunt 21,22,23,24,25,26,27
