Question

Prove that 27 to the 10th-9 to the 14th is divisible by 24

Answer 1

Answer:

$27^{10}-9^{14}\\\\=(3^3)^{10}-(3^2)^{14}\\\\=3^{30}-3^{28}\\\\=3^{(27+3)}-3^{(27+1)}\\\\=3^{27} \cdot3^3-3^{27}\cdot3^1\\\\=3^{27}(3^3-3^1)\\\\=3^{27}(27-3)\\\\=3^{27}\cdot24$

$3^{27}\cdot24$ is divisible by 24, hence proving that $27^{10}-9^{14}$ is divisibe by 24