Chứng tỏ rằng: a, 7 mũ 8 cộng 7 mũ 9 cộng 7 mũ 10 chia hết cho 57

* Hướng dẫn giải

a, $7^8+7^9+7^{10}=7^8.\left(1+7+7^2\right)$ = $7^8.57⋮57$

b, ${10}^{10}-{10}^9-{10}^8$ = ${10}^8.({10}^2-10-1)$ = ${10}^8.89⋮89$

c, ${64}^{10}-{32}^{11}-{16}^3$ = ${\left(2^6\right)}^{10}-{\left(25\right)}^{11}-{\left(2^4\right)}^{13}$ = $2^{60}-2^{55}-2^{52}=2^{52}\left(2^8-2^3-1\right)$

= $2^{52}.247=2^{52}.13.19⋮19$