Question

Answer 1

\(A=1+7+7^2+...+7^{11}\)

=>\(7A=7+7^2+7^3+...+7^{12}\)

=>\(6A=7^{12}-1\)

\(=\left(7^2-1\right)\left(7^2+1\right)\left(7^4+7^2+1\right)\left(7^4-7^2+1\right)\)

=>\(6A=50\cdot48\left(7^4+7^2+1\right)\left(7^4-7^2+1\right)⋮10\)

Answer 2

A = 1 + 7 + 7² + ... + 7¹¹

⇒ 7A = 7 + 7² + 7³ + ... + 7¹²

⇒ 6A = 7A - A

= (7 + 7² + 7³ + ... + 7¹²) - (1 + 7 + 7² + ... + 7¹¹)

= 7¹² - 1

Ta có:

7³ ≡ 3 (mod 10)

7¹² ≡ (7³)⁴ (mod 10) ≡ 3⁴ (mod 10) ≡ 1 (mod 10)

⇒ 7¹² - 1 ≡ 0 (mod 10)

⇒ (7¹² - 1) ⋮ 10

Vậy 6A ⋮ 10