\(7^{101}\equiv7\left(mod39\right)\)

\(13^{101}\equiv13\left(mod39\right)\)

\(19^{101}\equiv19\left(mod39\right)\)

\(\Rightarrow\left(7^{101}+13^{101}+19^{101}\right)\equiv7+13+19\left(mod39\right)\)

mà 7 + 13 + 19 = 39 chia hết cho 39 nên \(\Rightarrow7^{101}+13^{101}+19^{101}\)chia hết cho 39. ĐPCM

Sửa đề: Chứng minh (3¹⁰¹ + 7⁵³) ⋮ 10

Ta có:

3¹⁰¹ = 3.(3¹⁰)¹⁰

7⁵³ = (7⁷)⁷.7⁴

*) 3 ≡ 3 (mod 10)

3¹⁰ ≡ 9 (mod 10)

⇒ (3¹⁰)¹⁰ ≡ 9¹⁰ (mod 10) ≡ 1 (mod 10)

⇒ 3¹⁰¹ ≡ 3.(3¹⁰)¹⁰ (mod 10) ≡ 3.1 (mod 10) ≡ 3 (mod 10)

*) 7⁴ ≡ 1 (mod 10)

7⁷ ≡ 3 (mod 10)

⇒ (7⁷)⁷ ≡ 3⁷ (mod 10) ≡ 7 (mod 10)

⇒ 7⁵³ ≡ 7⁴.(7⁷)⁷ (mod 10) ≡ 1.7 (mod 10) ≡ 7 (mod 10)

⇒ 3¹⁰¹ + 7⁵³ ≡ 3 + 7 (mod 10) ≡ 10 (mod 10) ≡ 0 (mod 10)

Vậy (3¹⁰¹ + 7⁵³) ⋮ 10

\(1+7+7^2+7^3+...+7^{101}\\=(1+7)+(7^2+7^3)+(7^4+7^5)+...+(7^{100}+7^{101})\\=8+7^2\cdot(1+7)+7^4\cdot(1+7)+...+7^{100}\cdot(1+7)\\=8+7^2\cdot8+7^4\cdot8+...+7^{100}\cdot8\\=8\cdot(1+7^2+7^4+...+7^{100})\)

Vì \(8\cdot\left(1+7^2+7^4+...+7^{100}\right)⋮8\)

\(\Rightarrowđpcm\)

\(1+7+7^2+7^3+...+7^{101}\)

\(=\left(1+7\right)+7^2\left(1+7\right)+...+7^{100}\left(1+7\right)\)

\(=8\left(1+7^2+...+7^{100}\right)⋮8\)

\(P=\dfrac{1+19+\dfrac{19}{13}+\dfrac{19}{101}}{7+\dfrac{7}{13}+\dfrac{7}{19}+\dfrac{7}{101}}\)

\(=\dfrac{19\left(1+\dfrac{1}{3}+\dfrac{1}{19}+\dfrac{1}{101}\right)}{7\left(1+\dfrac{1}{13}+\dfrac{1}{19}+\dfrac{1}{101}\right)}=\dfrac{19}{7}\)

a) \(3^{10}+3^{11}+3^{12}\)

⇔ \(3^{10}\left(1+3+3^2\right)\)

⇔  \(3^{10}.13\) 

⇒   \(3^{10}.13\)  chia hết cho 13

a) \(3^{10}+3^{11}+3^{12}=3^{10}\left(1+3+3^2\right)=3^{10}\cdot13⋮13\)

b) \(5^{100}+5^{101}+5^{102}=5^{100}\left(1+5+5^2\right)=5^{100}\cdot31⋮31\)

Đặt A = 1 + 7 + 7² + ... + 7¹⁰¹

=> A = 7⁰ + 7¹ + ... + 7¹⁰¹

=> A = 7⁰ ( 1 + 7 ) + ... + 7¹⁰⁰ ( 1 + 7 )

=> A = 7⁰ . 8 + ... + 7¹⁰⁰ . 8

=> A = 8 . ( 7⁰ + ... + 7¹⁰⁰ ) chia hết cho 8 ( đpcm )

\(13^{101}-13=13\left(13^{100}-1\right)\)

Xét: \(169\equiv1\left(mod168\right)\Leftrightarrow169^{50}\equiv1\left(mod168\right)\Leftrightarrow13^{100}\equiv1\left(mod168\right)\)

\(\Leftrightarrow13^{100}-1\equiv0\left(mod168\right)\)<=>13¹⁰⁰-1 chia hết cho 168

=>13(13¹⁰⁰-1) chia hết cho 168=> đpcm

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