`a, A = 1 + 5 + 5^2 + .... + 5^179`

`=> A = (1 + 5) + (5^2 + 5^3) + ... + (5^178 + 5^179)`

`=> A = 6 + 5^2 (1 + 5) + ..... + 5^178 (1 + 5)`

`=> A = 6 + 5^2 . 6 + .... + 5^178 . 6`

`=> A = 6 . (5^2 + 5^4 + ... + 5^178 ) \vdots 6`

`=> A \vdots 6` ( `đpcm` )

`b, A = 1 + 5 + 5^2 + .... + 5^179`

`=> A = (1 + 5 + 5^2 ) + 5^3 . (1 + 5 + 5^2 ) + .... + 5^177 ( 1 + 5 + 5^2 )`

`=> A = 31 + 5^3 . 31 + .... + 5^177 . 31`

`=> A = 31 . (1 + 5^3 + .... + 5^77) \vdots 31`

`=> A \vdots 31`

Lại có `A \vdots 6` ( chứng minh ở câu a )

Mà `6` và `31` là 2 số nguyên tố cùng nhau
Nên `A \vdots 31 . 6 = 186`

Vậy `A \vdots 186` ( `đpcm` )

Lời giải chi tiết:

`a)` Ta có`:`

`\text{A}=1+5+5^2+...+5^179`

`\text{A}=(1+5)+(5^2+5^3)+...+(5^178+5^179)`

`\text{A}=1.(1+5)+5^2.(1+5)+...+5^178.(1+5)`

`\text{A}=1.6+5^(2).6+...+5^(178).6`

`\text{A}=6.(1+5^2+5^178)\vdots6`

Vậy `\text{A}\vdots6`

`b)`

Ta có`:` `\text{A}=1+5+5^2+...+5^179`

`\text{A}=(1+5+5^2+5^3+5^4+5^5)+(5^6+5^7+5^8+5^9+5^10+5^11)+...+(5^174+5^175+5^176+5^177+5^178+5^179)`

`\text{A}=1.(1+5+5^2+5^3+5^4+5^5)+5^6.(1+5+5^2+5^3+5^4+5^5)+...+5^174.(1+5+5^2+5^3+5^4+5^5)`

`\text{A}=1.3906+5^(6).3906+...+5^(174).3906`

`\text{A}=3906.(1+5^6+...++5^174)`

`\text{A}=186.21.(1+5^6+...++5^174)\vdots186`

Vậy `\text{A}\vdots186`