Đáp án+Giải thích các bước giải:

Ta có:

`A=a^11-a=a(a^10-1)=a(a^5-1)(a^5+1)`

`=(a-1)a(a+1)(a^4+a^3+a^2+a+1)(a^4-a^3+a^2-a+1)`

$(a-1)a(a+1)\vdots6$ vì là tích 3 số nguyên liên tiếp 

nên $A\vdots6$

`+)` $a\vdots11⇒A\vdots11$

`+) a≡1(`mod `11)=>a-1`$\vdots11⇒A\vdots11$

`+)a≡-1(`mod `11)=>a+1`$\vdots11⇒A\vdots11$

`+)a≡2(` mod `11)=>a^5+1≡2^5+1≡0(` mod `11)`

`+)a≡-2(`mod `11)=>a^5-1≡-2^5-1≡0(`mod `11)`

`+)a≡3(`mod `11)=>a^5-1≡3^5-1≡0(` mod `11)`

`+)a≡-3(`mod `11)=>a^5+1≡-3^5+1≡0(`mod `11)`

`+)a≡4(` mod `11)=>a^5-1≡4^5-1≡0(` mod `11)`

`+)a≡-4(`mod `11)=>a^5+1≡-4^5+1≡0(` mod `11)`

`+)a≡5(`mod `11)=>a^5-1≡5^5-1≡0(`mod `11)`

`+)a≡-5(` mod `11)=>a^5+1≡-5^5+1≡0(` mod `11)`

`+)a≡6(`mod `11)=>a^5+1≡6^5+1≡0(`mod `11)`

$⇒A\vdots11$`∀ainZ`

mà `(6;11)=1`

`=>`$A\vdots11.6=66$