因式定理追問

(x^10-1)=(x-1)(x^9+x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1)

Since x-1 is a factor of x^10-1. Sub. x=11, we have 10|11^10-1...(1)

On the other hand , the unit digit of 11^9, 11^8, ... 11 are 1 and so the unit digit of the sum of S=x^9+x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1 is 0. This implies that 10 | S. Combine the result with (1), we conclude that 100 is a factor of 11^10 -1.

x^10 - 1 = (x - 1)^2 Q(x) + ax + b
Put x = 1
0 = a + b ......(1)
Differentiate both sides,
10x^9 = (x - 1)^2 Q'(x) + 2( x - 1) Q(x) + a
Put x = 1
10 = a
From (1) b = - 10. That means
x^10 - 1 = (x - 1)^2 Q(x) + 10x - 10
(x^10 - 1) + (10 - 10x) = (x - 1)^2 Q(x)
Put x = 11
(11^10 - 1) + ( 10 - 110) = 100 Q(11)
(11^10 - 1) - 100 = 100 Q(11)
Since (11^10 - 1) - 100 is divisible by 100, 11^10 - 1 must be divisible by 100 or 100 is a factor of 11^10 - 1.

另解: 只作參考
11^10 - 1
= (10 + 1)^10 - 1
By binomial theorem
= {∑[n=0 to 10](10Cn)(10)^n} - 1
= {∑[n=2 to 10](10Cn)(10)^n} + (10C1)(10) + (10C0)(1) - 1
= {∑[n=2 to 10](10Cn)(10)^n} + 100
∑內最小n = 2, 即所有項都大或等於100的倍數
因此11^10 - 1 = {∑[n=2 to 10](10Cn)(10)^n} + 100是100的倍數