✔ 最佳答案
1)
For n = 1
7^1+4^1+1 = 12
which is divisible by 6
so it is true when n = 1
Assume it is true when n = k
i.e. 7^k + 4^k +1 is also divisible by 6
when n = k+1
7^(k+1) + 4^(k+1) + 1 = 7*7^k + 4*4^k + 1
= 7*7^k + 7*4^k + 7 - 3*4^k - 6
= 7(7^k + 4^k + 1) - 6 * 2 * 4^(k-1) - 6
as 7^k + 4^k + 1 is divisible by 6
so 7(7^k + 4^k + 1) - 6 * 2 * 4^(k-1) - 6 would also divisible by 6
as it is true when n = k +1 is it is true when n = k
so by mathematical induction, we have it is true for all positive integer
7^2006+4^2008 = 7*7^2006 - 6*7^2006 + 4*4^2007
= 7^2007 + 4^2007 + 1 - 6*7^2006 + 3*4*4^2006 - 1
= 7^2007 + 4^2007 + 1 - 6*7^2006 + 6*2*4^2006 - 1
as 7^2007 + 4^2007 + 1 is divisible by 6 (n = 2007)
hence the reminder of 7^2006+4^2008 divided by 6 would be 5 (not -1)
第2條~ 睇唔明個問題 2^3n 應該冇可能divisible by 7
3) 27^2007 = (3*9)^2007 = 3^2007*9^2007 = 3*3^2006*9^2007
= 3*9^(1003+2007) = 3*9^3010
= 3(9^3010 + 7) -21
as 9^n+7 is divisible by 8 for all posotive integer n,
so 3(9^3010 + 7) would also divisible by 8
hence the reminder of 27^2007 divided by 8 would be 3 (not -21)
2007-12-18 12:22:27 補充：
2^3n -1 is divisible by 7find the remainder when 2^200 is divided by 7 2^200 = 2^(3*66+2) = 4*2^(3*66) = 4*(2^3^66 -1) + 4由於 2^3n -1 is divisible by 7 for positive integer n所以 2^3^66-1 is also divisible by 7所以個reminder 係 4
