數學問題,數學問題,數學問題

7 + 7^2 + 7^3 + 7^4 + 7^5 + 7^6 + 7^7 + 7^8 + … + 7^2008
= (7 + 7^2 + 7^3 + 7^4) + 7^4 * (7 + 7^2 + 7^3 + 7^4) + 7^8 * (7 + 7^2 + 7^3 + 7^4) + … + 7^2004 * (7 + 7^2 + 7^3 + 7^4)
Since 7 + 7^2 + 7^3 + 7^4 = 7 + 49 + 343 + 2401 = 2800 is divisible by 100,
7 + 7^2 + 7^3 + 7^4 + 7^5 + 7^6 + 7^7 + 7^8 + … + 7^2008 is divisible by 100
Represent it as 100N
1 + 100N
= 1 + 7 + 7^2 + 7^3 + 7^4 + 7^5 + 7^6 + 7^7 + 7^8 + … + 7^2008
= (7^2009 – 1) / (7 – 1) ; use formula for sum of geometric sequence
= (7^2009 – 1) / 6
So 7^2009 – 1 = 6 + 600N
7^2009 = 7 + 600N
Therefore the remainder when 7^2009 is divided by 100 is 7

7^2009除100的餘數是多少
Sol
7^2009
=>[(7^4)^502]*7
=>[(2401)^502]*7
=>(01)^502*7
=>1*7
=>7

答案是 "7" ~~~~~~~~~