Polynomial

a) Find the remainder: ( x^2008+3)/ (x+1)
Using Remainder Theorem,
Put P(x) = x^2008+3
P(-1) = (-1)^2008 + 3 ~ Note: (-1)^even number = 1
P(-1) = 1 + 3
P(-1) = 4

The remainder is 4

b) If today is Saturday , what day of the week is the day after (6^2008+3) days?

6^2008 + 3 = (7 -1)^2008 + 3
Using Binomial Theorem, expand (7 -1)^(2008)
Then divide by 7 to find the remainder
(7 -1)^2008 = 7^2008 +2008/1*7^2007 * (-1) + … + …. (-1)^2008
= 7^2008 - 2008*7^2007 + … + …. + 1
All the terms are divisible by 7 except the last one
When (7 -1)^2008 is divided by 7, the remainder is + 1
When (7 -1)^2008 + 3 is divided by 7, the remainder is + 4
Saturday + 4 days = Wednesday

If today is Saturday, Wednesday is the day after (6^2008+3) days.

Comment:
You should put this question under 數學 section

If 6^(odd number) is divided by 7, the remainder is 6, provided the odd number > 1
If 6^(even number) is divided by 7, the remainder is 1
If you can write computer program in C, BASIC or FORTRAN, you can also confirm it besides using Binomial Theorem
There is a build-in function MOD in FORTRAN to find the remainder.