What would be the reminder when x^1001 - 1 is divided by x^4 + x^3 + 2x^2 + x + 1?

Question

find the middle term of the sequence formed by all three digits numbers which leave a reminder 3, when divided by 4. Also find the sum of all numbers on both sides of the middle term separately.

Answer 1

6

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x^1001-1 is divisible by x-1
if we divide x^4 + x^3 + 2x^2 + x + 1 by x-1
(x^4 + x^3 + 2x^2 + x + 1)/(x-1)=x3+2x2+4x+5 and a reminder of 6
in other words,
x^4 + x^3 + 2x^2 + x + 1=(x-1)*(x3+2x2+4x+5)+6
we should consider x>1, if x=1, we get zero