6
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
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.