elementary number theory - Remainder division using total digit

Find the remainder of the division of 1112131 41516171819 by 99
. (b) Find the remainder 111213141516171819 by 101

For a) I think I just need to sum up two digit such as 135 mod 99 =36
b.) sum up three digit each such as
1971 mod 101 then can i sum 197+1 mod 101?

Is this right?

Answer

$$N=19+18\times100+17\times100^2+16\times100^3+15\times100^4+14\times100^5+13\times100^6+12\times100^7+11\times100^8$$

a:$100≡1\ mod 99$ ⇒ $100^k≡1\mod 99$

⇒ $$N≡(11+12+13+14+15+16+17+18+19) mod 99 ≡135\ mod 99≡36 \ mod 99$$

b:$100≡ -1\ mod 101$ ⇒ $100^{2k}≡[(-1)^{2k}=1]\ mod 101$; $100^{2k+1}≡[(-1)^{2k+1}=-1]\ mod 101$

⇒ $$N≡(11-12+13-14+15-16+17-18+19) mod 101≡15\ mod 101$$