TITLE: 101102103104105106..............149150? What is the remainder when divided by 9?
QUESTION [13 upvotes]: 101102103104105106..............149150. What is the remainder when divided by 9?

My Approach
I think the remainder will be $0$.
Because I think all numbers are in multiplication and if I divide $108/9=0$ remainder, and thus multiplication of all these numbers will be $0$.
Can anyone guide me? Is my approach correct?

REPLY [5 votes]: Yet another possible shortcut is to notice that the sum of any 9 consecutive numbers is divisible by 9. Intuitively, one of the 9 numbers must be a multiple of 9, and the rest form symmetrical pairs around it whose mod 9 remainders "cancel out". Or, see Prove the sum of any $n$ consecutive numbers is divisible by $n$ (when $n$ is odd). for a more formal proof.
Since the mod 9 remainder of the given number is the same as that of the sum of the 50 consecutive numbers 101 + 102 + ... +150, 5 groups of 9 consecutive numbers can be dropped from that sum as having mod 9 remainders of 0. Choosing to drop the last 45 numbers leaves 101 + 102 + 103 + 104 + 105 which has a mod 9 remainder of 2.
[ EDIT ] To elaborate the "mod 9 remainder of the given number is the same as that of the sum of the 50 consecutive numbers 101 + 102 + ... +150".   
$10 ≡ 1\;(mod\;9)$ thus $10^2 = 10 * 10 ≡ 1 * 1 = 1\;(mod\;9)$ and by induction $10^n ≡ 1\;(mod\;9)$ for all $n >= 0$. The given number is $101 * 10^{148} + 102 * 10^{145} + ... + 149 * 10^3 + 150$, and since all factors $10^k ≡ 1\;(mod\;9)$, the sum simplifies $(mod\;9)$ to $101 + 102 + ... + 150$.