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?
$101102...149150=101(9999...9+1)+102(999...9+1)+...+149(999+1)+150$
$\equiv101+102+...+150\pmod{9}\equiv2+3+...+51\pmod{9}\equiv{(53)(50)\over2}=1325\equiv2\pmod{9}$
Assuming you mean this number:
101102103104105106107108109110111112113114115116117118119120121122123124125126127128129130131132133134135136137138139140141142143144145146147148149150
As you seem to know, the remainder when dividing by 9 is equal to the remainder why the sum of the digits is divided by 9. In this case the sum of the digits is 380. The sum of those digits is 11 so the remainder will be 2.
Since $10^n=(9+1)^n\equiv 1\pmod9$ we have $$ \begin{align} \underbrace{101102...150}_{150\text{-digit number}}&\equiv 101+102+...+150\\ &=50\cdot\frac{101+150}{2}\\ &=6275\\ &\equiv 6+2+7+5\\ &\equiv 2\pmod9 \end{align} $$
There's a bunch of different answers here. Looking at barak manos' idea is one way of doing it, but there is an arithmetical error.
Remember that the remainder mod 9 is invariant under taking sum of digits. That is, if we write a decimal expansion as $$ k = d_1d_2d_2\cdots d_{n-1}d_n $$ then $k \equiv \sum_{j=1}^n d_j \pmod 9$. So let's look at the digit sum in question. We have a concatenation of the numbers 101 through to 150. The first digit 1 occurs 50 times, so this will add 50 to our total. The remaining digits are the digit sums of 1 through to 50.
As is stated in the other answer, the ones column will produce each of the numbers 1 through 9 exactly five times; this will add $5(1 + \cdots + 9) = 225$ to our digit sum.
The tens column will add 0 nine times, 1 ten times, 2 ten times, 3 ten times 4 ten times, and 5 once. This adds up to $10(1 + 2 + 3 + 4) + 5 = 105$.
So our digit sum will be $$ 50 + 225 + 105 = 380 $$ Applying digit sum again we get $3 + 8 = 11$. So in the end we have $$ k \equiv 380 \equiv 11 \equiv 2 \pmod 9 $$
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 $101 + 102 + ... + 150$.
Here's another approach which allows us to determine the $\pmod9$ of any consecutive sum with at most four additions and at most 3 $\pmod x$ evaluations. Quite efficient!
For any three consecutive numbers, the remainder of their sum and nine will be the same as the next three consecutive (because each element in the triplet is three more than that same position in the previous triplet, and $3+3+3=9$).
We know our three numbers are $101+102+103=306\equiv0\pmod9$. Thus each consecutive triplet will also be $\equiv0\pmod9$. Evaluating the remaining elements (in this case 2) results in elements $$149+150=299\equiv2\pmod9$$.
If the number of elements was divisible by three, we could omit the last part because there would be no remaining elements. If our triplet was a non-zero remainder, we would have to add that to the sum of the remaining elements before modding. (We could add it to the resulting mod but if it goes over 9 we would have to mod again which would be inefficient...)
If you're wondering where the third $\pmod x$ evaluation is, we have to do a $n \pmod 3$ to determine how many elements are remaining.
The summation of digits at hundreds place is $50$ then the summation of digits at units place is $45.5=225$ notice the last digits they are $1,2,3,4,5,6,7,8,9,0 $ they are repeated $5$ times And summation of digits at tens place is $10+20+30+40+5=105$ so summation of all is $50+105+225=380$ so remainder is $2$. Hope this hepls you.
In base $10$, you can sum the digits in an iterative way. E.g.
193 -> 13 -> 4 (193 is not divisible by 9).
198 -> 18 -> 9 (198 is divisible by 9).
Due to this base is $10$, you can even add the number by parts and "reduce" them later. Since I could reduce a number like this:
101102 -> 101 + 102 -> 203 -> 5 (not divisible by 9)
I could make it in a bigger scale. Lets just explain why we could reduce like that:
101102 = 101*999 + 101 + 102 -> 101 + 102
(we discard the first term since it is already divisible by 9)
And we can do that for any order of digits, like:
101102103 = 101*999999 + 101 + 102*999 + 102 + 103 -> 101 + 102 + 103
(the *999999 and *999 terms were discarded since they are already divisible by 9)
So we can reduce your big number to:
101 + ... + 150 = 251 * 50 / 2 = 6275 -> 20 -> 2
The remainder is $2$.
