101102103104105106..............149150? What is the remainder when divided by 9?

$101102...149150=101(9999...9+1)+102(999...9+1)+...+149(999+1)+150$

$\equiv101+102+...+150\pmod{9}\equiv2+3+...+51\pmod{9}\equiv{(53)(50)\over2}=1325\equiv2\pmod{9}$


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 $$ \begin{align} \underbrace{101102...150}_{150\text{-digit number}}&\equiv 101+102+...+150\\ &=50\cdot\frac{101+150}{2}\\ &=6275\\ &\equiv 6+2+7+5\\ &\equiv 2\pmod9 \end{align} $$

Tags:

Elementary Number Theory

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