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?
Best Answer
$101102...149150=101(9999...9+1)+102(999...9+1)+...+149(999+1)+150$
$\equiv101+102+...+150\pmod{9}\equiv2+3+...+51\pmod{9}\equiv{(53)(50)\over2}=1325\equiv2\pmod{9}$