What is the sum R of all 3 digit numbers that leave a remainder of '2' when divided by 3?
Answers:
A) 897
B) 164,850
C) 164,749
D) 149,700
E) 156,720
The series is =&gt; 101+...998   Number of terms =&gt; 300   sum =&gt; 300/2 * [101+998]   hence sum R=&gt; 164850 i.e. option 164,850
<result>B</result>
What is the sum R of all 3 digit numbers that leave a remainder of '2' when divided by 3?