Find the sum of all natural numbers in 300and 500 leaving reminder 2 on division by 7
Answers
Answer:
11658
Step-by-step explanation:
If the numbers are to be strictly between 300 and 500, then their sum, S, will be:
S = 304 (i.e. 7 x 43 + 3) + 311 + 318 + … + 493 = 11158. However, the upper limit of
500 is such a number (7 x 71 +3). If you meant to include this, then S will be 11658.
Obviously, you or anyone with a calculator could have found this result by brute force, but what fun is that? Let’s throw a little math at the problem and recognize S as the sum of terms of an arithmetic series that begins 3, 10, … , 304, 311, … 493, 500, … . We are not concerned with any terms before 304 or after 500, so let’s just denote the terms of interest as 304, {304 + 7(1)}, {304 + 7(2)}, … , {304 + 7(27)}. This way
S = 304 (28) + 7 (1 + 2 + 3 + … + 27) = 8512 + 7(27 x 28/2) = 8512 + 2646 + 11158, if
we stop the series at 493 or S = 11658 if we include 500. Hope this helped!