Question

Find the sum of all even integersbetween 100 and 800 divisible by 6

Answer 1

N = 16k+7 (k:integer) 

N=100 ==> k=93/16 =5.81 ==> k = 6 
====> N = 16*6 + 7 = 103 
N=800 ==> k=793/16 =49.56 ==> k = 49 
====> N = 16*49 + 7 = 791 

from 6(included) to 49(included) : total numbers = 44 
it is an arithmetic progression and the sum 
= (103+791)*44/2 = 894*22 = 19668 
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if it is divisible by 7
I think this is the answer