What is the remainder when the sum $1 + 7 + 13 + 19 + \cdots + 253 + 259$ is divided by $6$?
+36915 Remainder = 2
The sum of the sequence is 5270 5270 / 6 = 878 R 2
+128408 1 + 7 + 13 + 19.........+ 253 + 259
(1 + 259) + ( 7 + 253) + .......
We have ( 259 -1 ) / 6 + 1 = 44 terms
22 pairs of them each sum to 260
So
22 * 260 = 5720
An even number whose digits sum to a multiple of 3 will be divisible by 6
So
5718 digit sum = 21 = multiple of 3 thus divisible by 6
So
5720 / 6 leaves a remainder of 2
+874 1+7+13+...+259
floor(259/6)=43
44(1)+43(6)=44+43*6=44(mod 6) = 2 (mod 6)
remainder is 2.
