Question 476158
Yes. Set a_1 = 8, a_k = k-1 for k > 1. To simplify things, we can actually set a_1 = 0 since this will not change the remainder mod 8. The sums will be 0, 1, 3, 6, 10, 15, 21, 28 (all triangular numbers except for 0). These numbers mod 8 are 0, 1, 3, 6, 2, 7, 5, 4, all different.


Note that you can actually prove that *[tex \LARGE a_1 \equiv 0 (mod 8)] by taking the differences between successive terms like this:


*[tex \LARGE (a_1 + a_2) - (a_1) = a_2 \not\equiv 0 (mod 8)]


*[tex \LARGE (a_1 + a_2 + a_3) - (a_1 + a_2) = a_3 \not\equiv 0 (mod 8)]

.
.
.
*[tex \LARGE (a_1 + ... + a_8) - (a_1 + ... + a_7) = a_8 \not\equiv 0 (mod 8)]


Since a_2, ..., a_8 cannot be 0 mod 8, it follows that a_1 = 8.