Question 476158: Does there exist a permutation a1, a2, . . . , a8 of the numbers 1, 2, . . . , 8, such that the 8 numbers a1, a1+a2, . . . , a1+a2+...+a8 all leave a di ффerent remainder when divided
by 8?
Answer by richard1234(7193)   (Show Source):
You can put this solution on YOUR website! 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 ) by taking the differences between successive terms like this:
 - (a_1) = a_2 \not\equiv 0 (mod 8))
 - (a_1 + a_2) = a_3 \not\equiv 0 (mod 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.
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