Totally logical : There's only two possible positions, basing on the corners method, for the first 3 of R1 : beginning in C1 or in C7. If it begins in C7, by testing the consequences of each position of the 4 in C4 with the consequences on R5 and/or R6 and, consequently, on R3, you obtain an impossibility for C3 and C5. So the only position is to begin in C1. By symmetry, the last 3 of R1 ends in C15. After this, you can fulfill R5 and R6. And the rest will come logically to the unique solution.