Why is x/10 optimized with an unnecessary shift when x has a restricted range?

Question

I have this function

long long int divideBy10(long long int a){
    return a / 10;
}

it's compiled to:

        mov     rax, rdi
        movabs  rcx, 7378697629483820647
        imul    rcx
        mov     rax, rdx
        shr     rax, 63
        sar     rdx, 2
        add     rax, rdx
        ret

if I add __builtin_assume(a > 0);

it's compiled to

        mov     rax, rdi
        movabs  rcx, -3689348814741910323
        mul     rcx
        mov     rax, rdx
        shr     rax, 3
        ret

The code is more efficient because it doesn't have to worry about negative signs. Now if I add __builtin_assume(a < 10000); I would've expected it to be compiled to one multiplication without shifts. but that's not the case.

I thought maybe the compiler only tracks if the number is positive or negative but

long long int noBranch(long long int a){
    __builtin_assume(a < 400);
    if( a < 500){
        return a;
    }
    return 0;
}

compiles to just a move instruction so the compiler is able to track these bounds.

Why are compilers not able to optimize the shift away? (this happens both in GCC and Clang).

This is the code, that I expected to be produced:

long long int d10(long long int a){
    __builtin_assume(a > 0);
    __builtin_assume(a < 10'000);
    __uint128_t e = a;
    e *= 0x199999999999999A;
    return e >> 64;

}

d10(long long):                                # @d10(long long)
        mov     rax, rdi
        movabs  rcx, 1844674407370955162
        mul     rcx
        mov     rax, rdx
        ret

Edit: I'm aware that you need a shift to be able to cover the whole range of long long int. I'm asking why can't compiler optimize the shift away if the range is restricted.