Isn't there a shortcut?

Note that $128=2^{7}$ . For a number to be divisible by $2^{n}$ , the last $n$ digits have to be divisible by $2^{n}$ .

The multiple of $2^{7}$ closest to $999,999,999,999,987$ is

$999,999,999,999,987+13=1,000,000,000,000,000$

since $0000000$ is divisible by $128$ .

Therefore, $999,999,999,999,987 \equiv -13 \pmod{128} \equiv \boxed {115} \pmod{128}$