Delete the number!

Without loss of generosity, suppose $x>y>z$ . We are essentially sorting the tuple $(x,y,z)$ and find the 100-th number.

When $x<9$ there are only ${9\choose 3} = 84<100$ combinations of $(x,y,z)$ while $x\leq 9$ has ${10\choose 3}=120>100$ combinations. Hence, $x=9$ .

Next, we want to find the $100-84=16$ -th tuple $(y,z)$ . When $y<6$ , there are only ${6\choose 2}=15<16$ combinations of $(y,z)$ while $y\leq 6$ has ${7\choose 2}=21>16$ combinations. Hence, $y=6$ .

Finally, $z$ is the $16-15=1$ -st number, which is simply 0. So the answer is $2^9+2^6+2^0=577$ .

I can write the generalised solution, but there are too many interesting problems from @Áron Bán-Szabó that are waiting for me to solve!