1. Let $a, b, c, d$ be integers, and $a<2b, b<3c, c<4d$. Given $d<100$, then the maximum possible value of $a$ is ( ).
(A) 2367
(B) 2375
(C) 2391
(D) 2399
$-.1 .(\mathrm{A})$.
Then, if $d=99$, we have $c<4 \times 99=395$. Taking $c=395$, we get $b<3 \times 395=1185$. Taking $b=1184$, we get $a<2368$. Therefore, the greatest integer for $a$ is 2367.