Cheating a multiple choice test, part 2

05AB1E, 19 18 16 bytes

Code:

žH+b¦2äøC’c‰±’sè

Uses the CP-1252 encoding. Try it online!

Explanation:

First, we add 65536 to the number (žH is a constant defined to 65536), which is also 10000000000000000 in binary. This is to pad the number with zeroes. Let's take the number 14234 as an example. 14234 + 65536 is equal to 79770. Which in binary is:

10011011110011010

We remove the first character, resulting in:

0011011110011010

We split the string into two pieces using 2ä:

00110111, 10011010

After that, we zip the array with ø:

01, 00, 10, 11, 01, 10, 11, 10

Converting them back into decimal (using C) results in:

1, 0, 2, 3, 1, 2, 3, 2

Now, we only need to index it with the string cbad. The compressed version for this string is ’c‰±’, which can also be tested here. Finally, we get the characters at the index of the above array. For the above example, this results in:

1, 0, 2, 3, 1, 2, 3, 2
b  c  a  d  b  a  d  a

JavaScript (ES6), 55 48 bytes

f=(n,i=8)=>i--?"CBAD"[n>>i&1|n>>i+7&2]+f(n,i):''

console.log(f(14234)); // BCADBADA
console.log(f(38513)); // ABBDCAAB
console.log(f(0));     // CCCCCCCC
console.log(f(120));   // CBBBBCCC
console.log(f(65535)); // DDDDDDDD
console.log(f(39253)); // ABCDABCD

Non-recursive version (55 bytes)

Using a regular expression, we can do:

n=>"76543210".replace(/./g,i=>"CBAD"[n>>i&1|n>>+i+7&2])

Python 2, 53 bytes

f=lambda n,k=8:k*'_'and f(n/2,k-1)+'CBAD'[n>>7&2|n&1]

Test it on Ideone.