Smallest integers after N divisible by 2, 3, and 4
Python 2, 32 bytes

lambda n:[n+2&amp;-2,n/3*3+3,n+4&amp;-4]

Bit arithmetic for 2 and 4, modular arithmetic for 3.
I found four 7-byte expressions for the next multiple of k above n but none shorter:
n-n%k+k
~n%k-~n
n/k*k+k
~n/k*-k

Any gives 34 bytes when copies for k=2,3,4, and 33 bytes if combined:
[n/2*2+2,n/3*3+3,n/4*4+4]
[n/k*k+k for k in 2,3,4]

But, 2 and 4 are powers of 2 that allow bit tricks to zero out the last 1 or 2 bytes.
n+2&amp;-2
n+4&amp;-4

This gives 6 bytes (instead of 7) for getting the next multiple, for 32 bytes overall, beating the for k in 2,3,4.
Unfortunately, the promising-looking n|1+1 and n|3+1 have the addition done first, so incrementing the output takes parentheses.