How to find the nearest multiple of 16 to my given number n If I'm given any random $n$ number. What would the algorithm be to find the closest number (that is higher) and a multiple of 16. 
Example $55$
Closest number would be $64$
Because
$16*4=64$
Not
$16*3=48$ because its smaller than $55$.
 A: If you are expressing the number in binary format, you could throw out the last 4 bits and add one and multiply by 16.  This does assume that given a multiple of 16, the number desired is strictly higher.  If in the case where n is a multiple of 16 the answer should be n, then you'd have to check first if the last 4 bits are all zero.
So, in the case of 55 which is 110111 in base 2, this would then becomes 11 in base 2 which is 3 and then adding one gives 4 which times 16 produces 64.

There are Bitshift operators in C that could be used so you could have a function that takes in a parameter then performs the following series of operations(using Rn's suggestion):
int a;
a = n-1;
a = a >> 4; /* which is the same as dividing by 16. */
a = a + 1;
return a << 4; /* which is the same as multiplying by 16 */

A: As you are surely trying to do this in a computer program, try the following C expression:  $(x|15)+1$. This will always increase, even if $x$ is already a multiple of $16$.
Or try $((x-1)|15)+1$ if you don't want to increase the number if it is already a multiple of $16$.
A: Use $16\lceil\frac{n}{16}\rceil$ to find the smallest multiple of $16$ not smaller than $n$, where the ceiling function $\lceil x\rceil$ denotes the smallest integer not smaller than $x$.
Use $16\lfloor\frac{n}{16}\rfloor+16$ to find the smallest multiple of $16$ larger than $n$, where the floor function $\lfloor x\rfloor$ denotes the largest integer not larger than $x$.
A: A summary of some of the other code, plus a few related items.
Find the nearest multiple of M
For a given number X:
( x - ( x % m ) )

At or above a given number X*:
( ( x - 1 ) | ( m - 1 ) ) + 1

Below a given number X*:
max( 0, ( ( x - 1 ) | ( m - 1 ) ) + 1 - m )

Test if X is a multiple of M
( x % m == 0 )

Round X to the nearest multiple of M
Standard rounding:
( m == 0 ) ? x : floor( ( x / m ) + 0.5 ) * m

Down to the nearest multiple (floor, toward -inf):
( m == 0 ) ? x : floor( x / m ) * m

Up to the nearest multiple (ceil, toward +inf):
( m == 0 ) ? x : ceil( x / m ) * m

Legend: C-like syntax: | bitwise OR, % modulo, x?y:z ternary ("if x then y else z"), == comparison.
* Generalized for any 2.
More Info: math.stackexchange.com/questions/1798899
A: Using & as bitwise AND, let a = n & 15, then n - a + ((a+15) & 16) is what you are looking for (it can be generalized for any $2^k$).
I hope this helps ;-)
A: Other answers and comments saying that the bitwise or approach works in the general case are wrong.
Consider $x = 1000, m = 3$
( 1000 | (3-1) ) + 1 == 1003
1003 % 3 != 0

This method works in the general case while keeping us type stable (i.e. no need for float division):
x - x % m

