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I have a number $n$ in the range $1$ - $255$. What I'm trying to do is split $n$ into the shortest list of numbers $1$ -$16$ that add up to $n$. For example, let's say $n$ is $32$. Then, we could simply use $16$ and $16$. However, if $n$ is $33$, it would be $16, 16, 1$.

Is there any simple and fast algorithm to accomplish this? I'm not even sure what to google when researching this, so even keywords and hints are a help. Thank you!

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  • $\begingroup$ partitions perhaps. $\endgroup$
    – user451844
    Commented Jul 7, 2017 at 18:32
  • $\begingroup$ Upon reading this, I realize that I have worded it absolutely wrong. I'm going to flag this for moderator attention as the question is trivial in it's current state - which is not what I am trying to ask. $\endgroup$
    – Nico A
    Commented Jul 7, 2017 at 18:38
  • $\begingroup$ Just edit the question and say what you mean. $\endgroup$
    – lhf
    Commented Jul 7, 2017 at 18:39
  • $\begingroup$ Is $(16,16,1)$ a list of length $2$ or $3$ in your book? $\endgroup$
    – Rodrigo de Azevedo
    Commented Jul 7, 2017 at 18:59

2 Answers 2

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So basically you are asking what the fastest way to compute the quotient and remainder of $n$ divided by $16$ is? Sounds more like a computer science question ...

In fact, if you have the number in binary (you seem to like binary, given your $255$, and $16$...), it's easy:

Say you have $158$, which is $10011110_2$

Then the first four bits ($1001=9_{10}$) gives you the quotient, and the last four bits ($1110=14_{10}$) the remainder. So you need $9$ $16$'s and $1$ $14$: $16,16,16,16,16,16,16,16,16,14$

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Write $n = 16q + r$ with $0 \le r < 16$. Then use $16$ exactly $q$ times and $r$ once, a total of $q+1$ terms.

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