0

As a part of a programming problem I was solving, we were required to find the first offset of a range at which the number is a odd multiple of another number. For e.g: Take the range $100$ to $120$. Only the odd multiples here are taken for the offsets - so offset 0 corresponds to 101, 1 to 103 and k to $100+2*k+1$ in general. Say we need to find the first odd multiple of 3 in this range - that would be 105. This corresponds to an offset of 2. So, I need to compute a function which can output these offsets, given the initial/ final number ($100/120$ here) in the range and the dividing number (3 in this case).

I found a solution here which says the function should be $(-1/2 × (L + 1 + Pk))\mod Pk$ , where L:start of range, Pk: dividing number (3 in our e.g). I'm unable to understand how this function is the right answer. Could some one please shed some light here?

Note: The above function is explained to some extent by the author here, in the comments - but I still don't get it.

Community
  • 1
user1968919
  • 9
  • 3
  • Improve formatting using $ sign – Semsem Mar 21 '14 at 11:11
  • $$p\mid \ell+2x+1\iff !\bmod p!:\ 2x\equiv -\ell-1\equiv -\ell-1-p\iff x\equiv \dfrac{-\ell-1-p}2 =: k$$ But the least natural $\equiv k\pmod{! p}$ is $,k\bmod p,,$ which yields the claim. $\ \ $ – Bill Dubuque Apr 09 '25 at 19:50
  • The solution of your congruence in the prior comment is a special case of general methods for solving congruences - see the linked dupe. $\ \ $ – Bill Dubuque Apr 09 '25 at 21:22

1 Answers1

0

Most programming languages have a "mod" operator to calculate the remainder of a division of two integers. Therefore let's assume that $L$ and $n$ are positive integers and let's define the numbers $x$ and $y$ by $$\begin{align} x&:=(L+n-1)\mod 2n\\ y&:=L+2n-1-x \end{align}$$ We see that $L+n-1-x$ is an even multiple of $n$ and so $n+L+n-1-x=y$ is an odd multiple of $n$. Also, since $0 \le x <= 2n-1$, we see that $L \le y \le L+2n-1$, so $y$ must be the odd multiple of $n$ we're looking for (if $y$ does not exceed the range).

To calculate the "offset" of $y$ in the range, we have to solve $L+2k+1=y$ for $k$, so $$ 2k+1=2n-1-x \Leftrightarrow k=n-1-\frac{x}{2} $$ This can only be an integer if $x$ is even, for example if $L$ even and $n$ odd (like in your example).

To apply this to your example ($L=100$, $n=3$), we have $L+n-1=102\equiv0\pmod6$, so $x=0$ and $y=100+6-1-x=105$, the offset $k$ is $k=n-1-\frac{0}{2}=3-1=2$.

Wolfgang Kais
  • 1,430
  • 8
  • 7