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How do I solve this equation

$L = D + [D:4]$

  • L is a known integer obtained previously.
  • D is an integer.
  • $[D:4]$ Is the quotient part of (D/4)

At first glance I did not know how to solve this equation as I have never seen this type in my calculus studies. After searching online I found that this is modular arithmetic for two reasons:

  1. The original relation is a congruence equation (Zeller's Rule) dealing with cyclical calendar numbers and I understand that modular arithmetic deals with cyclical integers.
  2. $D/4=quotient(d/4)+remainder (d/4)$ and since $D (mod 4)=remainder(d/4)$ that also means we are dealing with modular arithmetic.

    $∴ [D:4]=D/4-D (mod 4)$

    $∴ L=5/4 D-D (mod 4) ……………….. (1)$


how to go further with eq. (1)?

Although I read the modular arithmetic rules and practiced a little but I wasn’t sure if I was going the right path. I tried to eliminate the D (mod 4) part by multiplying it with its inverse according to the following rule:

Calculate $A * D (mod 4)$ for $A$ values 0 through (4-1), The modular inverse of $D (mod 4)$ is the “A” value that makes $A * D (mod 4) = 1$ and only the numbers that share no prime factors with 4 have a modular inverse $(mod 4)$*

From this point I can obtain an inverse (I think) but It makes no sense to me.

Can anyone help please.

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If $D = 4q + r$, you've already acheived $L = \frac{5D}{4} + r = 5q + \frac{9r}{4} \implies \frac{9r}{4} = L - 5q \in \Bbb N$. Since $\rm{gcd}(9,4) = 1$, $4|r\, \land \, 0 \leq r < 4 \implies r = 0$. So $D = \frac{4L}{5}$, given the restrictions.

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  • $\begingroup$ why $D=4q+r$ and not $ D=4q+4r$ $\endgroup$ – Ali_R4v3n Dec 15 '18 at 19:21
  • $\begingroup$ I'm assuming the Euclidean division of $D$ by $4$, with quotient $q$ and remainder $r$. $\endgroup$ – Lucas Henrique Dec 15 '18 at 21:00

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