solve modular arithmetic equation

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 $D \pmod 4=remainder(D /4)$ that also means we are dealing with modular arithmetic.

   
   
   

   $∴ [D:4]=D/4-D \pmod 4$

   
   
   

   $∴ L=5/4 D-D \pmod 4 ……………….. (1)$

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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 $D \pmod 4$ part by multiplying it with its inverse according to the following rule:

Calculate $A \cdot D \pmod 4$ for $A$ values $0$ through $(4-1)$, the modular inverse of
$\pmod 4$ is the $A$ value that makes $A \cdot D\equiv 1\pmod 4$ and only the numbers that share no prime factors with $4$ have a modular inverse $\pmod 4$

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

Can anyone help please.