# 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 $$D (mod 4)=remainder(d/4)$$ 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.

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