I'm not sure this what you're after, but here it is anyway.
Do you know about modular arithmetic?
Clock arithmetic is a more descriptive term for the aforementioned.
What time is $11:00 + 17 \text{ hours}$? The computation goes as follows: $11+17=28$. Now from $28$ you subtract $24$ as many times as you need until you get a number between $1$ and $24$. So $28-24=4$. Another way of doing this is dividing $28$ by $24$ and keeping the remainder. The remained is what you're after. You basically just need to know the euclidean algorithm to present this.
Now the real presentation (which is an application of modular arithmetic).
What I'll present below is a way of computing the day of the week of any given day of the month of any year. There are several of these. You can find'em here.
I like the following one better because it doesn't require you to memorize too many things. In case it looks complicated, don't give up. It's easy and anyone can do it provided they can do a simple calculations in their heads. With practice you'll nail it and do it easily. Also it actually helps you on day-to-day life.
Firstly we give each month a number, which I'll call month code or MC for short.
The codes go as follows:
$$\begin{align} &\text{January: } &1\, &\text{April: } &0\, &\text{July: } &0\, &\text{October: }&1\\
&\text{February: }&4\, &\text{May: }&2\, &\text{August: }&3\, &\text{November: }&4\\
&\text{March: }&4\, &\text{June: }&5\, &\text{September: }&6 \,&\text{December: }&6 \end{align}$$
There's an easy way to memorize this. The first column is $144=12^2$, the second is $25=5^2$ and the third is $6^2=36$. The fourth column you'll just have to remember.
We'll also need a code for each year and a code for each day of the week. The days of the week will be numbered $0,1,2,3,4,5,6$, respectively to Saturday, ..., Friday. Saturday can also be $7$. You can use both $0$ and $7$ because in our arithmetic $0=7$. These are the day codes (DC). Each day of the month will have it's own code DMC and it will be itself, for instance, today is the $25^{th}$ of April, therefore today's DMC is $25$.
The year codes (YC) are computed and not memorized.
To compute the day week of any date, just do DMC+MC+YC, then divide by $7$ and save the remainder. The remainder will be the DC of the day you're after.
For instance, we're in $2013$. This year's code is $1$ (trust me for now).
Let us find which day of the week it is today: $25+0+1=26=7\cdot 3+\color{green}5$
, therefore today is Thursday. And xmas will be on $25+6+1=32=7\cdot 4+\color{green}4$ therefore this year xmas will be on a Wednesday.
Now to find the YC. We take the year $1900$ and give it the YC $0$.
Instead of describing how to get the YC I will exemplify it:
Let's compute the YC of $2013$:
- $2013-1900=113$
- Find the quotient of the above divided by $4$ (because of leap years), $113=\color{green}{28}\cdot 4+1$. Save $28$.
- Add $113$ and $28$: $113+28=141$.
- Find the remainder of the above when divided by $7$: $141=20\cdot 7 +\color{green}1$. This is your YC.
Another example, for instance $1996$ (which with any luck is the year you were born):
- $1900-1996=96$
- $96=\color{green}{24}\cdot 4$
- $96+24=120$
- $120=17\cdot 7 + \color{green}1$
Crap, it's the same YC. Doesn't matter, you will get how it's done. Note that if you were indeed born in $1996$, then the day of the week of your birthday this year is the day of the week in which you were born.
Important: When you're computing the YC, check if it is a leap year, for if it is the YC will be what you get with the algorithm above only after the $29^{th}$ of February. For January and February you want to take whatever YC you get and subtract $1$. If it happens to be $0$, then $0-1=7-1=6$, because $0=7$.
Note: After having the YC of a certain year, it's easy to get the YCs for years on a relatively small neighborhood. You've already found that the YC of $2013$ is $1$. To get the YC for $2014$ just add $1$. To get the year code for $2015$ add $2$. To get the year code for $2016$ ($\color{red}{\text{danger - leap year}}$), add $3+1$ for March onwards and add $3+1-1$ for January and February.
Help: If anyone can prove why this algorithm works or tell me where to find a proof, I'd appreciate it.