Same Calendar Year

A normal year has 52 weeks and one additional day. This additional day is called an 'odd day' as it is the left over and will be carried forward to the subsequent week of the next year. A leap year has 2 odd days. Let 2007 starts with Monday. Then 2008 will start from Tuesday (add 1 odd day). Like this 2009 will start from Thursday (add 2 odd day as 2008 was a leap year) an so on ( 2010 - Fri, 2011 - Sat , 2012 - Sun, 2013 - Tue { 2 odd in 2012} , 2014- Wed, 2015- Thu, 2016 - Fri , 2017- Sun { again as 2016 is a leap year so add 2 odd day } , 2018 - Mon. Thus 2018 also starts from Monday and also a Gregorian calendar ( non leap year 365 day ) so calendar of 2007 and 2018 will be same.

Let Jan 1, 2007 Monday be day 1 or $D(2007)=1$ and $y$ years or $d$ days later the calendar is the same. Then Jan 1, 2007 $+y$ is $D(2007+y) = 1+d$ . If year $2007+y$ has the same calendar, then Jan 1, 2007 $+y$ has the same weekday Monday. Using modular arithmetic, we say that since $D(2007) = 1 \equiv 1 \text{ (mod 7)}$ $\implies D(2007+y) = 1 + d \equiv 1 \text{ (mod 7)}$ $\implies d \equiv 0 \text{ (mod 7)}$ .

We note that for normal year $365 \equiv 1 \text{ (mod 7)}$ and leap year $366 \equiv 2 \text{ (mod 7)}$ . And that 2008 is a leap year, we conclude that for $y \ge 2$ :

$\begin{aligned} d & \equiv \left( y + \left \lfloor \frac {\color{#3D99F6}{y-2}}4 \right \rfloor + \color{#3D99F6}{1} \right) \text{ (mod 7)} & \small \color{#3D99F6}{\text{The leap day is added at the end of the year, hence }-2} \end{aligned}$

$\begin{aligned} \implies y + \left \lfloor \frac {y-2}4 \right \rfloor + 1 & \equiv 0 \text{ (mod 7)} \\ y + \left \lfloor \frac {y-2}4 \right \rfloor + 1 & = 7\color{#3D99F6}{n} & \small \color{#3D99F6}{\text{where }n \text{ is a positive integer.}} \end{aligned}$

• For $n=1$ or $RHS=7$ , $y = 5$ $\implies LHS = 6$ and $y = 6$ $\implies LHS = 8$ , therefore, no solution.
• For $n=2$ or $RHS=14$ , $y = 11$ $\implies y + \left \lfloor \dfrac {y-2}4 \right \rfloor + 1 = 11 + 2 + 1 = 14 = RHS$ . Therefore, $y=11$ and the year is $2007+11 = \boxed{2018}$ .

The year 2007 has 365 days and starts on a Monday. So, we need to identify what will be the next not leap-year that will start on a Monday. We know that every year we start one day after (Ex: 2008 starts on a Tuesday). On a leap-year, we advance​ two days. So, the next year that will starts on a Monday is: - 2008: Tuesday - 2009: Thursday - 2010: Friday - 2011: Saturday - 2012: Sunday - 2013: Tuesday - 2014: Wednesday - . - . - 2017: Sunday - 2018: Monday