Math in Christmas: The Present and the Future

For this problem, we use Zeller's congruence to determine the days of the week. According to Wikipedia, for Gregorian calendar, the formula to use is $h = \left(q + \left\lceil \dfrac{13(m + 1)}{5}\right\rceil + K + \left\lceil \dfrac{K}{4}\right\rceil + \left\lceil \dfrac{J}{4}\right\rceil - 2J\right) \bmod 7$ where

• $h$ is the day of the week, where $0 = \text{Saturday}, 1 = \text{Sunday}, \cdots, 6 = \text{Friday}$
• $q$ is the day of the month; here, $q = 25$
• $m$ is the month; since the month is December, $m = 12$
• $K$ is the year of the century, which is $\text{year }\bmod 100$
• $J$ is the zero-based century, which is $\lfloor \text{year}/100\rfloor$

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For 2016, $K = 16$ and $J = 20$ . Zeller's congruence shows that $h = 1$ , which implies $d_{\text{present}} = 2$ .

For 6102, $K = 2$ and $J = 61$ . Then, $h = 2$ , which implies $d_{\text{future}} = 3$ .

Thus, $\left|d_{\text{present}} - d_{\text{past}}\right| = \boxed{1}$ .

Because non-leap year have 365 days in one, the next New Year's day will be the day after the current (non-leap) year due to 365 = 1 mod 7. But remember that we also have leap years once every four years, and the exception for leap year that happens every 100 years (except for those divisible by 400).
Difference in days (mod 7)
= (6102-2016) + [(6102-2016)/4] - [(6102-2001)/100] + [(6102-2001)/400]
= (4086 + 1021 - 41 + 10) mod 7
= 5076 mod 7
= 1 mod 7
So the future's Xmas will come on the next day of our present one. Even when we know nothing about what days they are, the multiple choices which lack any other even number beside 0 is a big give away. Now we know our current Xmas will fall on a Sunday, the only even (prime) number in the index, and since we already know the next day will be the future Xmas, the answer is 3-2=1.