Henry Ernest Dudeney/Modern Puzzles/219 - A Calendar Puzzle/Solution

by : $219$

 * A Calendar Puzzle

Solution
The Gregorian calendar is such that the pattern of leap years repeats every $400$ years.

So $400$ years contains:
 * $400 \times 365$ days

plus:
 * an extra $4 \times 25$ days for the leap years in each century

minus:
 * $3$ days for the years divisible by $100$ and not $400$ which are not actually in fact leap years.

Thus there are $400 \times 365 + 4 \times 25 - 3 = 146 \, 097$ days in $400$ years.

We have that $146 \, 097 = 7 \times 20 \, 871$.

Hence the first day of the century in a given block of $400$ years is always the same.

We can look up the day on which $1$st January $2001$ falls, and see it is Monday.

Now, between $2001$ and $2100$, $2101$ and $2200$, and $2201$ and $2300$ there are $100 \times 365 + 24$ days, that is, $36 \, 524$ days.

This is congruent to $5$ modulo $7$.

So subsequent first days of the century after $2001$ are:
 * $2101$: Saturday ($5$ days after Monday)
 * $2201$: Thursday ($5$ days after Saturday)
 * $2301$: Tuesday ($5$ days after Thursday)

and then, as we have seen, $2401$ starts on the same day as $2001$.

So the centuries begin on Monday, Saturday, Thursday, Tuesday, Monday, ... and so on.

That is, the century can never begin on Wednesday, Friday or Sunday.