Henry Ernest Dudeney/Puzzles and Curious Problems/346 - A Leap Year Puzzle/Solution

by : $346$

 * A Leap Year Puzzle
 * The month of February in $1928$ contained five Wednesdays.
 * There is, of course, nothing remarkable in this fact, but it will be found interesting to discover
 * when was the last year and when will be the next year that had, and that will have, $5$ Wednesdays in February.

Solution
The next year after $1928$ that contains $5$ February Wednesdays was $1956$.

The one immediately before $1928$ was $1888$.

Proof
$5$ February Wednesdays can occur only in a leap year.

A leap year February contains $29$ days.

Hence such a February always has exactly one day of the week of which there are $5$.

This happens on $1$st, $8$th, $15$th, $22$nd and $29$th February.

So the puzzle is equivalent to determining on which years $29$th February falls on a Wednesday.

Except when the year is divisible by $100$ but not $400$, leap years occur every $4$ years.

Each consecutive set of $4$ years in the Gregorian calendar contains $3 \times 365 + 366 = 1461$ days.

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

So, on consecutive leap years, $29$th February happens on:
 * $(1): \quad$ Wednesday, Monday, Saturday, Thursday, Tuesday, Sunday, Friday.

Hence (apart from at the turn of the century), there are $4 \times 7 = 28$ years between consecutive years when the days coincide.

Hence after $1928$, the next one is $1956$.

$28$ years before $1928$ is $1900$.

But $1900$, as we have determined, is not a leap year.

If it were a leap year, then $28$ years before that, that is $1872$, would have had $5$ Wednesdays.

But there is a day less between $1872$ and $1900$ than there would be.

So $29$th February in $1872$ fell on a Thursday.

Hence, from the sequence of days in $(1)$ above, $29$th February falls on a Wednesday $4$ leap years later.

That is, $16$ years, which is $1888$.