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

Puzzles and Curious Problems by Henry Ernest Dudeney: $346$

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.

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

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

$5$ 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 $1888$.

$\blacksquare$