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Problem of the Week
Problem C and Solution
A Small Leap

Problem

Most people think of a year as \(365\) days, however it is actually slightly more than \(365\) days. To account for this extra time we use leap years, which are years containing one extra day.

The flowchart shown can be used to determine whether or not a given year is a leap year. Using the flowchart, we can conclude the following:

How many leap years are there between the years \(2000\) and \(2400\), inclusive?

An alternative format for the flowchart
follows.

Solution

From the flowchart we can determine that leap years are either

Note that we can simplify the second case to just multiples of \(400\), since any multiple of \(400\) will also be a multiple of \(4\) and \(100\).

First we notice that \(2000\) and \(2400\) are both multiples of \(400\), so they are both leap years. In fact, they are the only multiples of \(400\) between \(2000\) and \(2400\).

Next we count the multiples of \(4\) between \(2000\) and \(2400\). Writing out some of the first few multiples of \(4\) gives: \(2000,~2004,~2008,~2012,\ldots\).

We will look at the \(400\) numbers from \(2000\) to \(2399\) and ignore \(2400\) for the moment since we already know it’s a leap year. Since \(2000\) is a multiple of \(4\) and every fourth number after that is also a multiple of \(4\), it follows that \(\frac{1}{4}\) of the \(400\) numbers from \(2000\) to \(2399\) will be multiples of \(4\). Thus, there are \(\frac{1}{4} \times 400 = 100\) multiples of \(4\) between \(2000\) and \(2399\), inclusive.

However, we have included the multiples of \(100\), so we need to subtract these. These are \(2000,~2100,~2200,\) and \(2300\). Thus there are \(100-4=96\) multiples of \(4\) between \(2000\) and \(2399\), inclusive, that are not also multiples of \(100\).

Thus, in total, there are \(2+96=98\) leap years between \(2000\) and \(2400\), inclusive.