# 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:

• $$2018$$ was not a leap year because $$2018$$ is not divisible by $$4$$.

• $$2016$$ was a leap year because $$2016$$ is divisible by $$4$$, but not $$100$$.

• $$2100$$ will not be a leap year because $$2100$$ is divisible by $$4$$ and $$100$$, but not $$400$$.

• $$2000$$ was a leap year because $$2000$$ is divisible by $$4$$, $$100$$, and $$400$$.

How many leap years are there between the years $$2000$$ and $$2400$$, inclusive?

## Solution

From the flowchart we can determine that leap years are either

• multiples of $$4$$ that are not also multiples of $$100$$, or

• multiples of $$4,~100,$$ and $$400$$.

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 itâ€™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.