# Middle day in a year

• Jul 23rd 2007, 06:58 AM
DivideBy0
Middle day in a year
The middle date of the year in 2006 is:

(a) 29th June
(b) 30th June
(c) 1st July
(d) 2nd July
(e) 3rd July

We are meant to assume that 2006 is not a leap year. The problem came when I tried $\frac{365}{2}$ and didn't get an integer...
My answer so far is 2nd July....
• Jul 23rd 2007, 07:32 AM
topsquark
Quote:

Originally Posted by DivideBy0
The middle date of the year in 2006 is:

(a) 29th June
(b) 30th June
(c) 1st July
(d) 2nd July
(e) 3rd July

We are meant to assume that 2006 is not a leap year. The problem came when I tried $\frac{365}{2}$ and didn't get an integer...
My answer so far is 2nd July....

I'm going to use a much smaller "year" as a demonstration here.

Think of it this way, say the year has 9 days in it. What is the middle day of the year?

Well, we have days
1 2 3 4 5 6 7 8 9
By inspection it looks like the middle day is 5. But $5 \neq \frac{9}{2}$. So how do we approach this?

Well, what you did amounts to solving $2n= 365$, which gives a noninteger day. The reason for this is that 365 isn't even. But even then the middle day of an even list has no middle! For example
1 2 3 4 5 6 7 8
What is the middle number? There isn't one!

What this amounts to is that it is better to solve the equation $2n - 1 = 365$. This will give you the correct day number. You can easily verify that this gives you the correct solution when the list has 9 elements.

-Dan