Q26 from AMC 2012 Slim took a long road trip across Australia over a number of days($x>1$).She travelled a total of 2012 km.On the first day,she travelled a whole number of kilometers and each subsequent day she travelled 1 more kilometer than the day before.What is the largest number of distance could she have travelled on the first day?
Here's what I tried:
Let the number of days be $x$ and the distance she travelled on the first day be   $y-\frac{x}{2}$
So,
$2012= y-\frac{x}{2} +y-\frac{x}{2}+1+...y...+y+\frac{x}{2}-1$
$2012=xy -\frac{x}{2} $
$4024=2xy-x $
Okay,so that's where i got stuck at already.
Is my method and concept correct?
 A: Suppose that on the first day, Slim traveled $k$ kilometers. Then on the next day, Slim would travel $k+1$ kilometers. And then $k + 2$ on the next.
After $d$ days, Slim travels a total of $k + (k+1) + \cdots + (k+d-1)$ kilometers. So you are trying to find solutions to
$$ \begin{align}
2012 &= k + (k+1) + \cdots + (k+d-1)\\
&= kd + 1 + 2 + \cdots + d-1 \\
&= kd + \frac{d(d-1)}{2}.
\end{align}$$
This can be rewritten as
$$ 4024 = 2kd + d(d-1) =d(2k + d - 1).$$
We want to find the largest integer $k$ so that there is some integer $d$ such that the above equality holds. This is now a factoring question: what is the largest $k$ so that $4024$ splits into two factors that are $2k - 1$ apart?
Note that $4024 = 1 \cdot 4024 = 2 \cdot 2012 = 4 \cdot 1006 = 8 \cdot 503$, and these are all the factorizations we have. The only pair of factors that works are $503$ and $8$ (as this is the only pair in which one is even and the other is odd).
The difference is $503 - 8 = 495$, so that $2k-1 = 495 \implies 2k =  496 \implies k = 248$.
So Slim travelled $248$ kilometers on the first day, and took $8$ days in total to complete the journey.
A: The average of $n$ consecutive integers is an integer if $n$ is odd, or an integer plus a half if $n$ is even.  So $2012/x$ must be an integer if $x$ is odd, or an integer plus a half if $x$ is even.  Considering possible values of $x$ in ascending order:


*

*$2012/2$ is an integer but 2 is even.

*$2012/3$ is neither an integer nor an integer plus a half.

*$2012/4$ is an integer but 4 is even.

*$2012/5$, $2012/6$ and $2012/7$ are neither an integer nor an integer plus a half.

*$2012/8 = 251.5$ is an integer plus a half and 8 is even, as required.


Hence the smallest number of days is 8, and the largest distance travelled on the first day is $251.5 - ((8-1)/2) = 248$ km.
