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2007 AIME II Problems/Problem 5

Problem

The graph of the equation $9x+223y=2007$ is drawn on graph paper with each square representing one unit in each direction. How many of the $1$ by $1$ graph paper squares have interiors lying entirely below the graph and entirely in the first quadrant?

Solution

Solution 1

There are $223 \cdot 9 = 2007$ squares in total formed by the rectangle with edges on the x and y axes and with vertices on the intercepts of the equation, since the intercepts of the lines are $(223,0),\ (0,9)$.

Count the number of squares that the diagonal of the rectangle passes through. Since the two diagonals of a rectangle are congruent, we can consider instead the diagonal $y = \frac{223}{9}x$. This passes through 8 horizontal lines ($y = 1 \ldots 8$) and 222 vertical lines ($x = 1 \ldots 222$). At every time we cross a line, we enter a new square. Since 9 and 223 are relatively prime, we don’t have to worry about crossing an intersection of a horizontal and vertical line at one time. We must also account for the first square. This means that it passes through $222 + 8 + 1 = 231$ squares.

The number of non-diagonal squares is $2007 - 231 = 1776$. Divide this in 2 to get the number of squares in one of the triangles, with the answer being $\frac{1776}2 = 888$.

Solution 2

Count the number of each squares in each row of the triangle. The intercepts of the line are $(223,0),\ (0,9)$.

In the top row, there clearly are no squares that can be formed. In the second row, we see that the line $y = 8$ gives a $x$ value of $\frac{2007 - 8(223)}{9} = 24 \frac 79$, which means that $\lfloor 24 \frac 79\rfloor = 24$ unit squares can fit in that row. In general, there are

$\sum_{i=0}^{8} \lfloor \frac{223i}{9} \rfloor$

triangles. Since $\lfloor \frac{223}{9} \rfloor = 24$, we see that there are more than $24(0 + 1 + \ldots + 8) = 24(\frac{8 \times 9}{2}) = 864$ triangles. Now, count the fractional parts. $\lfloor \frac{0}{9} \rfloor = 0, \lfloor \frac{7}{9} \rfloor = 0, \lfloor \frac{14}{9} \rfloor = 1,$ $\lfloor \frac{21}{9} \rfloor = 2, \lfloor \frac{28}{9} \rfloor = 3, \lfloor \frac{35}{9} \rfloor = 3,$ $\lfloor \frac{42}{9} \rfloor = 4, \lfloor \frac{49}{9} \rfloor = 5, \lfloor \frac{56}{9} \rfloor = 6$. Adding them up, we get $864 + 1 + 2 + 3 + 3 + 4 + 5 + 6 = 888$.

Solution 3

From Pick's Theorem, $\frac{2007}{2}=\frac{233}{2}-\frac{2}{2}+\frac{2I}{2}$. In other words, $2I=1776$ and I is $888$.

Do you see why we simply set $I$ as the answer as well? That's because every interior point, if moved down and left one (southwest direction), can have that point and the previous point create a unit square. For example, $(1, 1)$ moves to $(0, 0)$, so the square of points ${(0, 0), (1, 0), (1, 1), (0, 1)}$ is one example. This applies, of course, for $888$ points.

Solution 4

We know that the number of squares intersected in an $m\times{n}$ rectangle is $m + n -\mbox{gcd}(m,n)$. So if we apply that here, we get that the number of intersected squares is:

$9 + 223 - 1 = 231$.

Now just subtract that from the total number of squares and divide by 2, since we want the number of squares below the line.

So,

$\frac{2007 - 231}{2} = \frac{1776}{2} = \fbox{888}$

Solution 5 (slower)

This is very elementary solution that doesn't require any tricks. First establish what the line looks like.

We'll solve for each square by rows, since there are only 8, starting from the bottom, where $y=1$.

When $y=1$ $x=198.22$. When you picture this in your mind, it becomes clear you have to floor this number to get the largest integer value of x when $y=1$. So for the first row alone, there are 198 squares that are entirely beneath the line.

$y=2$ gives $x=174.44$, rounded to 173.

Continue this process to $y=8$, getting numbers 198, 173, 148, 128, 99, 74, 49, and 24.

$198+173+148+128+99+74+49+24=\box{888}$ (Error compiling LaTeX. ! Missing number, treated as zero.).