2008 iTest Problems/Problem 18

Problem

Find the number of lattice points that the line $19x+20y = 1909$ passes through in Quadrant I.

Solution 1

Solve for $y$ to get \[y = \frac{1909-19x}{20}.\] In order for $y$ to be an positive integer, $1909-19x$ must be a multiple of 20 greater than $0$, so $x \le 100$. This means that the ones digit of $1909-19x$ is $0$ and the tens digit of $1909-19x$ is even.

The ones digit of $1909-19x$ is $0$ when the last digit of $x$ is $1$, so the available options are $1, 11, 21 \cdots 91$. However, since $1909-19x=1909-20x+x$, the tens digit must be odd. Thus, the only values that work are $11$, $31$, $51$, $71$, and $91$, so there are only $\boxed{5}$ lattice points in the first quadrant.

Solution 2 (Modular Arithmetic)

As in Solution 1, we rearrange the equation to get \[y = \frac{1909 - 19x}{20}.\] This means $1909 - 19x$ must be positive and divisible by 20, and we know $x$ is an integer, so we set it congruent to $0 \pmod{20}$ and simplify from there: \[\begin{array}{rll} 1909 - 19x \equiv & 0  & \pmod{20} \\ 9 + x \equiv      & 0  & \pmod{20} \\ x \equiv          & -9 & \pmod{20} \\ x \equiv          & 11 & \pmod{20}. \end{array}\] Since our lattice points are in the first quadrant, $x$ is positive, so we can start listing off our solutions: $x = 11, 31, 51, 71, 91, 111...$. Noticing that $19(111) \approx 19(101) = 1919 > 1909$, we conclude that $111$ is too large, and so our solutions are $11, 31, 51, 71,$ and $91$, for a total of $\boxed{5}$ lattice points.

See Also

2008 iTest (Problems)
Preceded by:
Problem 17
Followed by:
Problem 19
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