Difference between revisions of "2008 iTest Problems/Problem 18"

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In order for <math>y</math> to be an positive integer, <math>1909-19x</math> must be a multiple of 20 greater than <math>0</math>, so <math>x \le 100</math>.  This means that the ones digit of <math>1909-19x</math> is <math>0</math> and the tens digit of <math>1909-19x</math> is even.   
 
In order for <math>y</math> to be an positive integer, <math>1909-19x</math> must be a multiple of 20 greater than <math>0</math>, so <math>x \le 100</math>.  This means that the ones digit of <math>1909-19x</math> is <math>0</math> and the tens digit of <math>1909-19x</math> is even.   
  
The ones digit of <math>1909-19x</math> is <math>0</math> when the last digit of <math>x</math> is <math>1</math>, so the available options are <math>1, 11, 21 \cdots 91</math>.  However, since <math>1909-19x=1909-20x+x</math>, the tens digit must be odd.  Thus, the only values that work are <math>11</math>, <math>31</math>, <math>51</math>, <math>71</math>, and <math>91</math>, so there are only <math>5</math> lattice points in the first quadrant.
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The ones digit of <math>1909-19x</math> is <math>0</math> when the last digit of <math>x</math> is <math>1</math>, so the available options are <math>1, 11, 21 \cdots 91</math>.  However, since <math>1909-19x=1909-20x+x</math>, the tens digit must be odd.  Thus, the only values that work are <math>11</math>, <math>31</math>, <math>51</math>, <math>71</math>, and <math>91</math>, so there are only <math>\boxed{5}</math> lattice points in the first quadrant.
  
 
==See Also==
 
==See Also==

Latest revision as of 14:28, 22 June 2018

Problem

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

Solution

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.

See Also

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