Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(Added modular arithmetic solution)
m
 
Line 6: Line 6:
  
 
Solve for <math>y</math> to get
 
Solve for <math>y</math> to get
−
<cmath>y = \frac{1909-19x}{20}</cmath>
+
<cmath>y = \frac{1909-19x}{20}.</cmath>
 
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.   
  

Latest revision as of 09:29, 23 June 2022

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
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 • 51 • 52 • 53 • 54 • 55 • 56 • 57 • 58 • 59 • 60 • 61 • 62 • 63 • 64 • 65 • 66 • 67 • 68 • 69 • 70 • 71 • 72 • 73 • 74 • 75 • 76 • 77 • 78 • 79 • 80 • 81 • 82 • 83 • 84 • 85 • 86 • 87 • 88 • 89 • 90 • 91 • 92 • 93 • 94 • 95 • 96 • 97 • 98 • 99 • 100
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2008_iTest_Problems/Problem_18&oldid=175231"