Difference between revisions of "2008 iTest Problems/Problem 18"
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Find the number of lattice points that the line <math>19x+20y = 1909</math> passes through in Quadrant I. | Find the number of lattice points that the line <math>19x+20y = 1909</math> passes through in Quadrant I. | ||
− | ==Solution== | + | ==Solution 1== |
Solve for <math>y</math> to get | Solve for <math>y</math> to get | ||
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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. | 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. | ||
+ | |||
+ | ==Solution 2 (Modular Arithmetic)== | ||
+ | As in Solution 1, we rearrange the equation to get | ||
+ | <cmath>y = \frac{1909 - 19x}{20}.</cmath> | ||
+ | This means <math>1909 - 19x</math> must be positive and divisible by 20, and we know <math>x</math> is an integer, so we set it congruent to <math>0 \pmod{20}</math> and simplify from there: | ||
+ | <cmath>\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}</cmath> | ||
+ | Since our lattice points are in the first quadrant, <math>x</math> is positive, so we can start listing off our solutions: <math>x = 11, 31, 51, 71, 91, 111...</math>. Noticing that <math>19(111) \approx 19(101) = 1919 > 1909</math>, we conclude that <math>111</math> is too large, and so our solutions are <math>11, 31, 51, 71,</math> and <math>91</math>, for a total of <math>\boxed{5}</math> lattice points. | ||
==See Also== | ==See Also== |
Revision as of 11:57, 14 June 2022
Problem
Find the number of lattice points that the line passes through in Quadrant I.
Solution 1
Solve for to get In order for to be an positive integer, must be a multiple of 20 greater than , so . This means that the ones digit of is and the tens digit of is even.
The ones digit of is when the last digit of is , so the available options are . However, since , the tens digit must be odd. Thus, the only values that work are , , , , and , so there are only lattice points in the first quadrant.
Solution 2 (Modular Arithmetic)
As in Solution 1, we rearrange the equation to get This means must be positive and divisible by 20, and we know is an integer, so we set it congruent to and simplify from there: Since our lattice points are in the first quadrant, is positive, so we can start listing off our solutions: . Noticing that , we conclude that is too large, and so our solutions are and , for a total of lattice points.
See Also
2008 iTest (Problems) | ||
Preceded by: Problem 17 |
Followed by: Problem 19 | |
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