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 | ||
− | <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. | ||
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== |
Latest revision as of 08:29, 23 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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