Difference between revisions of "2011 AMC 10B Problems/Problem 24"
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== Problem == | == Problem == | ||
− | A lattice point in an <math>xy</math>-coordinate system is any point <math>(x, y)</math> where both <math>x</math> and <math>y</math> are integers. The graph of <math>y = mx +2</math> passes through no lattice point with <math>0 < x \le 100</math> for all <math>m</math> such that <math>1 | + | A lattice point in an <math>xy</math>-coordinate system is any point <math>(x, y)</math> where both <math>x</math> and <math>y</math> are integers. The graph of <math>y = mx +2</math> passes through no lattice point with <math>0 < x \le 100</math> for all <math>m</math> such that <math>\frac{1}{2} < m < a</math>. What is the maximum possible value of <math>a</math>? |
<math> \textbf{(A)}\ \frac{51}{101} \qquad\textbf{(B)}\ \frac{50}{99} \qquad\textbf{(C)}\ \frac{51}{100} \qquad\textbf{(D)}\ \frac{52}{101} \qquad\textbf{(E)}\ \frac{13}{25}</math> | <math> \textbf{(A)}\ \frac{51}{101} \qquad\textbf{(B)}\ \frac{50}{99} \qquad\textbf{(C)}\ \frac{51}{100} \qquad\textbf{(D)}\ \frac{52}{101} \qquad\textbf{(E)}\ \frac{13}{25}</math> | ||
− | ==Solution== | + | == Solution 1== |
− | We see that for the graph of <math>y=mx+2</math> to not pass through any lattice points | + | For <math>y=mx+2</math> to not pass through any lattice points with <math>0<x\leq 100</math> is the same as saying that <math>mx\notin\mathbb Z</math> for <math>x\in\{1,2,\dots,100\}</math>, or in other words, <math>m</math> is not expressible as a ratio of positive integers <math>s/t</math> with <math>t\leq 100</math>. Hence the maximum possible value of <math>a</math> is the first real number after <math>1/2</math> that is so expressible. |
+ | |||
+ | For each <math>d=2,\dots,100</math>, the smallest multiple of <math>1/d</math> which exceeds <math>1/2</math> is <math>1,\frac23,\frac34,\frac35,\dots,\frac{50}{98},\frac{50}{99},\frac{51}{100}</math> respectively, and the smallest of these is <math>\boxed{\textbf{(B)}\frac{50}{99}}</math>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | We see that for the graph of <math>y=mx+2</math> to not pass through any lattice points, the denominator of <math>m</math> must be greater than <math>100</math>, or else it would be canceled by some <math>0<x\le100</math> which would make <math>y</math> an integer. By using common denominators, we find that the order of the fractions from smallest to largest is <math>\text{(A), (B), (C), (D), (E)}</math>. We can see that when <math>m=\frac{50}{99}</math>, <math>y</math> would be an integer, so therefore any fraction greater than <math>\frac{50}{99}</math> would not work, as substituting our fraction <math>\frac{50}{99}</math> for <math>m</math> would produce an integer for <math>y</math>. So now we are left with only <math>\frac{51}{101}</math> and <math>\frac{50}{99}</math>. But since <math>\frac{51}{101}=\frac{5049}{9999}</math> and <math>\frac{50}{99}=\frac{5050}{9999}</math>, we can be absolutely certain that there isn't a number between <math>\frac{51}{101}</math> and <math>\frac{50}{99}</math> that can reduce to a fraction whose denominator is less than or equal to <math>100</math>. Since we are looking for the maximum value of <math>a</math>, we take the larger of <math>\frac{51}{101}</math> and <math>\frac{50}{99}</math>, which is <math>\boxed{\textbf{(B)}\frac{50}{99}}</math>. | ||
+ | |||
+ | ==Solution 3== | ||
+ | We want to find the smallest <math>m</math> such that there will be an integral solution to <math>y=mx+2</math> with <math>0<x\le100</math>. We first test A, but since the denominator has a <math>101</math>, <math>x</math> must be a nonzero multiple of <math>101</math>, but it then will be greater than <math>100</math>. We then test B. <math>y=\frac{50}{99}x+2</math> yields the solution <math>(99,52)</math> which satisfies <math>0<x\le100</math>. Checking the answer choices, we know that the smallest possible <math>a</math> must be <math>\frac{50}{99}\implies\boxed{\textbf{(B)}}</math> | ||
+ | |||
+ | |||
+ | == Solution 4 == | ||
+ | |||
+ | Notice that for <math>y=\frac{1}{2}x+2=\frac{50}{100}x+2</math>, <math>x=99</math> is one of the integral values of <math>x</math> such that the value of <math>\frac{50}{100}x</math> is the closest to its next integral value. | ||
+ | |||
+ | |||
+ | Thus the maximum value for <math>a</math> is the value of <math>m</math> when the equation <math>y=99m+2</math> goes through its next lattice point, which occurs when <math>m=\frac{b}{99}</math> for some positive integer <math>b</math>. | ||
+ | |||
+ | |||
+ | Finding the common denominator, we have <cmath>\frac{50}{100}=\frac{4950}{9900}, \frac{b}{99}=\frac{100b}{9900}</cmath> Since <math>a>\frac{1}{2}</math>, the smallest value for <math>b</math> such that <math>100b>4950</math> is <math>b=50</math>. | ||
+ | |||
+ | Thus the maximum value of <math>a</math> is <math>\frac{50}{99}.\boxed{\mathrm{(B)}}</math> | ||
+ | |||
+ | ~ Nafer | ||
+ | |||
+ | == Solution 5 (MAA.org) == | ||
+ | https://www.maa.org/sites/default/files/pdf/CurriculumInspirations/CB095_A-Line-through-Lattice-Points.pdf | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2011|ab=B|num-a=25|num-b=23}} | {{AMC10 box|year=2011|ab=B|num-a=25|num-b=23}} | ||
+ | [[Category:Introductory Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 10:31, 11 September 2021
Contents
Problem
A lattice point in an -coordinate system is any point where both and are integers. The graph of passes through no lattice point with for all such that . What is the maximum possible value of ?
Solution 1
For to not pass through any lattice points with is the same as saying that for , or in other words, is not expressible as a ratio of positive integers with . Hence the maximum possible value of is the first real number after that is so expressible.
For each , the smallest multiple of which exceeds is respectively, and the smallest of these is .
Solution 2
We see that for the graph of to not pass through any lattice points, the denominator of must be greater than , or else it would be canceled by some which would make an integer. By using common denominators, we find that the order of the fractions from smallest to largest is . We can see that when , would be an integer, so therefore any fraction greater than would not work, as substituting our fraction for would produce an integer for . So now we are left with only and . But since and , we can be absolutely certain that there isn't a number between and that can reduce to a fraction whose denominator is less than or equal to . Since we are looking for the maximum value of , we take the larger of and , which is .
Solution 3
We want to find the smallest such that there will be an integral solution to with . We first test A, but since the denominator has a , must be a nonzero multiple of , but it then will be greater than . We then test B. yields the solution which satisfies . Checking the answer choices, we know that the smallest possible must be
Solution 4
Notice that for , is one of the integral values of such that the value of is the closest to its next integral value.
Thus the maximum value for is the value of when the equation goes through its next lattice point, which occurs when for some positive integer .
Finding the common denominator, we have Since , the smallest value for such that is .
Thus the maximum value of is
~ Nafer
Solution 5 (MAA.org)
See Also
2011 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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 | ||
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