Difference between revisions of "2011 AMC 10B Problems/Problem 24"
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<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, 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>(A), (B), (C), (D), (E)</math>. We can see that when <math>m=\frac{50}{99}</math>, <math>y</math> | + | 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> could 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 largest 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}} | {{MAA Notice}} |
Latest revision as of 20:45, 19 March 2023
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
,
could 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 largest 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 | ||
All AMC 10 Problems and Solutions |
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