AMC 10 · 2011 · #24
Grade 8 arithmeticA lattice point in an xy-coordinate system is any point (x,y) where both x and y are integers. The graph of y=mx+2 passes through no lattice point with 0<x≤100 for all m such that 21<m<a. What is the maximum possible value of a?
Pick an answer.
AMC 10 2011 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: The line $y = mx + 2$ should miss every lattice point whose $x$-coordinate satisfies $1 \le x \le 100$, and it must do so for every slope $m$ in the open interval $\left(\tfrac{1}{2}, a\right)$. Find the largest $a$ for which this stays true.
Givens: A lattice point is a point $(x,y)$ with both coordinates integers; The line is $y = mx + 2$; The slope $m$ ranges over the open interval $\left(\tfrac{1}{2}, a\right)$; "No lattice point" must hold for all integer $x$ with $1 \le x \le 100$; Answer choices: (A) $\tfrac{51}{101}$, (B) $\tfrac{50}{99}$, (C) $\tfrac{51}{100}$, (D) $\tfrac{52}{101}$, (E) $\tfrac{13}{25}$
Unknowns: The maximum possible value of $a$
Understand
Restated: The line $y = mx + 2$ should miss every lattice point whose $x$-coordinate satisfies $1 \le x \le 100$, and it must do so for every slope $m$ in the open interval $\left(\tfrac{1}{2}, a\right)$. Find the largest $a$ for which this stays true.
Givens: A lattice point is a point $(x,y)$ with both coordinates integers; The line is $y = mx + 2$; The slope $m$ ranges over the open interval $\left(\tfrac{1}{2}, a\right)$; "No lattice point" must hold for all integer $x$ with $1 \le x \le 100$; Answer choices: (A) $\tfrac{51}{101}$, (B) $\tfrac{50}{99}$, (C) $\tfrac{51}{100}$, (D) $\tfrac{52}{101}$, (E) $\tfrac{13}{25}$
Plan
Primary tool: #14 Extreme Principle
Secondary: #16 Change Focus / Count the Complement, #4 Introduce a Variable, #3 Eliminate Possibilities
The whole question is a boundary hunt, so Tool #14 (Extreme Principle) is primary: the largest safe $a$ is pinned by the single closest "bad" slope sitting just above $\tfrac{1}{2}$. To even see those bad slopes, use Tool #16 (Change Focus) — instead of chasing lattice points, translate "line hits a lattice point" into "the slope equals a fraction $k/x$ with $x \le 100$." Tool #4 (Introduce a Variable) names that fraction so we can measure how far above $\tfrac{1}{2}$ it sits. Finally Tool #3 (Eliminate Possibilities) checks the five answer choices against the boundary we find, catching the even-denominator trap $\tfrac{51}{100}$.
Execute — Answer: B
8.F.A.3 Step 1 Reframe as a slope-is-a-fraction question
- On $y = mx + 2$, at an integer $x$ the height is $y = mx + 2$.
- Since $2$ is already an integer, $y$ is an integer exactly when $mx$ is an integer.
- So the line passes through a lattice point with $1 \le x \le 100$ precisely when $mx = k$ for some integer $k$, i.e.
- when the slope equals a fraction $m = \tfrac{k}{x}$ with denominator $x \le 100$.
- Avoiding lattice points means avoiding all such fractions.
💡 Adding the whole number $2$ never changes whether a height is a whole number, so only the $mx$ part matters.
6.EE.B.6 Step 2 Name the target value of a
- We need the open interval $\left(\tfrac{1}{2}, a\right)$ to contain none of the forbidden fractions $\tfrac{k}{x}$ (with $x \le 100$).
- Let $f$ be the smallest such fraction that is strictly greater than $\tfrac{1}{2}$.
- Then any $a \le f$ keeps the interval clean, and pushing $a$ past $f$ would swallow $f$ and hit a lattice point.
- So the maximum $a$ equals that first forbidden fraction above $\tfrac{1}{2}$.
💡 The widest safe interval stops right at the nearest obstacle, so find the nearest obstacle above $\tfrac{1}{2}$.
5.NF.A.1 Step 3 Gap above one-half for a fixed denominator
- Fix a denominator $x$.
- The smallest numerator making $\tfrac{k}{x} > \tfrac{1}{2}$ is the smallest integer $k$ with $k > \tfrac{x}{2}$.
- The distance of that fraction above $\tfrac{1}{2}$ is $\tfrac{k}{x} - \tfrac{1}{2} = \tfrac{2k - x}{2x}$.
- Because $\tfrac{k}{x} > \tfrac{1}{2}$, the numerator $2k - x$ is a positive integer, so the gap is at least $\tfrac{1}{2x}$.
💡 Rewriting over the common denominator $2x$ turns "how close to a half" into a plain integer numerator $2k-x$.
2.OA.C.3 Step 4 Odd denominators get closest
- To make the gap $\tfrac{2k - x}{2x}$ tiny we want the numerator $2k - x$ to equal $1$.
- But $2k - x = 1$ means $x = 2k - 1$, which is odd.
- If $x$ is even, then $2k - x$ is even, hence at least $2$, doubling the smallest gap to $\tfrac{2}{2x} = \tfrac{1}{x}$.
- So odd denominators can hug $\tfrac{1}{2}$ twice as tightly as even ones.
💡 A numerator of exactly $1$ needs an odd bottom; even bottoms are forced to skip to $2$.
4.NF.A.2 Step 5 Push the denominator to its maximum
- With $x$ odd the gap $\tfrac{1}{2x}$ shrinks as $x$ grows, so take the largest odd denominator allowed: $x = 99$.
- The smallest $k$ with $k > \tfrac{99}{2}$ is $k = 50$, giving the fraction $\tfrac{50}{99}$ with gap $\tfrac{1}{198}$.
- The best even competitor is $x = 100$: $\tfrac{51}{100}$ with gap $\tfrac{1}{100} = \tfrac{1.98}{198}$, which is larger.
- Since $\tfrac{1}{198}$ beats every other denominator's gap, $\tfrac{50}{99}$ is the closest forbidden fraction above $\tfrac{1}{2}$.
💡 Smaller unit-fraction gaps come from bigger denominators, so run the odd denominator all the way up to $99$.
7.NS.A.1 Step 6 Confirm the boundary and the answer
- The interval $\left(\tfrac{1}{2}, \tfrac{50}{99}\right)$ contains no fraction with denominator $\le 100$, so every slope in it misses all lattice points; taking $a = \tfrac{50}{99}$ is safe because the interval is open and excludes $\tfrac{50}{99}$ itself.
- Any larger $a$ would admit $m = \tfrac{50}{99}$, whose line passes through $(99, 52)$.
- Checking the choices: $\tfrac{51}{100}$, $\tfrac{52}{101}$, and $\tfrac{13}{25}$ all exceed $\tfrac{50}{99}$, so they would let a lattice point in; $\tfrac{51}{101} \approx 0.50495$ is safe but smaller than $\tfrac{50}{99} \approx 0.50505$, so it is not the maximum.
- The largest valid value is $\tfrac{50}{99}$, which is (B).
💡 The best $a$ sits exactly at the nearest forbidden fraction, and the open interval lets us include that endpoint value as the cap.
8.F.A.3 On $y = mx + 2$, at an integer $x$ the height is $y = mx + 2$. Since $2$ is alre 6.EE.B.6 We need the open interval $\left(\tfrac{1}{2}, a\right)$ to contain none of the 5.NF.A.1 Fix a denominator $x$. The smallest numerator making $\tfrac{k}{x} > \tfrac{1}{2 2.OA.C.3 To make the gap $\tfrac{2k - x}{2x}$ tiny we want the numerator $2k - x$ to equa 4.NF.A.2 With $x$ odd the gap $\tfrac{1}{2x}$ shrinks as $x$ grows, so take the largest o 7.NS.A.1 The interval $\left(\tfrac{1}{2}, \tfrac{50}{99}\right)$ contains no fraction wi Review
Reasonableness: All five choices cluster just above $\tfrac{1}{2}$, which fits: the answer must be a fraction barely larger than $0.5$. Numerically $\tfrac{50}{99} \approx 0.50505$, and the only choice below it is $\tfrac{51}{101} \approx 0.50495$ — a safe but sub-maximal value, exactly the kind of near-miss the problem plants. The tempting wrong answer $\tfrac{51}{100} = 0.51$ is the best even-denominator fraction; our parity step is precisely what rules it out, since the odd denominator $99$ hugs $\tfrac{1}{2}$ twice as tightly. The winning slope $\tfrac{50}{99}$ does hit the lattice point $(99, 99\cdot\tfrac{50}{99}+2) = (99, 52)$, confirming it is genuinely the first obstacle.
Alternative: Mediant / Farey view: among fractions with denominator $\le 100$, the neighbor of $\tfrac{1}{2}$ on the right is found by solving $2k - x = 1$ with $x$ as large as possible, i.e. the fraction $\tfrac{k}{x}$ closest to $\tfrac{1}{2}$ from above in the Farey sequence $F_{100}$. That neighbor is $\tfrac{50}{99}$, since $\left|\tfrac{50}{99} - \tfrac{1}{2}\right| = \tfrac{1}{2 \cdot 99}$ is the minimal possible cross-difference. This reproduces (B) without checking each denominator separately.
CCSS standards used (min grade 8)
8.F.A.3Interpret the equation y = mx + b as defining a linear function (Reading $y = mx + 2$ as a line of slope $m$ and seeing that it hits a lattice point exactly when the slope equals a fraction $k/x$.)6.EE.B.6Use variables to represent numbers and write expressions to solve problems (Naming the forbidden slopes $k/x$ and defining the target $a$ as the smallest such fraction above $1/2$.)5.NF.A.1Add and subtract fractions with unlike denominators (Computing the gap $\tfrac{k}{x} - \tfrac{1}{2} = \tfrac{2k-x}{2x}$ over the common denominator $2x$.)2.OA.C.3Determine whether a group of objects has an odd or even number (The parity argument that $2k-x=1$ forces $x$ odd, so even denominators can only reach a gap of $1/x$.)4.NF.A.2Compare two fractions with different numerators and different denominators (Comparing candidate gaps $\tfrac{1}{198}$ (from $\tfrac{50}{99}$) against $\tfrac{1}{100}$ (from $\tfrac{51}{100}$) to pick the closest fraction.)7.NS.A.1Apply and extend understanding of addition and subtraction to rational numbers (Ordering the answer choices around $\tfrac{50}{99}$ to confirm it is the maximum valid $a$.)
⭐ Turn "the line dodges all lattice points" into "the slope isn't any fraction $k/x$ with $x \le 100$," then find the fraction that hugs $\tfrac{1}{2}$ most tightly from above — an odd denominator wins, and $x=99$ gives $\tfrac{50}{99}$.
⭐ Turn "the line dodges all lattice points" into "the slope isn't any fraction $k/x$ with $x \le 100$," then find the fraction that hugs $\tfrac{1}{2}$ most tightly from above — an odd denominator wins, and $x=99$ gives $\tfrac{50}{99}$.
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