AMC 10 · 2024 · #3
Grade 8 arithmeticProblem
For how many integer values of x is ∣2x∣≤7π
Modified Problem in Certain China Testpapers
For how many integer values of x is ∣2x∣≤6π
Solution for Certain China Testpapers
Use the fact that π≈3.14, and thus you can get 6π≈18.84. We could easily see that the answer is {−9,−8,...,8,9}⟹(C) 19
~RULE101
Pick an answer.
AMC 10 2024 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: Count how many integers $x$ make the inequality $|2x| \leq 7\pi$ true.
Givens: The inequality $|2x| \leq 7\pi$; $\pi \approx 3.14159$ is irrational, slightly more than $\tfrac{22}{7}$; $x$ must be an integer (positive, negative, or zero); Answer choices: (A) $16$, (B) $17$, (C) $19$, (D) $20$, (E) $21$
Unknowns: The number of integer values of $x$ that satisfy $|2x| \leq 7\pi$
Understand
Restated: Count how many integers $x$ make the inequality $|2x| \leq 7\pi$ true.
Givens: The inequality $|2x| \leq 7\pi$; $\pi \approx 3.14159$ is irrational, slightly more than $\tfrac{22}{7}$; $x$ must be an integer (positive, negative, or zero); Answer choices: (A) $16$, (B) $17$, (C) $19$, (D) $20$, (E) $21$
Plan
Primary tool: #13 Convert to Algebra
Secondary: #2 Make a Systematic List
The givens are an algebraic absolute-value inequality plus an irrational bound — Tool #13 (Convert to Algebra) is the natural lead. Translate $|2x| \leq 7\pi$ into $-7\pi \leq 2x \leq 7\pi$, then divide by $2$ to get $-\tfrac{7\pi}{2} \leq x \leq \tfrac{7\pi}{2}$. Once we have a clean numerical interval, Tool #2 (Make a Systematic List) finishes the job: $\tfrac{7\pi}{2} \approx 10.996$, so the integers $-10, -9, \ldots, 10$ are exactly the ones inside the band — list them in order and count.
Execute — Answer: E
6.NS.C.7 Step 1 - Unfold the absolute value.
- The inequality $|2x| \leq 7\pi$ says the distance from $2x$ to $0$ is at most $7\pi$, which means $2x$ sits between $-7\pi$ and $7\pi$.
💡 Absolute value is just "distance from $0$ on the number line" — the Grade 6 definition that makes $|A| \leq B$ collapse into the compound inequality $-B \leq A \leq B$.
7.EE.B.4 Step 2 Solve for $x$ alone by dividing the compound inequality by the positive constant $2$, which keeps both directions of the inequality intact.
💡 Dividing a compound inequality by a positive number is the same algebra move as for a regular equation — a Grade 7 inequality-solving step.
8.NS.A.2 Step 3 - Estimate the bound numerically.
- Use $\pi \approx 3.14159$, so $7\pi \approx 21.991$ and $\tfrac{7\pi}{2} \approx 10.996$.
- The interval for $x$ is therefore approximately $[-10.996, 10.996]$.
💡 Approximating an irrational like $\pi$ with a nearby rational to pin down its size is exactly the Grade 8 "rational approximations of irrational numbers" standard.
6.NS.C.6 Step 4 - List the integers inside the interval $[-10.996, 10.996]$.
- Since $10.996 < 11$, the integer $11$ is excluded; since $10 < 10.996$, the integer $10$ is included.
- Symmetry covers the negative side.
💡 Walking along the number line and listing each integer that sits inside a given interval is the Grade 6 "number line" idea, applied symmetrically around $0$.
2.OA.A.1 Step 5 - Count the integers in the list.
- From $-10$ up to $10$ inclusive, the count is $10 - (-10) + 1$, or equivalently $10$ negatives, $10$ positives, plus $0$.
💡 Counting how many whole steps it takes to walk from one end of a list to the other is a Grade 2 add/subtract word-problem skill — "last minus first plus one".
6.NS.C.7 Unfold the absolute value. The inequality $|2x| \leq 7\pi$ says the distance fro 7.EE.B.4 Solve for $x$ alone by dividing the compound inequality by the positive constant 8.NS.A.2 Estimate the bound numerically. Use $\pi \approx 3.14159$, so $7\pi \approx 21.9 6.NS.C.6 List the integers inside the interval $[-10.996, 10.996]$. Since $10.996 < 11$, 2.OA.A.1 Count the integers in the list. From $-10$ up to $10$ inclusive, the count is $1 Review
Reasonableness: Boundary check: does $x = 11$ really fail? $|2 \cdot 11| = 22$ and $7\pi \approx 21.99 < 22$, so yes $11$ fails. Does $x = 10$ really pass? $|2 \cdot 10| = 20$ and $20 < 21.99$, so yes $10$ passes. The interval $[-10, 10]$ has $21$ integers, matching (E).
Alternative: Tool #6 (Guess and Check) on the borderline integers: the question is essentially "how many integer pairs $\pm n$ satisfy $2n \leq 7\pi$, plus $x = 0$?". Test $n = 10$: $20 < 7\pi \approx 21.99$, pass. Test $n = 11$: $22 > 21.99$, fail. So allowed $n$ are $1$ through $10$, giving $10$ positives, $10$ negatives, and $0$ — total $21$.
CCSS standards used (min grade 8)
2.OA.A.1Solve one- and two-step word problems using addition and subtraction within 100 (Counting the integers from $-10$ to $10$ using "last minus first plus one" addition.)6.NS.C.6Understand a rational number as a point on the number line (Picturing the interval $[-10.996, 10.996]$ on the number line so the integers inside can be listed in order.)6.NS.C.7Understand ordering and absolute value of rational numbers (Rewriting $|2x| \leq 7\pi$ as the compound inequality $-7\pi \leq 2x \leq 7\pi$ via the "distance from zero" interpretation.)7.EE.B.4Use variables to represent quantities and construct simple equations and inequalities (Dividing the compound inequality by $2$ to isolate $x$ as $-\tfrac{7\pi}{2} \leq x \leq \tfrac{7\pi}{2}$.)8.NS.A.2Use rational approximations of irrational numbers to compare their size (Approximating $\pi \approx 3.14159$ to estimate $\tfrac{7\pi}{2} \approx 10.996$ and decide which integers fall inside the interval.)
⭐ This AMC 10 problem only needs Grade 8 rational approximations of $\pi$ (plus Grade 6 absolute value) — once you know $\tfrac{7\pi}{2} \approx 10.996$, you just count $-10$ through $10$ on the number line!
⭐ This AMC 10 problem only needs Grade 8 rational approximations of $\pi$ (plus Grade 6 absolute value) — once you know $\tfrac{7\pi}{2} \approx 10.996$, you just count $-10$ through $10$ on the number line!
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