AMC 10 · 2019 · #9
Grade 8 arithmeticProblem
The function f is defined by f(x)=⌊∣x∣⌋−∣⌊x⌋∣for all real numbers x, where ⌊r⌋ denotes the greatest integer less than or equal to the real number r. What is the range of f?
Pick an answer.
AMC 10 2019 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: Define $f(x) = \lfloor|x|\rfloor - |\lfloor x \rfloor|$ for every real $x$. Here $\lfloor r \rfloor$ is the greatest integer $\le r$. What is the set of values $f$ can take (its range)?
Givens: $f(x) = \lfloor|x|\rfloor - |\lfloor x \rfloor|$; $x$ is any real number; $\lfloor r \rfloor$ = greatest integer at most $r$; Choices: (A) $\{-1, 0\}$, (B) the nonpositive integers, (C) $\{-1, 0, 1\}$, (D) $\{0\}$, (E) the nonnegative integers
Unknowns: The range of $f$
Understand
Restated: Define $f(x) = \lfloor|x|\rfloor - |\lfloor x \rfloor|$ for every real $x$. Here $\lfloor r \rfloor$ is the greatest integer $\le r$. What is the set of values $f$ can take (its range)?
Givens: $f(x) = \lfloor|x|\rfloor - |\lfloor x \rfloor|$; $x$ is any real number; $\lfloor r \rfloor$ = greatest integer at most $r$; Choices: (A) $\{-1, 0\}$, (B) the nonpositive integers, (C) $\{-1, 0, 1\}$, (D) $\{0\}$, (E) the nonnegative integers
Plan
Primary tool: #9 Solve an Easier Related Problem
Secondary: #5 Look for a Pattern, #6 Guess and Check, #3 Eliminate Possibilities
The formula has $|\cdot|$ and $\lfloor\cdot\rfloor$ that interact differently depending on whether $x$ is positive, negative, integer, or non-integer (Tool #9: try one easy value from each kind of $x$). Tool #5 spots that within each kind, the answer is the same (no surprises from changing the size of $x$). Tool #6 plugs concrete numbers — $2.5, -2.5, -3, 0.5$ — and Tool #3 then narrows the five answer choices to the one matching set of observed outputs.
Execute — Answer: A
6.NS.C.7 Step 1 - Case 1: $x \ge 0$ (positive or zero).
- Then $|x| = x$ and $\lfloor x \rfloor \ge 0$, so $|\lfloor x \rfloor| = \lfloor x \rfloor$.
- Hence $f(x) = \lfloor x \rfloor - \lfloor x \rfloor = 0$.
- Try $x = 2.5$: $\lfloor 2.5\rfloor = 2$, $|2| = 2$, $f = 2 - 2 = 0$.
- Try $x = 7$: $f = 7 - 7 = 0$.
- Always $0$.
💡 For nonnegative $x$, the absolute value bars do nothing, so both terms are the same.
6.NS.C.7 Step 2 - Case 2: $x$ is a negative integer (e.g., $x = -3$).
- Then $|x| = -x \ge 1$ is a nonneg integer, so $\lfloor|x|\rfloor = |x|$.
- And $\lfloor x \rfloor = x$ (already integer), so $|\lfloor x \rfloor| = |x|$.
- Therefore $f(x) = |x| - |x| = 0$.
💡 Integers shrug off the floor function — both pieces match.
7.NS.A.3 Step 3 - Case 3: $x$ is a negative non-integer.
- Try $x = -2.5$.
- Then $|x| = 2.5$ and $\lfloor 2.5 \rfloor = 2$.
- But $\lfloor -2.5 \rfloor = -3$ (floor rounds DOWN, away from zero for negatives), so $|\lfloor x \rfloor| = 3$.
- Therefore $f(-2.5) = 2 - 3 = -1$.
💡 Floor of a negative non-integer drops to the next integer down — that extra step makes $f = -1$.
7.NS.A.3 Step 4 - Pattern check: try other negative non-integers.
- $x = -0.5$: $\lfloor 0.5\rfloor - |\!-\!1| = 0 - 1 = -1$.
- $x = -10.2$: $\lfloor 10.2 \rfloor - |\!-\!11| = 10 - 11 = -1$.
- Always $-1$.
- So negative non-integers contribute exactly the value $-1$.
💡 Negative fractional part always shifts the floor down by exactly one — same gap every time.
8.F.A.1 Step 5 - Combine all cases: outputs seen are $0$ (Cases 1 and 2) and $-1$ (Case 3).
- So the range is $\{-1, 0\}$.
- Eliminate the other choices: (B), (C), and (E) include values like $-2, 1$, or $-3$ that we never see; (D) misses $-1$.
💡 Only two outputs ever appear, so the range has exactly those two values.
6.NS.C.7 Case 1: $x \ge 0$ (positive or zero). Then $|x| = x$ and $\lfloor x \rfloor \ge 6.NS.C.7 Case 2: $x$ is a negative integer (e.g., $x = -3$). Then $|x| = -x \ge 1$ is a n 7.NS.A.3 Case 3: $x$ is a negative non-integer. Try $x = -2.5$. Then $|x| = 2.5$ and $\lf 7.NS.A.3 Pattern check: try other negative non-integers. $x = -0.5$: $\lfloor 0.5\rfloor 8.F.A.1 Combine all cases: outputs seen are $0$ (Cases 1 and 2) and $-1$ (Case 3). So th Review
Reasonableness: The difference $\lfloor|x|\rfloor - |\lfloor x \rfloor|$ measures how much the floor function disagrees with itself depending on order with absolute value. For $x \ge 0$ these commute; for $x$ a negative integer they still match; only for negative non-integers does floor go one extra integer down, producing a gap of exactly $1$. So $f$ takes only $0$ or $-1$, never anything else — $\{-1, 0\}$ is the full range, matching (A).
Alternative: Tool #2 (Systematic List): tabulate $f(x)$ for $x = -2.5, -2, -1.5, -1, -0.5, 0, 0.5, 1, 1.5$. The outputs $-1, 0, -1, 0, -1, 0, 0, 0, 0$ make the range $\{-1, 0\}$ visually obvious.
CCSS standards used (min grade 8)
6.NS.C.7Understand ordering and absolute value of rational numbers (Computing $|x|$ for positive, zero, and negative $x$ and comparing it to $\lfloor x \rfloor$.)7.NS.A.3Solve real-world problems involving the four operations with rational numbers (Evaluating $f$ at negative non-integers like $-2.5$ and noticing the floor of a negative non-integer drops one extra integer.)8.F.A.1Understand that a function is a rule that assigns exactly one output to each input (Treating $f$ as a function and asking which set of outputs (its range) appears as $x$ varies.)
⭐ This AMC 10 problem only needs Grade 8 function thinking you already know: split into positive vs. negative vs. integer cases. For $x \ge 0$ and negative integers, $f = 0$; for negative non-integers (like $-2.5$), the floor drops one extra integer making $f = -1$. So the range is $\{-1, 0\}$.
⭐ This AMC 10 problem only needs Grade 8 function thinking you already know: split into positive vs. negative vs. integer cases. For $x \ge 0$ and negative integers, $f = 0$; for negative non-integers (like $-2.5$), the floor drops one extra integer making $f = -1$. So the range is $\{-1, 0\}$.
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