AMC 10 · 2019 · #9
Grade 8 arithmeticProblem
The function is defined by for all real numbers , where denotes the greatest integer less than or equal to the real number . What is the range of ?
Pick an answer.
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 $\{-1, 0\}$.