AMC 10 · 2020 · #16
Easy mode Grade 5Problem
Picture a long stick that goes from 0 to n, where n is a whole number bigger than 4.
Bela and Jenn take turns putting marks on the stick. Bela goes first. On his first turn, he can mark any point between 0 and n.
After that, here is the rule for every turn: the new mark must be more than 1 unit away from every mark already on the stick (whether it was placed by Bela or by Jenn).
A player who cannot place a new mark loses the game.
If both players play as well as possible, who will win?
Pick an answer.
AMC 10 2020 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: Bela and Jenn take turns placing real numbers on the closed interval $[0, n]$, where $n > 4$ is a fixed integer. Bela goes first and may pick anything in $[0, n]$. After that, each new pick must be more than $1$ unit away from every previously chosen number. The first player who can't move loses. Decide who wins with best play.
Givens: The playing field is the interval $[0, n]$ with $n$ a fixed integer and $n > 4$; Bela moves first; players alternate; Every new pick must be at distance $> 1$ from every previously chosen number; A player who cannot pick a legal number loses; Choices: (A) Bela always wins, (B) Jenn always wins, (C) Bela iff $n$ odd, (D) Jenn iff $n$ odd, (E) Jenn iff $n > 8$
Unknowns: Which player has a forced win, and under what condition on $n$
Understand
Restated: Bela and Jenn take turns placing real numbers on the closed interval $[0, n]$, where $n > 4$ is a fixed integer. Bela goes first and may pick anything in $[0, n]$. After that, each new pick must be more than $1$ unit away from every previously chosen number. The first player who can't move loses. Decide who wins with best play.
Givens: The playing field is the interval $[0, n]$ with $n$ a fixed integer and $n > 4$; Bela moves first; players alternate; Every new pick must be at distance $> 1$ from every previously chosen number; A player who cannot pick a legal number loses; Choices: (A) Bela always wins, (B) Jenn always wins, (C) Bela iff $n$ odd, (D) Jenn iff $n$ odd, (E) Jenn iff $n > 8$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #9 Solve an Easier Related Problem, #3 Eliminate Possibilities
Tool #1 (Diagram): drawing the segment $[0, n]$ instantly suggests a left-right symmetry around the midpoint $n/2$. That symmetry is the key — if Bela starts at the center, every move by Jenn has a mirror image still available for Bela. Tool #9 (Easier Problem): test small concrete cases $n = 5$ and $n = 6$ on a number line to confirm the mirror strategy works no matter the parity of $n$. Tool #3 (Eliminate): the answer choices split on parity of $n$ — once two small cases (one odd, one even) both go to Bela, choices (B), (C), (D), (E) all die and (A) remains.
Execute — Answer: A
4.G.A.3 Step 1 - Draw the segment $[0, n]$ on a number line and mark its midpoint $m = n/2$.
- The interval is left-right symmetric about $m$: a point $x$ in $[0, n]$ has a mirror partner $n - x$, also in $[0, n]$, at the same distance from $m$.
💡 A segment has a line of symmetry through its middle — every point has a partner.
4.G.A.3 Step 2 - Bela's strategy: on turn 1 he picks $m = n/2$ (the exact center).
- After that, whenever Jenn picks some $x$, Bela picks its mirror $n - x$.
💡 Start at the symmetry center, then copy the opponent across the middle.
5.G.A.1 Step 3 - Why the mirror move is always legal for Bela.
- The set of forbidden points (within distance $1$ of any prior pick) is itself symmetric about $m$ after every Bela move: the center pick is its own mirror, and each Jenn pick $x$ is paired with Bela's previous reply $n - x$.
- So if Jenn just placed $x$ legally, the mirror $n - x$ is at least $1$ unit from every earlier pick (because the earlier picks form a symmetric set) and at distance $|n - x - x| = |n - 2x|$ from $x$ — and this distance is $> 1$ as long as $x \ne m$, which is exactly the case since the center $m$ was Bela's first pick and is already forbidden to Jenn.
💡 Symmetric forbidden zone in $\Rightarrow$ symmetric forbidden zone out — Bela's mirror move stays legal.
4.G.A.3 Step 4 - Conclusion: whenever Jenn has a legal move, Bela has its mirror as his legal move.
- So Bela is never the first to run out — Jenn must lose first.
- Bela wins for every $n > 4$, no parity condition.
💡 If you can always copy, you never run out before your opponent.
5.G.A.1 Step 5 - Sanity-check the strategy on small cases with Tool #9 (Easier Problem).
- For $n = 5$: Bela picks $2.5$.
- The remaining legal zone is $[0, 1.5) \cup (3.5, 5]$.
- Whatever Jenn picks there, $5 - x$ is in the other piece and stays legal until Jenn is squeezed out first.
- For $n = 6$: Bela picks $3$.
- Legal zone is $[0, 2) \cup (4, 6]$.
- Same mirror logic — Bela wins.
- Both parities work.
💡 Try two small cases — odd and even — to see the parity does not matter.
2.OA.C.3 Step 6 - Eliminate the other choices (Tool #3).
- Since Bela wins for $n = 5$ (odd) and $n = 6$ (even), the answer cannot be (B) Jenn always wins, (D) Jenn iff $n$ odd, or (E) Jenn iff $n > 8$.
- And (C) Bela iff $n$ odd is killed by the $n = 6$ win.
- Only (A) Bela always wins survives.
💡 One odd win and one even win knock out every parity-dependent choice.
4.G.A.3 Draw the segment $[0, n]$ on a number line and mark its midpoint $m = n/2$. The 4.G.A.3 Bela's strategy: on turn 1 he picks $m = n/2$ (the exact center). After that, wh 5.G.A.1 Why the mirror move is always legal for Bela. The set of forbidden points (withi 4.G.A.3 Conclusion: whenever Jenn has a legal move, Bela has its mirror as his legal mov 5.G.A.1 Sanity-check the strategy on small cases with Tool #9 (Easier Problem). For $n = 2.OA.C.3 Eliminate the other choices (Tool #3). Since Bela wins for $n = 5$ (odd) and $n Review
Reasonableness: The mirror strategy is the classic symmetry argument for two-player games on a symmetric board, and it gives the first mover an automatic edge because they can claim the unique fixed point (the center). The answer (A) Bela always wins matches the intuition that moving first plus a symmetry to exploit is a forced win.
Alternative: Tool #9 (Easier Problem) alone: simulate the game with only integer picks on $n = 5, 6, 7, 8$ using a number line (Tool #1). In every case Bela wins by starting at the middle integer (or near it) and mirroring. Confirming four small cases with no parity exception again points to choice (A).
CCSS standards used (min grade 5)
2.OA.C.3Determine whether a group of objects has an odd or even number (Checking both an odd and an even value of $n$ to rule out parity-conditional choices.)4.G.A.3Recognize a line of symmetry for a two-dimensional figure (Identifying the segment $[0, n]$ as symmetric about its midpoint $n/2$ and building Bela's mirror strategy.)5.G.A.1Use a pair of perpendicular number lines forming a coordinate system (Reasoning about distances on the number line $[0, n]$ to verify the mirror move stays more than $1$ unit from every prior pick.)
⭐ This AMC 10 problem only needs Grade 5 number-line symmetry you already know — Bela picks the middle and then mirrors Jenn across that middle. Because the segment is symmetric, every legal move by Jenn has a legal mirror reply for Bela, so Jenn always runs out first. The answer is $\textbf{(A)}$.
⭐ This AMC 10 problem only needs Grade 5 number-line symmetry you already know — Bela picks the middle and then mirrors Jenn across that middle. Because the segment is symmetric, every legal move by Jenn has a legal mirror reply for Bela, so Jenn always runs out first. The answer is $\textbf{(A)}$.
More like this
Same archetype — closest grade level first.
