AMC 8 · 2005 · #24
Grade 6 arithmeticProblem
A certain calculator has only two keys [+1] and [x2]. When you press one of the keys, the calculator automatically displays the result. For instance, if the calculator originally displayed "9" and you pressed [+1], it would display "10." If you then pressed [x2], it would display "20." Starting with the display "1," what is the fewest number of keystrokes you would need to reach "200"?
Pick an answer.
AMC 8 2005 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: A calculator has only two keys: $[+1]$ (add $1$) and $[\times 2]$ (double). The display starts at $1$. What is the fewest number of keystrokes needed to make the display read $200$?
Givens: Starting display $= 1$; Available keys: $[+1]$ and $[\times 2]$; Target display $= 200$; Answer choices: (A) $8$, (B) $9$, (C) $10$, (D) $11$, (E) $12$
Unknowns: The minimum number of keystrokes that takes the display from $1$ to $200$
Understand
Restated: A calculator has only two keys: $[+1]$ (add $1$) and $[\times 2]$ (double). The display starts at $1$. What is the fewest number of keystrokes needed to make the display read $200$?
Givens: Starting display $= 1$; Available keys: $[+1]$ and $[\times 2]$; Target display $= 200$; Answer choices: (A) $8$, (B) $9$, (C) $10$, (D) $11$, (E) $12$
Plan
Primary tool: #11 Work Backwards
Secondary: #9 Solve an Easier Related Problem
Going forward from $1$, every keystroke gives two choices ($+1$ or $\times 2$), so the forward search branches wildly. Tool #11 (Work Backwards) flips it: the inverse of $[\times 2]$ is $[\div 2]$ (only legal when the number is even) and the inverse of $[+1]$ is $[-1]$. Going from $200$ down to $1$, $[\div 2]$ cuts the number in half but $[-1]$ only chips off $1$, so to be fastest we should divide whenever the number is even and subtract $1$ only when it is odd. That makes the backward path uniquely forced. Tool #9 (Easier Related Problem) handles the "why not fewer?" half: if we only had $[\times 2]$, $k$ keystrokes turn $1$ into $2^k$, and $2^7 = 128 < 200 < 256 = 2^8$, so $7$ keystrokes can't reach $200$ even with the strongest key. Combined with the forced backward count, $9$ is both achievable and minimum.
Execute — Answer: B
4.OA.B.4 Step 1 - Set up the reverse rules.
- The inverse of $[+1]$ is $[-1]$, and the inverse of $[\times 2]$ is $[\div 2]$.
- Halving is much faster than subtracting $1$, so whenever the current number is even, use $[\div 2]$; if it is odd, $[\div 2]$ is illegal, so we must do one $[-1]$ to make it even.
💡 Even-vs-odd is the Grade 4 factor idea: a number is divisible by $2$ exactly when it is even, so parity decides which inverse key is allowed.
4.OA.C.5 Step 2 Apply the rules starting at $200$ and count keystrokes until you reach $1$.
💡 Generating each new value from the previous one by a fixed rule is Grade 4 "number pattern from a rule." $9$ rule-applications take us from $200$ to $1$.
4.OA.C.5 Step 3 - Reverse the backward path to get a forward keystroke sequence from $1$ to $200$.
- Every $[\div 2]$ flips back to $[\times 2]$ and every $[-1]$ flips back to $[+1]$.
💡 A backward path of length $9$ becomes a forward path of length $9$ — same arrows, reversed direction.
6.EE.A.1 Step 4 - Show that no shorter path can work.
- Consider the easier problem in which only $[\times 2]$ exists: after $k$ keystrokes, the display is $2^k$.
- Since $2^7 = 128 < 200$, even seven of the strongest possible keystrokes can't reach $200$.
- With the real (weaker on average) rules, fewer than $8$ keystrokes is also impossible; and an $8$-keystroke path would have to land exactly on $2^8 = 256$ if it used all $[\times 2]$, so it cannot hit $200$ either.
- So the minimum is at least $9$, and our path of length $9$ achieves it.
💡 Replacing the rules with the simpler "only doubling" rule gives a clean lower bound using Grade 6 whole-number exponents — and that bound, combined with parity, forces $9$.
4.OA.C.5 Step 5 - Read off the answer.
- A path of $9$ keystrokes exists and no shorter path does, so the minimum number of keystrokes is $9$.
💡 Backward search plus a lower-bound check pins the answer between "can we?" and "can we do better?"
4.OA.B.4 Set up the reverse rules. The inverse of $[+1]$ is $[-1]$, and the inverse of $[ 4.OA.C.5 Apply the rules starting at $200$ and count keystrokes until you reach $1$. 4.OA.C.5 Reverse the backward path to get a forward keystroke sequence from $1$ to $200$. 6.EE.A.1 Show that no shorter path can work. Consider the easier problem in which only $[ 4.OA.C.5 Read off the answer. A path of $9$ keystrokes exists and no shorter path does, s Review
Reasonableness: Spot-check by walking the forward path one keystroke at a time: $1, 2, 3, 6, 12, 24, 25, 50, 100, 200$. That is $9$ arrows after the starting $1$, so $9$ keystrokes, landing on $200$ as required. Also, the path uses $7$ presses of $[\times 2]$ and $2$ presses of $[+1]$, and the doublings alone multiply $1$ by $2^7 = 128$; the two $[+1]$ presses then nudge the value up to $200$, which is consistent with $200 = 128 + \text{(some adjustment)}$ produced by the timing of the two $[+1]$s. The matching answer choice is (B).
Alternative: Tool #5 (Look for a Pattern) using binary. Write $200$ in binary: $200 = 128 + 64 + 8 = 11001000_2$. Reading the binary digits left to right, each digit costs one $[\times 2]$ (a shift) plus an extra $[+1]$ when the digit is $1$. There are $8$ digits, so $8$ shifts, but the very first shift on the leading $1$ is replaced by the starting $1$ itself — that saves one $[\times 2]$. So total keystrokes $= (\text{digits} - 1) + (\text{number of } 1\text{'s} - 1) = 7 + 2 = 9$. Same answer (B), reached via the binary representation.
CCSS standards used (min grade 6)
4.OA.B.4Find all factor pairs for a whole number; determine whether a given whole number is prime or composite (Using even/odd (divisibility by $2$) to decide whether the inverse key $[\div 2]$ is legal at each backward step.)4.OA.C.5Generate a number or shape pattern that follows a given rule (Generating the backward sequence $200, 100, 50, 25, 24, 12, 6, 3, 2, 1$ by applying the rule "halve if even, else subtract $1$" at each step.)6.EE.A.1Write and evaluate numerical expressions involving whole-number exponents (Comparing $200$ to $2^7 = 128$ and $2^8 = 256$ to prove no $7$- or $8$-keystroke sequence can reach $200$.)
⭐ Running the calculator in reverse turns this AMC 8 problem into a Grade 4 pattern rule ("halve if even, else minus $1$") — and a Grade 6 powers-of-$2$ check confirms $9$ keystrokes really is the fewest.
⭐ Running the calculator in reverse turns this AMC 8 problem into a Grade 4 pattern rule ("halve if even, else minus $1$") — and a Grade 6 powers-of-$2$ check confirms $9$ keystrokes really is the fewest.
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