Competition · AMC preparation · step 4 of 4
AMC 8 · 2005 · #24
Grade 6 arithmeticPick an answer.
AMC 8 2005 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Going forward from 1, every keystroke gives two choices (+1 or × 2), so the forward search branches wildly. Tool #11 (Work Backwards) flips it: the inverse of [ × 2] is [ ÷ 2] (only legal when the number is even) and the inverse of [+1] is [-1]. Going from 200 down to 1, [ ÷ 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 [ × 2], k keystrokes turn 1 into 2^k, and 2⁷ = 128 ＜ 200 ＜ 256 = 2⁸, so 7 keystrokes can't reach 200 even with the strongest key. Combined with the forced backward count, 9 is both achievable and minimum.
Set up the reverse rules
Work in reverse: inverse of [+1] is [-1], of [ × 2] is [ ÷ 2]. Halving beats subtracting, so divide when even, subtract 1 when odd.
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.B.4Work BackwardsWork backward from 200
Apply the rule from 200 downward: 200, 100, 50, 25, 24, 12, 6, 3, 2, 1 — reaching 1 in 9 keystrokes.
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.
Running the calculator backward from 200 — halving the number whenever it is even and subtracting one only when it is odd — traces the shortest route back down to 1.
▸ Why?
Searching backward from 200 is a fair stand-in for the forward puzzle: every forward key has one inverse that exactly undoes it, and a shortest backward route matches a shortest forward route step for step.
▸ Why?
Halving undoes doubling and subtracting one undoes adding one, because division reverses multiplication and subtraction reverses addition.
▸ Why?
Each backward step pairs with exactly one forward step and back again, so the up route and the down route always have the same number of steps.
▸ Why?
Among the two legal moves, halving is the one to prefer at every chance, because it shrinks the number far faster than subtracting one does.
▸ Why?
Doubling builds a number by adding it to itself, so its inverse halving strips away a whole half of the value in one move, while subtracting one peels off only a single unit.
▸ Why?
Halving is only allowed when the number is even, since an even number is exactly two equal whole-number groups that split cleanly in two, whereas an odd number would not divide into whole halves.
Flip the path forward
Flip the path: each [ ÷ 2] becomes [ × 2], each [-1] becomes [+1], giving the forward route 1→2→3→6→12→24→25→50→100→200.
A backward path of length 9 becomes a forward path of length 9 — same arrows, reversed direction.
4.OA.C.5Work BackwardsRule out shorter paths
Why not fewer? With only [ × 2], k presses give 2^k, and 2⁷ = 128 ＜ 200 ＜ 256 = 2⁸, so seven doublings fall short of 200.
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.
6.EE.A.1Solve An Easier Related ProblemRead off the answer
A 9-keystroke path exists and nothing shorter works, so the fewest 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.C.5Work BackwardsRunning 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.
- Set up the reverse rules
- Work backward from 200
- Flip the path forward
- Rule out shorter paths
- Read off the answer
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