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.
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 BackwardsApply 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.
4.OA.C.5Work BackwardsFlip 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 BackwardsWhy 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 ProblemA 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.